Which of the following is an important intermediate in the mechanism of the reaction
in the box?​

Which Of The Following Is An Important Intermediate In The Mechanism Of The Reactionin The Box?

Answers

Answer 1

Answer:

Explanation:

.


Related Questions

Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm  L / mole  K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.

Answers

Answer:

1. 0.98 atm

Explanation:

The following data were obtained from the question:

Mass of unknown liquid (m) = 0.3175 g

Temperature (T) = 99 °C

Pressure (P) = 748.2 mmHg

Volume (V) = 145.0 mL

Gas constant (R) = 0.08206 atm.L/Kmol

1. Determination of the pressure in atm.

760 mmHg = 1 atm

Therefore,

748.2 mmHg = 748.2/760 = 0.98 atm

Therefore, the pressure in atm is 0.98 atm.

If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?

Answers

The Average atomic mass of phosphorus is 29.9.

What is Average atomic mass ?

The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.

Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.

An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.

The versions of an element with different neutrons have different masses and are called isotopes.

The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,

Average atomic mass of P = ∑(Isotope mass) (its abundance)

∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)

Abundance of isotope = % of the isotope / 100.

∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.

Hence , The Average atomic mass of phosphorus is 29.9.

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Draw the Lewis structure of acetaldehyde (CH₃CHO) and then choose the appropriate pair of molecular geometries of the two central atoms.
a) tetrahedral/trigonal pyramidal.
b) trigonal planar / linear.
c) tetrahedral / trigonal planar.
d) trigonal/tetrahedral.
e) planar /trigonal planar.

Answers

Given that,

Draw the Lewis structure of acetaldehyde (CH₃CHO).

We know that,

The Lewis structure shows the number of electrons around an atom.

According to structure,

We need to find the molecular geometries of the two central atoms

Using molecular geometries

For first central atom,

Number of bond pair = 2

Here, double bond to O count as single bond

The number of lone pair is zero.

The geometry is Trigonal planar.

For second central atom,

Number of bond pair = 4

The number of lone pair is zero.

The geometry is tetrahedral

Hence, The molecular geometries of the two central atoms are trigonal planar and tetrahedral.

(d) is correct option.

The central carbon atoms in acetaldehyde have a tetrahedral geometry and a trigonal planar geometry respectively.

Acetaldehyde has two central carbon atoms. The Lewis structure of acetaldehyde shows the arrangement of electrons around the atoms in the compound. The lone pairs are shown as dots while the bond pairs are represented using a single dash.

The first central carbon atom in acetaldehyde has a tetrahedral geometry while the second central carbon atom in acetaldehyde has a trigonal planar geometry.

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What is the percent composition of sulfur in SO2?

OA

32.1 g

(32.1 g + 16.0 g + 16.0 g)

x 100%

32.1 g

O B.

x 100%

(16.0 g + 16.0 g)

O c.

(32.1 g + 16.0 g + 16.0 g)

X 100%

32.1 g

(32.1 g)(6.02 x 102)

O D.

100 g

Answers

Answer:

5 .04 percent.

Explanation:

it is also known as sulfur dioxide. so it's 5.0.4 percent.

Your question is incomplete. Read below to find the content.

0.4 % is the percent composition of sulfur in SO2.

How is SO2 prepared from sulfur?

In the laboratory, sulfur dioxide is prepared by the reaction of metallic sulfite or a metallic bisulfite with dilute acid. For example, a reaction between the dilute sulphuric acid and sodium sulfite will result in the formation of SO2.

What do you mean by percentage composition?

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. Here, the quantity is measured in terms of grams of the elements present.

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The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?

Answers

Answer:

The peaks are registered from tetramethyl silane (TMS)

Explanation:

Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.

Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.

I hope this helped.

Fill in the blanks
1. the ____ atoms of the ____ groups and the hydrogen atoms attached to the ____ atoms form ____.
(answer choices; nitrogen, carbon, a-helices or b-pleated sheets, carbonyl, globular of firbrous structures, oxygen, amino, random coil, thiol)
2. In the a helix, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in _____. In the B-pleasted sheet, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in ____.
(answer choices; parallell sections of a long polypeptide chain, the next amino acid unit, carboxyl, the next turn of the helical chain, thiol, amino, carbonyl)

Answers

Answer:

1. oxygen; carbonyl; nitrogen; α-helices or β-pleated sheets

2. carbonyl; amino; the next turn of the helical chain; carbonyl; amino; parallel sections of a long polypeptide chain.

Explanation:

The secondary structure of proteins is of two major conformations, the α helix and β conformations which are very stable.

1. The oxygen atoms of the carbonyl groups and the hydrogen atoms attached to the nitrogen atoms form α-helices or β-pleated sheets.

2. In the α-helix, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in the next turn of the helical chain. In the β-pleated sheet, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in parallel sections of a long polypeptide chain.

18. Sucralose contains which two functional groups: (2 points)

A) benzene

B) halogen

C) carboxyl

D) hydroxy!

Answers

Answer:

The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.

Explanation:

An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.  

The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.  

To determine the concentration of chloride ion (Cl-) in a 100 mL sample of ground water, a chemist adds a large enough volume of AgNO3 solution to precipitate all Cl- as AgCl. The mass of the resulting precipitate is 93.9 mg. What is the chloride ion concentration in milligrams of chloride per liter of ground water

Answers

Answer:

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Explanation:

Hello,

In this case, we can represent the chemical reaction as:

[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:

[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Best regards.

The concentration of chloride ions in the groundwater sample is 230 mg/L.

Calculation of chloride ion concentration:

Based on the given information,

The mass of the resulting precipitate, that is, AgCl is 93.9 mg or 0.0939 g. The molar mass of AgCl is 143.2 g/mol.

Now the number of moles of AgCl precipitate can be calculated as,

n = Given mass/Molar mass

Now putting the values we get,

[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]

Thus, 6.5 ×  10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.

Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,

Mass = Mole × Molar mass

Mass = 6.5 × 10⁻⁴ moles  × 35.453 g/mol

Mass = 0.0230 g or 23 mg

The volume of the groundwater sample is 100 ml or 0.1000 L.

Now the concentration of the chloride ions in the sample given is,

C = 23 mg/0.1000 L

C = 230 mg/L

Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.

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Natural atom of the same element may have the same _________?
A)proton
B)neutron
C)electron
D)All​

Answers

Answer:B

Explanation:

Answer: i think it is c

Explanation: i checked my textbook.

6. Potassium hydrogen phthalate (KHP, KHC8H4O4) is also a good primary standard. 20 mL of NaOH was titrated with 0.600 M KHC8H4O4 solution. The data was graphed and the equivalence point was found when 15.5 mL of the standard 0.600 M KHP solution was added. The reaction equation is: a. What is the molar ratio of NaOH:KHC8H4O4? b. What is the molarity of the NaOH solution?

Answers

Answer:

a. 1

b. 0.465M NaOH

Explanation:

KHP reacts with NaOH as follows:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:

1/1 = 1

b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:

15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP

In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.

The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:

0.0093 moles NaOH / 0.020L =

0.465M NaOH

a. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.

The balanced equation for the reaction:

NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O

b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point

Molarity of KHP solution = 0.600 M

Volume of KHP solution = 15.5 mL = 0.0155 L

Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L

Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point

Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L

Molarity of NaOH solution ≈ 0.465 M

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If the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%, how can the cell obtain more sugar? {Hint: Sugar is polar and does not pass through the cell membrane easily.}
Select one:
a. Sugar can undergo facilitated diffusion through a channel protein, and no energy is required.
b. The cell must use active transport to pump the sugar in. Energy is required.
c. Sugar can diffuse straight through the phospholipid bilayer.
d. The cell CANNOT obtain more sugar. It is doomed.

Answers

The correct answer is A) Sugar can undergo facilitated diffusion through a channel protein, and no enlever his required.
!

The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.

FACILITATED DIFFUSION:

Facilitated diffusion is the process whereby molecules move across a cell through the help of carrier/channel proteins.

Facilitated diffusion is a type of passive transport and hence, does not require energy to occur. This is because movement occurs down a concentration gradient (high to low).

According to this question, the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%. This means that the sugar can travel down a concentration gradient across the cell membrane.

However, because sugar is a polar molecule and does not pass through the cell membrane easily, a carrier proteins is needed to aid its movement.

Therefore, the cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.

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A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?

Answers

Answer:

3 fluorine atoms will be present

Answer:

3

Explanation:

The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.

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A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is

Answers

Answer:

1. EF = PSCl₃; 2. MF = PSCl₃  

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

Draw the structure of 1,4-hexanediamine.

Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.

View Available Hint(s)

Answers

Answer:

1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.

Explanation:

1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.

The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.

Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].

1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.

The structure of 1,4-hexanediamine is shown below.

if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?

Answers

Answer:

The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.

In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.

Answers

Answer:

The correct answer is statement A.

Explanation:

In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.  

For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.  

The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.

What is TLC ?

In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.

The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.

TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.

Thus, option A is correct.

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A general chemistry student found a chunk of metal in the basement of a friend's house. To figure out what it was, she tried the following experiment. First she measured the mass of the metal to be 385.8 grams. Then she dropped the metal into a measuring cup and found that it displaced 17.8 mL of water. Calculate the density of the metal. Density = _______ g/mL Use the table below to decide the identity of the metal. This metal is most likely _________.
substances density g/cm3
water 1.00
aluminium 2.72
chromium 7.25
nickel 8.91
silver 10.50
lead 11.34

Answers

1. 21.67g/ml

2. aluminium

Explanation:

1. density = mass/volume

385.8/17.8= 21.67ml

2. 1g/ml=0.1g/cm^3

21.67g/ml = 2.167g/cm^3

..... substance is probably aluminium

1.  the density of the metal is 21.67g/ml

2.  This metal is most likely aluminum

The calculation is as follows;

1.

[tex]density = mass \div volume[/tex]

[tex]385.8\div 17.8= 21.67ml[/tex]

2.

1g/ml=0.1g/cm^3

So,  

21.67g/ml = 2.167g/cm^3

Therefore,  substance is probably aluminum

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Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.

Answers

Answer:

2NaOH(s) → Na₂O(s) + H₂O(g)

Hope that helps.

1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G

Answers

Answer:

The reaction given in the question is:  

2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)

The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:

ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.  

The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.  

Now the values of ΔH° and ΔS° are,  

ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)

ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)

ΔH°rxn = -1277.56 kJ/mole

ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)

ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15

ΔS°rxn = 313.11 J/mole/K

Now the formula for calculating ΔG°rxn is,  

ΔG°rxn = ΔH°rxn - TΔS°rxn

ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole

ΔG°rxn = -1370.86 kJ/mol

1. Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest.
A. HCl
B. H2S
C. HBr
D. BH3
2. Without consulting the table of acid-dissociation constants, match the following acids to the given Ka1 values.
1. H2S
2. H2SO3
3. H2SO4
A. Kal = 1.7 x 10^-7
B. Kal = 1.7 x 10^-2
C. Kal = very large

Answers

Answer:

ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!

Es urgente

Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.

Answers

Answer:

[tex]m_{air}=0.947g[/tex]

[tex]n_{O_2} =0.00686molO_2[/tex]

Explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]

Best regards.

Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.

Answers

Answer:

Its C I hopefully help you

Yes he have reason the answer is C

We wear cotton clothes in summer.

Answers

Answer:

we wear cotton clothes because it helps to cool us down and remove the excess heat that causes us to feel hot.

Answer:

[tex]\boxed{\mathrm{view \: explanation}}[/tex]

Explanation:

We wear cotton clothes in the summer beacuse cotton absorbs and removes body moisture caused by the sweat and allows better air circulation than fabric clothes.

Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.

Answers

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture

Answers

Answer:

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K →  23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K  . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)

2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations

Answers

Answer:

1. 0.138g of valium would be lethel in the woman

2. 125mg/min is the drip of the patient

Explanation:

1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.

A person that weighs 200lb requires:

200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)

90.72kg × (1.52mg / kg) =

137.9mg ≡

0.138g of valium would be lethel in the woman

2. The IV contains 1.5g = 1500mg/mL.

If the patient is receiving 5.0mL/h, its rate in mg/h is:

5.0mL/h × (1500mg/mL) = 7500mg/h

Now as 1h = 60min:

7500mg/h × (1h / 60min) =

125mg/min is the drip of the patient

Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.

Answers

Answer:

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Explanation:

Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),

Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of  48  electrons surrounding its nucleus.

The electronic configuration of Cadmium is;

Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

Cd2+ means that it has two less electrons, hence it's electron configuration is given as;

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)

Answers

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound

Answers

Answer:

This description shows a methyl group.

Explanation:

This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False

Answers

Answer:

The given statement is false.

Explanation:

However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.

So that the given is incorrect.

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