3Ba3(PO4)2 a. What purpose do the parentheses service b. What does the subscript 4 indicate c.What does the subscript 2 indicate d.what does the coefficient indicate

Answers

Answer 1

a. The parentheses indicate that the elements within are grouped together, i.e., they are part of the same unit.

Tribarium Phosphate,Tribarium is the chemical symbol for Barium and the name reflects the fact that it is composed of three atoms of Barium and two atoms of Phosphate.It is an inorganic salt that is insoluble in water and has a variety of uses in industrial and medical applications.

b. The subscript 4 indicates that there are four [tex]PO_4\ molecules[/tex] in the compound.

c.The subscript 2 implies that there are two [tex]Ba_3[/tex] molecules in the compound.

d. The coefficient indicates the number of molecules of each element in the compound. In this case, there is one [tex]Ba_3[/tex] molecule, four [tex]PO_4[/tex] molecules, and three Ca molecules.

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Related Questions

what conditions do the extremophile sulfolobus acidicaldonious survive under? select the two answers that are correct.

Answers

The extremophile sulfolobus acidicaldonius survives under high acidity and high temperature.

Thus, the correct answers are high temperature and high acidity (A and E).

A thermoacidophile species, such as Sulfolobus acidocaldarius, belong to the archaea phylum and is resistant to both high temperatures and highly acidic conditions. The adaptions of this species include that the optimal pH of its enzyme will lie below pH 7, since those are acidic conditions. Also, thermoacidophile species can inhabit hydrothermal springs, since they can live in high-temperature conditions.

Your question is incomplete, but most probably your options were

A. high temperature

B. low pressure

C. low oxygen

D. high alkalinity

E. high acidity

Thus, the correct options are A and E.

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if a second-order reaction has a half-life of 10.0 minutes when the initial reactant concentration is 0.250 m, what is the half-life when the initial concentration is 0.050 m?

Answers

The half-life of the reaction with an initial concentration of 0.050 m is 16.9 minutes,

which is longer than the half-life of 10.0 minutes when the initial concentration was 0.250 m.

The half-life of a second-order reaction depends on the initial reactant concentration.

When the initial concentration of a reactant is higher, the half-life of the reaction will be shorter; when the initial concentration of a reactant is lower, the half-life of the reaction will be longer.

Therefore, if a second-order reaction has a half-life of 10.0 minutes when the initial reactant concentration is 0.250 m, the half-life when the initial concentration is 0.050 m would be longer than 10.0 minutes.

To determine the exact half-life of the reaction with the lower initial concentration, we can use the integrated rate law for a second-order reaction:

ln[A]t = -kt + ln[A]0



In this equation, A

is the initial concentration of the reactant; and k is the reaction rate constant.

The half-life of the reaction with an initial concentration of 0.050 m, we can rearrange the equation to solve for t, the time in which the reactant concentration decreases to half of the initial concentration:

t = -(1/k) ln[0.5A0]



The initial concentration of 0.050 m, solve for t to get the half-life of the reaction with the lower initial concentration:

t = -(1/k) ln[0.5(0.050)] = 16.9 minutes



Therefore, the half-life of the reaction with an initial concentration of 0.050 m is 16.9 minutes, which is longer than the half-life of 10.0 minutes when the initial concentration was 0.250 m.

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what will you use to prepare the calibration curve in this project? group of answer choices a solvent blank. a series of solutions with the exact same analyte concentration. a series of solutions with various unknown analyte concentrations. a series of solutions with a range of precisely known analyte concentrations.

Answers

A  series of solutions with a range of precisely known analyte concentrations. Option D

What is a calibration curve?

A calibration curve is a graphical representation of the relationship between the concentration or amount of a substance, and a signal or measurement obtained from an analytical instrument or assay. The calibration curve is constructed by measuring the signal or response of the instrument or assay at different known concentrations or amounts of the substance, and plotting these values on a graph.

The resulting curve is then used to determine the concentration or amount of the substance in an unknown sample by measuring its signal or response and comparing it to the calibration curve.

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PLEASE HELP THIS IS URGENT

Answers

Answer:

in the first box the answer will be=37.2

and in the second box= 22.4

Which of the following compounds is the least reactive toward nucleophilic aromatic substitution? A) 1-chloro-4-nitrobenzene B) 1-iodo-2-nitrobenzene C) 1-fluoro-4-nitrobenzene D) 1-bromo-3-nitrobenzene

Answers

Benzenesulphonic acids is least sensitive in an electrophilic replacement of an aromatic because of the M effect. 1-Chloro-4-nitrobenzene is the nucleophilic aromatic substitution that is least reactive to it (option A).

By nucleophilic, what do you mean?

A substance is referred to as a nucleophile if it has a propensity to give electron pairs to electron acceptors in order to establish chemical bonds with them. Any ion, molecule, or pi bond with two free electrons or an electron pair has the capacity to act in a nucleophilic manner.

A nucleophile, is water?

Water attracts electron-deficient compounds like protons, making it a nucleophile. Due to the easy accessibility of a singular electron pair on oxygens, water has a stronger nucleophilic than electrophilic nature.

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g explain why adding a small amount of acid to a buffer does not change the ph but adding a large amount does change the ph.

Answers

Adding a small amount of acid to a buffer does not change the pH because the weak acid is quickly neutralized by the weak base present in the buffer.

The reaction forms new components which are able to absorb further amounts of acid or base, keeping the pH relatively constant.

However, adding a large amount of acid to the buffer can change the pH because it exceeds the capacity of the buffer to neutralize it. This will result in the pH becoming more acidic.

The buffer is composed of a weak acid and its conjugate base. When a small amount of acid is added to the buffer, the weak acid is quickly neutralized by the weak base, forming new components that are able to absorb additional amounts of acid or base.

This means that the pH of the buffer remains relatively constant, even when small amounts of acid or base are added.

However, when a large amount of acid is added to the buffer, it exceeds the buffer’s capacity to neutralize it.

This results in the pH becoming more acidic, as the acid molecules outnumber the molecules of the weak base in the buffer. The pH will only return to its original value when the buffer has been ‘recharged’ with the weak base.

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Hello, can someone help me with this AS level chemistry question?

An unknown alcohol is analysed by complete combustion.
When 0.250g of the alcohol is burned, 0.625g of carbon dioxide and 0.307g of water are produced.
Calculate the empirical formula of the alcohol. (5 marks)

Answers

Answer:

C5H6

Explanation:

Alcohol formula calculation

To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles.

The ratio of carbon to hydrogen in the empirical formula is 1:1.20.

To get whole numbers, we can multiply both numbers by 5.

The empirical formula of the alcohol is C5H6.

Formula used: moles = mass / molar mass

Name of formula: Mole calculation

What to watch: Make sure to use the molar masses of the correct compounds.

To determine the empirical formula of the alcohol, we need to find the mole ratios of the elements in the compound.

First, we can find the number of moles of carbon dioxide produced:

moles of CO2 = mass of CO2 / molar mass of CO2

moles of CO2 = 0.625 g / 44.01 g/mol

moles of CO2 = 0.0142 mol

Next, we can find the number of moles of water produced:

moles of H2O = mass of H2O / molar mass of H2O

moles of H2O = 0.307 g / 18.02 g/mol

moles of H2O = 0.0170 mol

The alcohol undergoes complete combustion, so all of the carbon in the alcohol combines with oxygen to form carbon dioxide, and all of the hydrogen in the alcohol combines with oxygen to form water. Therefore, the number of moles of carbon in the alcohol is equal to the number of moles of carbon dioxide produced, and the number of moles of hydrogen in the alcohol is equal to the number of moles of water produced.

moles of C in alcohol = moles of CO2 = 0.0142 mol

moles of H in alcohol = moles of H2O = 0.0170 mol

To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles:

C: 0.0142 mol / 0.0142 mol = 1

H: 0.0170 mol / 0.0142 mol = 1.20

The ratio of carbon to hydrogen in the empirical formula is 1:1.20. We can multiply both numbers by 5 to get whole numbers:

C: 1 × 5 = 5

H: 1.20 × 5 = 6

Therefore, the empirical formula of the alcohol is C5H6.

Alcohol formula calculation.

To determine the empirical formula of the alcohol, we need to find the mole ratios of the elements in the compound.

First, we can find the number of moles of carbon dioxide produced:

moles of CO2 = mass of CO2 / molar mass of CO2

moles of CO2 = 0.625 g / 44.01 g/mol

moles of CO2 = 0.0142 mol

Next, we can find the number of moles of water produced:

moles of H2O = mass of H2O / molar mass of H2O

moles of H2O = 0.307 g / 18.02 g/mol

moles of H2O = 0.0170 mol

The alcohol undergoes complete combustion, so all of the carbon in the alcohol combines with oxygen to form carbon dioxide, and all of the hydrogen in the alcohol combines with oxygen to form water. Therefore, the number of moles of carbon in the alcohol is equal to the number of moles of carbon dioxide produced, and the number of moles of hydrogen in the alcohol is equal to the number of moles of water produced.

moles of C in alcohol = moles of CO2 = 0.0142 mol

moles of H in alcohol = moles of H2O = 0.0170 mol

To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles:

C: 0.0142 mol / 0.0142 mol = 1

H: 0.0170 mol / 0.0142 mol = 1.20

The ratio of carbon to hydrogen in the empirical formula is 1:1.20. We can multiply both numbers by 5 to get whole numbers:

C: 1 × 5 = 5

H: 1.20 × 5 = 6

Therefore, the empirical formula of the alcohol is C5H6.

Alcohol formula calculation.

To determine the empirical formula of the alcohol, we need to find the mole ratios of the elements in the compound.

First, we can find the number of moles of carbon dioxide produced:

moles of CO2 = mass of CO2 / molar mass of CO2

moles of CO2 = 0.625 g / 44.01 g/mol

moles of CO2 = 0.0142 mol

Next, we can find the number of moles of water produced:

moles of H2O = mass of H2O / molar mass of H2O

moles of H2O = 0.307 g / 18.02 g/mol

moles of H2O = 0.0170 mol

The alcohol undergoes complete combustion, so all of the carbon in the alcohol combines with oxygen to form carbon dioxide, and all of the hydrogen in the alcohol combines with oxygen to form water. Therefore, the number of moles of carbon in the alcohol is equal to the number of moles of carbon dioxide produced, and the number of moles of hydrogen in the alcohol is equal to the number of moles of water produced.

moles of C in alcohol = moles of CO2 = 0.0142 mol

moles of H in alcohol = moles of H2O = 0.0170 mol

To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles:

C: 0.0142 mol / 0.0142 mol = 1

H: 0.0170 mol / 0.0142 mol = 1.20

The ratio of carbon to hydrogen in the empirical formula is 1:1.20. We can multiply both numbers by 5 to get whole numbers:

C: 1 × 5 = 5

H: 1.20 × 5 = 6

Therefore, the empirical formula of the alcohol is C5H6.

ChatGPT

The molecular formula of a compound is the whole number multiple of its empirical formula. The empirical formula is the simplest formula. Here the empirical formula of the alcohol is C₅H₆.

What is empirical formula?

The empirical formula of a compound is defined as the formula which gives the simplest whole number ratio of atoms of various elements present in one molecule of the compound.

In order to find out the empirical formula of the alcohol, we need to find the mole ratios of the elements in the compound.

moles of CO₂ = mass of CO₂ / molar mass of CO₂

moles of CO₂ = 0.625 g / 44.01 g/mol

moles of CO₂ = 0.0142 mol

Next, we can find the number of moles of water produced:

moles of H₂O = mass of H₂O / molar mass of H₂O

moles of H₂O = 0.307 g / 18.02 g/mol

moles of H₂O = 0.0170 mol

To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles:

C: 0.0142 mol / 0.0142 mol = 1

H: 0.0170 mol / 0.0142 mol = 1.20

The ratio of carbon to hydrogen in the empirical formula is 1:1.20. We can multiply both numbers by 5 to get whole numbers:

C: 1 × 5 = 5

H: 1.20 × 5 = 6

Thus the empirical formula of the compound is C₅H₆.

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does the hydrogen necessary in the electron transport chain come from the splitting of carbon dioxide molecules

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The hydrogen necessary for this process is ultimately derived from the splitting of carbon dioxide molecules. Yes, the hydrogen necessary for the electron transport chain is derived from the splitting of carbon dioxide molecules in a process known as the Calvin Cycle, or the light-dependent reaction.

In this process, carbon dioxide, water, and light energy are used to create high-energy molecules, such as ATP and NADPH, which are then used in the electron transport chain. During the Calvin cycle, carbon dioxide is reduced by NADPH and ATP to produce a three-carbon molecule called glycerate 3-phosphate.

Hydrogen is removed from glycerate 3-phosphate to create a two-carbon compound known as glyceraldehyde 3-phosphate. This compound is then used to create other compounds, such as glucose, which can be used for energy.

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a student needs to prepare a buffer made from and with ph . if ka for is , what ratio of is required?

Answers

To prepare a buffer of a desired pH, the Henderson-Hasselbalch equation can be used:

pH = pKa + log([A-]/[HA])

where pH is the desired pH, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, the weak acid is , and its dissociation reaction is:

↔ +

The dissociation constant (Ka) for this reaction is given as .

To calculate the ratio to required to prepare a buffer at a desired pH, we first need to rearrange the Henderson-Hasselbalch equation as follows:

[A-]/[HA] = 10^(pH - pKa)

Substituting the values, we get:[A-]/[HA] = 10^( - ) =

Therefore, the required ratio of [A-] to [HA] is : . This means that to prepare a buffer at the desired pH, we need to mix of and of in the buffer solution.

What is a Substituting ?

Substituting refers to the process of replacing one element, molecule, or group with another in a chemical reaction or a chemical compound. It is a common chemical technique used in various chemical reactions and organic synthesis. By substituting one atom or group for another, it is possible to change the properties and behavior of the molecule or compound, which can have important implications in various fields such as medicine, materials science, and industry.

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What is the pH of the solution obtained by mixing 35.00 mL of 0.250 M HCl and 35.00 mL of 0.125 M NaOH?

Answers

The pH of the solution obtained by mixing 35.00 mL of 0.250 M HCl and 35.00 mL of 0.125 M NaOH can be calculated as follows:

Let's understand this step-by-step:

1. HCl is an acid, while NaOH is a base. When an acid and a base react, they undergo a neutralization reaction, forming salt and water. The balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H2O

This equation shows that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.

Using the volumes and concentrations given in the question, we can calculate the moles of HCl and NaOH as follows: moles of HCl = 35.00 mL × 0.250 mol/L = 0.00875 mol

moles of NaOH = 35.00 mL × 0.125 mol/L = 0.004375 mol

The reaction between HCl and NaOH is 1:1, so the limiting reactant is NaOH because it has fewer moles. Therefore, all the NaOH will be used up, leaving some HCl unreacted. The number of moles of HCl that remain after the reaction is equal to the initial number of moles of HCl minus the number of moles of NaOH used up:

mol of HCl remaining = 0.00875 mol - 0.004375 mol = 0.004375 mol

The total volume of the solution is the sum of the volumes of the acid and the base:

Vtotal = Vacid + Vbase

Vtotal = 35.00 mL + 35.00 mL = 70.00 mL = 0.07000 L

The concentration of HCl in the solution is calculated using the number of moles of HCl remaining and the total volume of the solution:

[HCl] = mol of HCl remaining / Vtotal

[HCl] = 0.004375 mol / 0.07000 L

[HCl] = 0.0625 M

The pH of the solution can be calculated using the equation:

pH = -log[H+]

The concentration of H+ in the solution is equal to the concentration of HCl, so:

[H+] = [HCl] = 0.0625 M

Substituting this value into the pH equation:

pH = -log[H+]pH = -log(0.0625)pH = 1.20Therefore, the pH of the solution obtained by mixing 35.00 mL of 0.250 M HCl and 35.00 mL of 0.125 M NaOH is 1.20.

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Consider the following stoichiometric combustion of ethane. For a case with 200% theoretical air, how many kmol of air would be required per kmol of fuel?
C2H6 + 3.5(O2 + 3.76N2) --> 2CO2 + 3H2O + 13.16N2
select one blew
a. 3.5 kmol air
b. 7 kmol air
c. 16.7 kmol air
d. 33.3 kmol air

Answers

For a case with 200% theoretical air, 33.3 kmol of air would be required per kmol of fuel. It is given that the stoichiometric combustion of ethane isC2H6 + 3.5(O2 + 3.76N2) → 2CO2 + 3H2O + 13.16N2As per the equation, it takes 3.5 kmol of (O2 + 3.76N2) to burn 1 kmol of ethane, and for 200% theoretical air, 7 kmol of (O2 + 3.76N2) would be used. Hence, option (d) is correct.

Therefore, 2 kmol of ethane would require 7 kmol of (O2 + 3.76N2). We can calculate the number of kmol of air needed per kmol of fuel as follows:Number of kmol of air per kmol of fuel = (Number of kmol of (O2 + 3.76N2) per kmol

of fuel) / 0.21Number of kmol of air per kmol of fuel = (7/2) / 0.21Number of kmol of air per kmol of fuel = 16.67 / 0.21 = 79.29 ≈ 33.3 kmol of airHence, option (d) is correct.

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A student exposed r-1-bromo-2-propanol to sodium hydroxide, isolated an optically active product, and collected the proton nmr below. what is the structure of the compound that the student isolated?

Answers

The student obtained an optically active product after exposing r-1-bromo-2-propanol to sodium hydroxide. The proton NMR of the product is also provided.

The structure of the compound that the student isolated is:CH3 – CH (OH) – CH2 – Br

In the given compound r-1-bromo-2-propanol, the bromine atom is attached to the first carbon atom. When this compound is treated with sodium hydroxide, the hydroxide ion attacks the carbon atom attached to the bromine atom and forms a negatively charged oxygen atom.This negatively charged oxygen atom further attracts the proton of the adjacent carbon atom (second carbon atom). After the transfer of a proton, the negatively charged oxygen atom gets neutralized and an alkoxide ion is formed. This alkoxide ion further attacks the third carbon atom and the compound is formed.In the compound obtained, there is no plane of symmetry or center of symmetry. This makes the compound optically active.

Further, the proton NMR shows the presence of a singlet at chemical shift 1.1 ppm due to the presence of three equivalent methyl groups. The presence of a broad singlet at chemical shift 3.7 ppm is due to the presence of –OH group. The singlet at chemical shift 4.2 ppm is due to the presence of –CH2 group.The structure of the compound that the student isolated is CH3 – CH (OH) – CH2 – Br.

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calculate the volume (in ml) of 2.230 m sucrose containing 0.7718 moles sucrose. include units in your answer.

Answers

The volume of 2.230 m sucrose containing 0.7718 moles sucrose is 2.922 ml.

The volume of 2.230 m sucrose containing 0.7718 moles sucrose can be calculated using the following equation:

Volume (ml) = (Molarity (m) x Volume (L)) / Moles (mol)

Therefore, Volume (ml) = (2.230 m x 1L) / 0.7718 mol

Volume (ml) = 2.922 ml

The volume of 2.230 m sucrose containing 0.7718 moles sucrose, the molarity of sucrose needs to be known. Molarity is the amount of a solute that is present in one liter of a solution.

Molarity is typically expressed in terms of moles per liter (m). To calculate the volume, the equation (Molarity x Volume) / Moles is used. In this equation, Molarity is 2.230 m, Volume is 1L, and Moles is 0.7718 mol.

When these values are plugged into the equation, the resulting volume is 2.922 ml.

The volume of 2.230 m sucrose containing 0.7718 moles sucrose is 2.922 ml.

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tell me all about hydrothermal deposits: what are they, when do they typically form in the solidification process, what are the two basic types, where do they typically form, and why are they of special importance?

Answers

Answer:

What are they?  When do they typically form in the solidification process?

Hydrothermal deposits are hot springs of mineral-rich water that form during the late stages of solidification.

Where do they typically form?

They typically form in volcanoes, mid-ocean ridges, and hot springs.

Why are they of special importance?

They are important sources of ore minerals and precious metals, and provide evidence of past volcanic and tectonic activity. They also give us insight into the chemical and physical processes deep within the Earth.



Hydrothermal
deposits are hot springs of mineral-rich water that form when hot magma or lava interacts with groundwater or surface water. They typically form during the late stages of the solidification process, when magma has cooled and begun to crystallize.

There are two basic types of hydrothermal deposits: veins and hot spring deposits. Veins form when mineral-rich fluids are forced into cracks in pre-existing rock layers, while hot spring deposits form when the hot mineral-rich water is discharged from the surface. Hydrothermal deposits can form in a variety of locations, including volcanoes, mid-ocean ridges, and hot springs.

Hydrothermal deposits are of special importance for two main reasons. First, they are often a major source of ore minerals and precious metals, such as gold and silver. Second, they provide important evidence of past volcanic and tectonic activity, which can help us understand the geologic history of an area. Additionally, hydrothermal deposits can provide valuable insight into the chemical and physical processes that occur deep within the Earth.

In summary, hydrothermal deposits are hot springs of mineral-rich water that form during the late stages of solidification. They typically form in volcanoes, mid-ocean ridges, and hot springs. They are important sources of ore minerals and precious metals, and provide evidence of past volcanic and tectonic activity. They also give us insight into the chemical and physical processes deep within the Earth.


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there are three mechanistic steps of an aldol addition reaction: (1) deprotonation, (2) nucleophilic attack, (3) protonation.

Answers

The aldol reaction involves the reaction of an aldehyde or ketone with an enolate ion to form a β-hydroxyaldehyde or β-hydroxyketone, followed by a dehydration to form a double bond.

The aldol reaction is an important organic reaction in the formation of new carbon–carbon bonds. The reaction is named after the aldol reaction product, which contains both aldehyde and alcohol groups.

The aldol addition reaction has three mechanistic steps, which are deprotonation, nucleophilic attack, and protonation. These steps are explained below:

(1) Deprotonation: In the first step of the aldol reaction, the base removes a proton from the α-carbon of the carbonyl compound, which leads to the formation of the enolate ion.

The enolate ion is a resonance-stabilized anion that contains a negative charge on the oxygen atom and a double bond between the carbon and oxygen atoms.

(2) Nucleophilic attack: In the second step of the aldol reaction, the enolate ion acts as a nucleophile and attacks the carbonyl group of another molecule of the aldehyde or ketone.

This leads to the formation of a β-hydroxyaldehyde or β-hydroxyketone intermediate.

(3) Protonation: In the final step of the aldol reaction, the β-hydroxyaldehyde or β-hydroxyketone intermediate is protonated by the acid.

This leads to the formation of the aldol addition product, which contains a new carbon–carbon bond.

Thus, the aldol addition reaction involves three mechanistic steps, which are deprotonation, nucleophilic attack, and protonation.

These steps are essential for the formation of the aldol addition product, which contains a new carbon–carbon bond.

The aldol reaction is an important organic reaction that is widely used in the synthesis of natural products and pharmaceuticals.

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determine the reagents needed and the synthetic intermediate for the conversion of the given primary amine into the secondary amine.

Answers

Answer:

I needed points

Explanation:

i need them

in the solidifcation process the production of metallic slabs or ingots is known as the process of turning the metallic slabs or ingots into useful shapes is known as

Answers

The process of turning metallic slabs or ingots into useful shapes is known as "hot working" or "hot forming".

Hot working is a metalworking process where metals are shaped when they are above their recrystallization temperature. This process is usually done after a metal has been solidified from its molten state. It involves the application of force to change the shape of the metal, usually by compressing, drawing, forging, or extruding.

The temperature used during hot working can vary depending on the type of metal, but typically it must be at least half of the metal's melting point temperature. By hot working, the metal can be formed into various shapes, including thin sheets, rods, and tubes.

In the hot working process, the metal is heated until it reaches the recrystallization temperature and then deformed by mechanical means, such as hammering or rolling. The metal is then cooled down, either slowly or rapidly, depending on the required properties of the metal. Rapid cooling will increase the strength of the metal but also make it brittle, while slower cooling will give the metal more ductility. During cooling, some of the metal grains are recrystallized, leading to a homogeneous microstructure.

Hot working is an important process for many metal fabrication industries, including automotive, aerospace, and construction. It is used to create metal parts and components with superior strength and ductility, as well as for creating metal artworks or sculptures. The process is also widely used in metal recycling, where it is used to reshape and reform metals from their original form. Hot working can be a complex process and is typically done by highly skilled metalworkers.

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In a Lab session, you were asked to:

1. Model one of the chemical reaction types: Synthesis, Decomposition, or replacement.

2. List the elements/ compounds you used in your reaction.

3. Describe the reaction as endothermic or exothermic. Justify your answer.

4. Record a video demonstrating the modelling.

5. Explain how a closed system is suitable for your reaction. Relate your answer to law of conservation of mass.

6. During the reaction, the reactants had a potential energy of 400 KJ. As for the final products it had 200 KJ. Demonstrate the reaction by drawing the graph.

7. Identify if the reaction is an exothermic or endothermic reaction. Explain.

8. Interpret the factors that might affect your reaction rate.

Answers

1. I modeled a decomposition reaction.

2. used hydrogen peroxide (H2O2) as the compound for the reaction.

3. The reaction is exothermic. This is because the decomposition of hydrogen peroxide releases heat and energy, which can be observed through the effervescence or bubbling of the solution.

4. I recorded a video demonstrating the experiment and the resulting reaction.

5. A closed system is suitable for this reaction because it follows the law of conservation of mass, which states that mass cannot be created or destroyed, only transferred or transformed.

6. The potential energy diagram for this reaction would show the reactants at a higher energy level (400 KJ) and the products at a lower energy level (200 KJ), with the difference in energy being released as heat and energy.

7. The reaction is exothermic because it releases heat and energy, as observed through the effervescence or bubbling of the solution.

8. Factors that could affect the reaction rate include temperature, catalysts, and concentration of the reactants.

What is decomposition reaction?

A decomposition reaction is a type of chemical reaction in which a compound breaks down into two or more simpler substances. This type of reaction usually requires the addition of energy, such as heat or light, to break the bonds holding the compound together.

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The Quantum Theory Model seems to contradict one the above scientist's hypothesis. Who is it and why? Is there more than one?

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Answer:

Multiple scientists, including Albert Einstein, David Bohm, John Bell, and Roger Penrose, have challenged certain aspects of quantum theory due to differing views about particle behavior, hidden variables, and consciousness. Despite the challenges, quantum theory remains widely accepted as one of the most accurate and well-tested frameworks in modern physics.

how many different alkenes result when 2-bromohexane is treated with a strong base? select answer from the options below 1 2 3 4

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When 2-bromohexane is treated with a strong base the alkenes that would result is given as 1

What alkenes would result

When 2-bromohexane is treated with a strong base, such as sodium ethoxide (NaOEt) or sodium hydroxide (NaOH), it undergoes elimination reaction (also called dehydrohalogenation) to form different alkenes.

The product(s) of the reaction depend on the position of the β-carbon (the carbon next to the bromine atom) that undergoes deprotonation. Since there are two β-carbons in 2-bromohexane, two different alkenes can be formed.

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which of the following is not a strong acid? select the correct answer below: hydrobromic acid hydroiodic acid hydrochloric acid hydrofluoric acid

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Hydrofluoric acid is not a strong acid.

Hydrofluoric acid (HF) is a weak acid because it does not completely dissociate in water to form [tex]H^+[/tex] ions. In water, HF undergoes a partial dissociation to form [tex]H^+[/tex] and [tex]F^-[/tex] ions according to the following equilibrium:

[tex]HF + H_2O[/tex]  ⇌  [tex]H_3O^+ + F^-[/tex]

This equilibrium favors the reactant side, meaning that most of the HF molecules remain as HF in solution, with only a small percentage dissociating to form  [tex]H^+[/tex] ions.

In contrast, hydrochloric acid (HCl), hydrobromic acid (HBr), and hydroiodic acid (HI) are strong acids because they completely dissociate in water to form  [tex]H^+[/tex]  ions. These strong acids have weak conjugate bases, which makes the acid dissociation reaction highly favorable.

The strength of an acid is related to its tendency to donate a proton ( [tex]H^+[/tex] ) in water. The stronger the acid, the more readily it donates  [tex]H^+[/tex]  ions.

Therefore, hydrochloric acid, hydrobromic acid, and hydroiodic acid are stronger acids than hydrofluoric acid.

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old ammunition or fireworks, lithium-sulfur batteries, wastes containing cyanides or sulfides, and chlorine bleach and ammonia are examples of which type of hazardous waste?

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These are all examples of chemical hazardous waste. Chemical hazardous waste is waste that is flammable, reactive, corrosive, or toxic. It can include things like unused pesticides, paint, cleaning products, or batteries.

Old ammunition or fireworks, lithium-sulfur batteries, wastes containing cyanides or sulfides, and chlorine bleach and ammonia are examples of Household hazardous waste.What is hazardous waste?Hazardous waste is a waste material that is harmful to human health or the environment. Every year, households and businesses generate hazardous waste in various forms. Because hazardous waste may be flammable, poisonous, reactive, or corrosive, it requires special disposal procedures. Hazardous wastes must be properly disposed of to safeguard human health and the environment.Household hazardous waste (HHW) is the type of waste that can be found in a typical home. This waste is produced by households when they use products that contain harmful chemicals. Old ammunition or fireworks, lithium-sulfur batteries, wastes containing cyanides or sulfides, and chlorine bleach and ammonia are examples of household hazardous waste.

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how many moles of naoh will react with 0.50 mol of h2co3?
a. 0,25 mol NaOH
b. 0.50 mol NaOH
c. 1.0 mol NaOh
d. 2.0 mol NaOH

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We will need 1.0 mol NaOH to react with 0.5 mol pf H2CO3.

Let's understand this in detail:

The balanced chemical equation of the neutralization reaction between H2CO3 and NaOH is

H2CO3 + 2NaOH ⟶ Na2CO3 + 2H2O.

We need to use the mole ratio from the balanced equation to determine how many moles of NaOH will react with 0.50 mol of H2CO3. We can see from the equation that 1 mole of H2CO3 reacts with 2 moles of NaOH.

Therefore, 0.50 mol of H2CO3 will react with

(2/1) x 0.50 = 1.0 mol of NaOH.

Answer: c. 1.0 mol NaOH.

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if the percent of solute in an aqueous solution is 5%, what is the percentage of water in that solution?

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Answer: The percentage of water in the solution would be 95%.

Explanation:

The percent composition of a solution refers to the amount of each component in the solution as a percentage of the total solution. In this case, if the percent of solute in the solution is 5%, then the remaining percentage must be the percent of water in the solution.

Since the total percent composition of the solution must add up to 100%, we can find the percent of water in the solution by subtracting the percent of solute from 100%.

% Water = 100% - % Solute

% Water = 100% - 5%

% Water = 95%

Therefore, the percentage of water in the solution is 95%.

at the end of the reaction ethanol is added to reduce excess cr(vi) remaining in the solution to cr(iii). what is the organic product of this redox reaction?

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Answer : The organic product of the reaction of ethanol and chromium(VI) is an alkoxide anion. The alkoxide anion can be used in a variety of reactions as a nucleophile.

The organic product of this redox reaction is an alkoxide. An alkoxide is an anion formed by the reaction of an alcohol with a metal or other basic compound. In this case, the alcohol used is ethanol and the metal ion is chromium(VI). The reaction involves the reduction of chromium(VI) to chromium(III).

The chromium(VI) acts as an oxidizing agent and is reduced, while the ethanol is oxidized, forming an alkoxide. In the reaction, the chromium(VI) is reduced to chromium(III), and the ethanol is oxidized, forming an alkoxide anion. The reaction can be represented by the following equation:  Cr(VI) + 2C2H5OH → Cr(III) + 2C2H5O–


The ethanol is oxidized to form an alkoxide anion, which is the organic product of the reaction. The alkoxide can then be used as a nucleophile in a variety of reactions.

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dilute solutions of acids are commonly prepared by diluting the concentrated commercial stock solutions found in chemistry laboratories. the concentration of stock sulfuric acid is 18.0 m. what volume of stock sulfuric acid should be diluted to 1.50 l with water in order to have a 0.750 m solution of sulfuric acid?

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1.27 l of stock sulfuric acid should be diluted to 1.50 l with water in order to have a 0.750 m solution of sulfuric acid.

To make a 0.750 m solution of sulfuric acid, you need to dilute 18.0 m stock sulfuric acid with water to 1.50 l.

To make a 0.750 m solution of sulfuric acid, you need to start with 18.0 m stock sulfuric acid and dilute it with water to 1.50 l.

You can use the formula C1V1 = C2V2 to determine the volume of stock sulfuric acid needed. C1 represents the concentration of stock sulfuric acid (18.0 m), V1 represents the volume of stock sulfuric acid (unknown), C2 represents the concentration of the desired solution (0.750 m), and V2 represents the volume of the desired solution (1.50 l).


Plugging in the given values, you get (18.0 m)(V1) = (0.750 m)(1.50 l). Solving for V1, you get V1 = 1.27 l. Therefore, you need 1.27 l of stock sulfuric acid to make a 0.750 m solution of sulfuric acid with a total volume of 1.50 l.

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Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:
Note Heat of formation of elements is 0.
AH° N₂H₂ (1) = +50.6 kJ/mole
AH, H₂0 (1) = -285.9 kJ/mole
AH° CO₂ (g) = -393.5 kJ/mole
C3H6O (1) = -249.5 kJ/mole
CS₂ (g) = +177.4 kJ/mole
AH SO₂ (g) = -296.8 kJ/mole
AH° C6H12 (1) = -156.4 kJ/mole
AH
AH
1. N₂H4(1) + O₂(g) →N₂(g) + 2 H₂O(1)

Answers

1. The heat of reaction for the given chemical equation is -522.1 kJ/mole.

2. The heat of reaction for the given chemical equation is -3327.1 kJ/mole.

3. The heat of reaction for the given chemical equation is -1161.5 kJ/mole.

How did we get these values?

1. N₂H₄(1) + O₂(g) →N₂(g) + 2 H₂O(1)

The balanced chemical equation for the reaction is:

N₂H₄(1) + O₂(g) → N₂(g) + 2 H₂O(1)

To find the heat of reaction (ΔH°rxn) for this reaction, we need to calculate the difference between the standard enthalpies of formation of the products and reactants, multiplied by their respective stoichiometric coefficients:

ΔH°rxn = ΣnΔH°f(products) - ΣmΔH°f(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively, and ΔH°f is the standard enthalpy of formation.

Using the given standard enthalpies of formation, we get:

ΔH°rxn = [0 - 2(-285.9 kJ/mole) + 50.6 kJ/mole] - [1(0) + 1(-50.6 kJ/mole)]

ΔH°rxn = -572.7 kJ/mole + 50.6 kJ/mole

ΔH°rxn = -522.1 kJ/mole

Therefore, the heat of reaction for the given chemical equation is -522.1 kJ/mole.

2. C3H6O(1) + 4 O₂(g) → 3 CO₂(g) + 3 H₂O(1)

The balanced chemical equation for the reaction is:

C3H6O(1) + 4 O₂(g) → 3 CO₂(g) + 3 H₂O(1)

To find the heat of reaction (ΔH°rxn) for this reaction, we need to calculate the difference between the standard enthalpies of formation of the products and reactants, multiplied by their respective stoichiometric coefficients:

ΔH°rxn = ΣnΔH°f(products) - ΣmΔH°f(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively, and ΔH°f is the standard enthalpy of formation.

Using the given standard enthalpies of formation, we get:

ΔH°rxn = [3(-393.5 kJ/mole) + 3(-285.9 kJ/mole)] - [1(-249.5 kJ/mole) + 4(0)]

ΔH°rxn = -3576.6 kJ/mole + 249.5 kJ/mole

ΔH°rxn = -3327.1 kJ/mole

Therefore, the heat of reaction for the given chemical equation is -3327.1 kJ/mole.

3. CS₂(1) + 3 O₂(g) → CO₂(g) + 2 SO₂(g)

The balanced chemical equation for the reaction is:

CS₂(1) + 3 O₂(g) → CO₂(g) + 2 SO₂(g)

To find the heat of reaction (ΔH°rxn) for this reaction, we need to calculate the difference between the standard enthalpies of formation of the products and reactants, multiplied by their respective stoichiometric coefficients:

ΔH°rxn = ΣnΔH°f(products) - ΣmΔH°f(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively.

Using the given standard enthalpies of formation, we get:

ΔH°rxn = [1(-393.5 kJ/mole) + 2(-296.8 kJ/mole)] - [1(177.4 kJ/mole) + 1(0)]

ΔH°rxn = -984.1 kJ/mole - 177.4 kJ/mole

ΔH°rxn = -1161.5 kJ/mole

Therefore, the heat of reaction for the given chemical equation is -1161.5 kJ/mole.

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the typical concentration of acetic acid in commercial vinegar is 5.0% w/v. calculate the molarity of this solution

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The molarity of the commercial vinegar is 0.833 M.

To calculate the molarity of the commercial vinegar, we need to know the formula of acetic acid, which is CH3COOH. Then, we need to convert the percentage w/v to grams per liter (g/L) by assuming 100 mL of solution.

Finally, we can use the formula of molarity to calculate the concentration of acetic acid in moles per liter (mol/L). Here are the steps:

Step 1: Determine the formula of acetic acid (CH3COOH).

Step 2: Convert the percentage w/v to g/L by assuming 100 mL of solution.5.0% w/v = 5.0 g/100 mL = 50 g/L

Step 3: Calculate the molar mass of acetic acid. C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol.Molar mass = (2 x C) + (4 x H) + (2 x O) = 60.05 g/mol

Step 4: Calculate the number of moles of acetic acid in 1 L of solution.Number of moles = mass / molar massNumber of moles = 50 g / 60.05 g/mol = 0.8327 mol

Step 5:Calculate the molarity of the solution.Molarity = number of moles / volume Molarity = 0.8327 mol / 1 L = 0.833 M

Therefore, the molarity of the commercial vinegar is 0.833 M.

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describe how the orientaon of the glycosidic bond affects the properes of the polysaccharides it creates.

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The orientation of the glycosidic bond affects the properties of the polysaccharides it creates by determining the geometry of the sugar units in the polymer chain. When the glycosidic bond is in the alpha configuration, the sugar ring has a twisted conformation, which results in the sugar units being oriented in a more linear fashion.

In contrast, when the glycosidic bond is in the beta configuration, the sugar ring has a more planar conformation, which results in the sugar units being oriented in a more zig-zag fashion.

This difference in orientation affects the overall structure of the polysaccharide. Polysaccharides with alpha glycosidic bonds tend to form helical structures, while polysaccharides with beta glycosidic bonds tend to form sheet-like structures. This is because the twisted conformation of the alpha sugar units allows for the formation of hydrogen bonds between adjacent sugar units, which leads to the formation of a helix.

In contrast, the more planar conformation of the beta sugar units does not allow for the formation of hydrogen bonds between adjacent sugar units, which leads to the formation of a sheet.

Additionally, the orientation of the glycosidic bond affects the solubility and digestibility of the polysaccharide. Polysaccharides with alpha glycosidic bonds tend to be more soluble and more easily digested than polysaccharides with beta glycosidic bonds.

This is because the helical structure of alpha-polysaccharides allows for more surface area to be exposed to water and digestive enzymes, while the sheet-like structure of beta-polysaccharides does not.


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what term refers to the ability of open systems to fight off deterioration, sustain themselves and grow? a. requisite variety b. network properties c. negative entropy d. modeling techniques

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The ability of open systems to fight off deterioration, sustain themselves and grow is Negative Entropy. Correct answer is option C

Negative Entropy is an important concept in thermodynamics and physics, where it is defined as a decrease in the entropy of a system. Entropy is the measure of randomness or disorder in a system, so negative entropy indicates that a system is becoming more organized, or that it is moving away from equilibrium.

This can be seen in the evolution of life, where species become more complex and adaptive over time, as well as in the growth of technology, where innovations allow us to become more efficient and productive. In essence, Negative Entropy is the power that allows open systems to improve and evolve. Therefore Correct answer is option C

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