What is the pH of 6.00 M H2CO3 if it has 7% dissociation? SHOW YOUR WORK!!!

Answers

Answer 1

3.2  is the pH of 6.00 M H[tex]_2[/tex]CO[tex]_3[/tex] if it has 7% dissociation. pH is a numerical indicator of how acidic or basic aqueous and other liquid solutions are.

pH is a numerical indicator of how acidic or basic aqueous and other liquid solutions are. The word translates to the measurements of the hydrogen ion concentration and is used frequently in chemistry, biology, especially agronomy.

The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per litre, making it neutral (neither acid nor alkaline).

pH = -log[H⁺]      

7% of 6.00

0.42

pH = -log[0.42]      

pH = 3.2

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Related Questions

What was the practical effect of not having a legal identity

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The inability to exercise their civil and political rights is hampered by a lack of legal identity.

What is legal identity?

Legal identity is described as the fundamental aspects of an individual's identity, such as name, gender, place of birth, and date of birth, which are conferred through registration and the issue of a certificate by an authorized civil registration body following the occurrence of birth.

The individual's identity includes his or her family name, surname, date of birth, gender, and nationality.

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which of the following is an example of a lewis acid-base reaction? select the correct answer below: an ammonium ion combines with a chloride ion to form a salt. metallic potassium reacts violently in oxygen to form potassium oxide. zinc(ii) replaces copper(ii) as the cation in a chloride salt. a molecule of ammonia combines with a proton to form the complex ion known as ammonium.

Answers

The correct answer to the question is the fourth option, where a molecule of ammonia combines with a proton to form the complex ion known as ammonium. This is an example of a Lewis acid-base reaction, where ammonia acts as a Lewis base and accepts a proton from a Lewis acid, forming a new compound. The other options do not involve Lewis acid-base reactions. For example, the first option involves an ionic bond formation between an ammonium ion and a chloride ion, and the second option involves a redox reaction between metallic potassium and oxygen. The third option involves a single displacement reaction between zinc and copper ions in a chloride salt.

A Lewis acid is a molecule or an ion that can accept an electron pair, while a Lewis base is a molecule or an ion that can donate an electron pair. In a Lewis acid-base reaction, a Lewis acid accepts an electron pair from a Lewis base to form a new compound.

Among the given options, the fourth option is an example of a Lewis acid-base reaction. In this option, a molecule of ammonia (NH3) acts as a Lewis base and accepts a proton (H+) from a Lewis acid to form the complex ion known as ammonium (NH4+). Here, the proton acts as a Lewis acid by accepting the electron pair from the nitrogen atom in ammonia.
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write a balanced equation for the complete oxidation of propanoic acid. express your answer as a chemical equation. omit states of matter.

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A balanced equation for the complete oxidation of propanoic acid.

2CH₃CH₂COOH + 7O₂ =  6CO₂ + 6H₂O

Oxidation of acids gives carbon dioxide and water as the product.

Because it is already in a high oxidation state in propanoic  acid, further oxidation removes the carboxyl carbon as carbon dioxide.

Depending on the reaction conditions, the oxidation state of the remaining organic structure may be higher, lower or unchanged.

Thus, the balanced equation can be written as -

2CH₃CH₂COOH + 7O₂ =  6CO₂ + 6H₂O

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TRUE/FALSECellular respiration involves oxidation of organic molecules and an associated release of energy, some of which is stored in the bonds of ATP

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TRUE. Cellular respiration is a metabolic process that involves the oxidation of organic molecules, typically glucose, and the release of energy in the form of ATP. The process can be divided into three stages: glycolysis, the citric acid cycle, and oxidative phosphorylation.

During glycolysis, glucose is broken down into two molecules of pyruvate, and a small amount of ATP is produced. In the citric acid cycle, pyruvate is further broken down into carbon dioxide, and more ATP is produced. Finally, in oxidative phosphorylation, the electrons produced during the oxidation of glucose are passed through a series of electron carriers, which ultimately results in the production of a large amount of ATP. The oxidation of organic molecules during cellular respiration involves the removal of electrons and the associated release of energy. Some of this energy is captured in the form of ATP, which is used to power cellular processes such as muscle contraction, biosynthesis, and active transport. Therefore, it is true that cellular respiration involves oxidation of organic molecules and an associated release of energy, some of which is stored in the bonds of ATP.

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calculate [h3o ], [clo4–], and [oh–] in an aqueous solution that is 0.125 m in hclo4(aq) at 25 °c.

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We have calculated [H3O+], [Clo4-], and [OH-] in the given aqueous solution.

To start with, we need to understand the concept of acid dissociation constant (Ka) and how it relates to the concentrations of various species in an aqueous solution. For hclo4(aq), the Ka value is very high, indicating that it is a strong acid.
Now, let's look at the equation for the dissociation of hclo4(aq):
HClO4 + H2O ⇌ H3O+ + ClO4-According to this equation, one mole of hclo4(aq) reacts with one mole of water to form one mole of H3O+ ions and one mole of ClO4- ions. Since the solution is 0.125 M in hclo4(aq), the initial concentration of hclo4 is also 0.125 M.At equilibrium, let the concentration of H3O+ ions be x. Therefore, the concentration of ClO4- ions will also be x. The initial concentration of hclo4(aq) is 0.125 M, but some of it will dissociate to form H3O+ and ClO4- ions. Let's assume that a fraction 'α' of hclo4(aq) has dissociated. Therefore, the concentration of H3O+ ions and ClO4- ions at equilibrium will be x = α and [Clo4-] = α, respectively. Using the equation for Ka, we can write:
Ka = [H3O+][Clo4-]/[hclo4(aq)]Substituting the values we have obtained so far, we get:
Ka = α^2/(0.125-α)Since Ka is a constant for a given acid at a particular temperature, we can solve this equation to obtain the value of α, which is the fraction of hclo4(aq) that has dissociated. Once we know α, we can easily calculate the concentrations of H3O+ and ClO4- ions:
[H3O+] = [Clo4-] = αFinally, we can use the equation for Kw (the ion product constant for water) to calculate the concentration of OH- ions:
[tex]Kw = [H3O+][OH-] = 1.0 * 10^{-14}[H3O+] = α, \\so[OH-] = 1.0 * 10^{-14}/[H3O+][/tex]Therefore, we have calculated [H3O+], [Clo4-], and [OH-] in the given aqueous solution.

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Ethers can be cleaved using ______.A. NaOH or KOH B. H2SO4 or HCl C. HI or HBr D. HNO3 or H2SO4

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Ethers can be cleaved using a variety of reagents, but the most common methods involve either acid or base-catalyzed cleavage. Option A, NaOH or KOH, refers to base-catalyzed cleavage, also known as "Williamson ether synthesis." In this reaction, an ether is reacted with a strong base to form an alkoxide ion, which is then protonated to form an alcohol and an alkyl halide. This method is commonly used to synthesize alcohols from ethers.

Option B, H2SO4 or HCl, refers to acid-catalyzed cleavage. In this reaction, the ether is protonated by the acid to form an oxonium ion, which can then undergo nucleophilic attack by a variety of nucleophiles, such as water or an alcohol, to form two different products. This method is commonly used to prepare alcohols from ethers, but can also be used to form carbocations or other intermediates.

Option C, HI or HBr, refers to a specialized method of cleaving ethers called "acid-catalyzed hydrolysis." In this reaction, the ether is protonated by the strong acid, followed by nucleophilic attack by the halide ion. This method is commonly used to convert ethers into halohydrins.

Option D, HNO3 or H2SO4, is not commonly used to cleave ethers, but can be used to oxidize them to ketones or aldehydes. Overall, the choice of reagent used to cleave ethers depends on the specific desired products and reaction conditions.

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All processes either require energy or give off energy. Sort the examples by whether they are exothermic or endothermic
a mammal metabolizing food, reaction in a chemical "hot pack" for warming hand in winter, burning candle, setting off a firework, water evaporating, the reaction is a chemical "cold pack" for treating injuries, ice melting, photosynthesis plants using sunlight to turn CO2 into plant material

Answers

Exothermic processes release energy, while endothermic processes absorb energy.



Exothermic:
1. A mammal metabolizing food
2. Reaction in a chemical "hot pack" for warming hands in winter
3. Burning candle
4. Setting off a firework

Endothermic:
1. Water evaporating
2. The reaction in a chemical "cold pack" for treating injuries
3. Ice melting
4. Photosynthesis: plants using sunlight to turn CO2 into plant material

In summary, exothermic processes give off energy, while endothermic processes require energy.

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A 50.00 ml sample of 0.0897 m hcl is titrated with 0.111 m naoh. determine the ph of the solution after addition of 40.00 ml of naoh group of answer choices 3.301 1.301 2.301 3.201 no correct answer

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The pH of the solution after the addition of 40.00 mL of NaOH is approximately 3.301.

To determine the pH of the solution after the addition of 40.00 mL of NaOH, first, we need to find the moles of HCl and NaOH involved in the reaction. The reaction between HCl and NaOH is a neutralization reaction, which can be represented as follows:

HCl + NaOH → NaCl + H2O

Now, let's calculate the moles of both reactants:

Moles of HCl = (0.0897 mol/L) * (50.00 mL) * (1 L/1000 mL) = 0.004485 mol
Moles of NaOH = (0.111 mol/L) * (40.00 mL) * (1 L/1000 mL) = 0.00444 mol

Next, determine the moles of HCl remaining after the reaction:

Moles of HCl remaining = Moles of HCl - Moles of NaOH = 0.004485 - 0.00444 = 0.000045 mol

Now, calculate the concentration of the remaining HCl in the solution:

[HCl] = Moles of HCl remaining / Total volume of solution
[HCl] = 0.000045 mol / (50.00 mL + 40.00 mL) * (1 L/1000 mL) = 0.000045 mol / 0.090 L = 0.0005 mol/L

Finally, determine the pH of the solution using the formula:

pH = -log10[H+]
pH = -log10(0.0005)

pH ≈ 3.301

So, the pH of the solution after the addition of 40.00 mL of NaOH is approximately 3.301.

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How do polar aprotic solvents favor SN2 reactions?

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Polar aprotic solvents favor SN2 reactions because they weaken the nucleophile, stabilize the transition state, and do not solvate the leaving group.

In SN2 reactions, the nucleophile attacks the substrate at the same time as the leaving group departs, resulting in a single concerted step. Polar aprotic solvents, such as DMSO and acetone, do not have available hydrogen atoms to form hydrogen bonds with the nucleophile.

This weakens the nucleophile and increases its reactivity. Additionally, polar aprotic solvents are good at stabilizing the transition state because they can solvate the ions that form during the reaction. Finally, these solvents do not solvate the leaving group, which prevents the formation of a stable intermediate and favors the SN2 pathway.

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writes the reaction that extracts lidocaine from the toluene layer into the aqueous layer. then write the reaction used to recover lidocaine from the aqueous layer.

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The extraction of lidocaine from the toluene layer into the aqueous layer can be achieved by adding a suitable aqueous solution, such as a dilute acid or base, to the mixture and then shaking it to allow for partitioning of the lidocaine between the two layers.

The reaction that takes place can be represented as follows:

Toluene layer (organic phase):
Lidocaine + H+ (aq) --> Lidocaine-H+ (aq)

Aqueous layer (aqueous phase):
Lidocaine-H+ (aq) --> Lidocaine (aq) + H+ (aq)

To recover the lidocaine from the aqueous layer, a suitable organic solvent such as toluene can be added to the mixture and shaken to allow for partitioning of the lidocaine into the organic phase. The reaction that takes place can be represented as follows:

Aqueous layer (aqueous phase):
Lidocaine (aq) + Toluene (org) --> Lidocaine (org) + Toluene (aq)

Organic layer (toluene phase):
Lidocaine (org) can now be isolated by evaporation of the toluene solvent.

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the nuclear fission of u-235 can occur by different pathways. if 2 neutrons is/are absorbed by u-235 to produce 6 neutrons and br-87, identify other nucleus that is produced using the boxes and pull-down menu provided.

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Nuclear fission of U-235 can occur through various pathways when neutrons are absorbed. In the given scenario, U-235 absorbs 2 neutrons, producing 6 neutrons and Br-87.

To identify the other nucleus produced, we need to balance the mass and atomic numbers on both sides of the reaction. Before fission, we have: - U-235 (mass number 235, atomic number 92) - 2 neutrons (mass number 2, atomic number 0) After fission, we have: - 6 neutrons (mass number 6, atomic number 0) - Br-87 (mass number 87, atomic number 35) - Unknown nucleus (mass number X, atomic number Y) To balance the mass numbers: 235 + 2 = 87 + 6 + X X = 144 To balance the atomic numbers: 92 = 35 + Y Y = 57 Thus, the unknown nucleus has a mass number of 144 and an atomic number of 57, making it Lanthanum-144 (La-144). So, the nuclear fission of U-235 with 2 absorbed neutrons produces 6 neutrons, Br-87, and La-144.

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What effect do noncompetitive inhibitors have on the Vmax of an enzyme?

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Noncompetitive inhibitors have a distinct effect on the Vmax of an enzyme.

Vmax is the maximum rate of reaction an enzyme can achieve when all its active sites are saturated with substrate. Noncompetitive inhibitors bind to a site on the enzyme that is not the active site, known as an allosteric site. This binding results in a conformational change in the enzyme's structure that makes it less effective in converting substrate into product.

The binding of the noncompetitive inhibitor to the allosteric site reduces the enzyme's activity, but it does not affect the affinity of the enzyme for the substrate. Therefore, the Km value remains unchanged, and the enzyme-substrate complex formation remains the same. Moreover, the presence of the noncompetitive inhibitor reduces the number of active enzymes available to catalyze the reaction. This decrease in the number of active sites reduces the Vmax of the enzyme, meaning that the maximum rate of reaction the enzyme can achieve is reduced.

In summary, noncompetitive inhibitors bind to an allosteric site on the enzyme and cause a conformational change in its structure, which decreases its activity. The decrease in activity results in a reduction of the Vmax of the enzyme, which reduces the maximum rate of reaction it can achieve.

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One of the chemicals used in this experiment is SnCl2 (aq). What is its name?

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One of the chemicals used in this experiment is SnCl² (aq). The name of this compound is Tin(II) chloride.

Tin(II) chloride, also known as stannous chloride, is an inorganic compound consisting of one tin (Sn) atom and two chloride (Cl) atoms. It is a white crystalline solid that is soluble in water, which is why it's given in aqueous form (aq) in the experiment. Tin(II) chloride has a wide range of applications in various industries. It is used as a reducing agent in the manufacturing of other tin compounds, as well as a mordant in the textile industry for dyeing processes.

Additionally, it plays a significant role in electroplating, where it serves as a tin plating electrolyte. Furthermore, Tin(II) chloride is employed as a catalyst in the production of plastic polymers, and it can also be used as a food additive for the purpose of color retention. In summary, SnCl² (aq) or Tin(II) chloride is a versatile chemical compound with numerous applications in various industries. One of the chemicals used in this experiment is SnCl² (aq). The name of this compound is Tin(II) chloride.

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write a balanced reaction equation for the bromination of stilbene using pyridinium tri-bromide.find the limiting reagent in the following procedure.

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The balanced reaction equation for the bromination of stilbene using pyridinium tribromide is: C14H12 + Br2 → C14H10Br2 In this reaction, stilbene (C14H12) reacts with bromine (Br2) to form dibromostilbene (C14H10Br2).

The balanced reaction equation for the bromination of stilbene using pyridinium tri-bromide is: Stilbene + 3 Br2 + Pyridinium tribromide → 2,3,4,5-tetrabromostilbene + Pyridinium bromide In the given procedure, the amount of stilbene is not mentioned, so we cannot determine the limiting reagent directly.

To find the limiting reagent in a given procedure, you will need to compare the stoichiometric ratios of the reactants to determine which one will be completely consumed first during the reaction.

To do this, you'll need the amounts (usually in moles or grams) of both stilbene and pyridinium tribromide used in the procedure.

However, we can assume that the amount of pyridinium tri-bromide is in excess, and the limiting reagent will be the reactant that is completely consumed first.

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what minimum volume of 0.259 m potassium iodide solution is required to completely precipitate all of the lead in 180.0 ml of a 0.120 m lead (ii) nitrate solution? what minimum volume of 0.259 potassium iodide solution is required to completely precipitate all of the lead in 180.0 of a 0.120 lead nitrate solution? 41.7 ml 167 ml 83.4 ml 334 ml

Answers

The minimum volume of 0.259 M potassium iodide solution required to completely precipitate all of the lead in 180.0 ml of a 0.120 M lead (II) nitrate solution is 167 ml.

To solve this problem, we need to use the balanced chemical equation for the reaction between potassium iodide and lead (II) nitrate:

Pb(NO3)2 + 2KI → PbI2 + 2KNO3

From the equation, we can see that 1 mole of lead (II) nitrate reacts with 2 moles of potassium iodide to form 1 mole of lead (II) iodide, which is the precipitate we want to form.

We can use this information to calculate the number of moles of lead (II) nitrate in 180.0 ml of a 0.120 M solution:

moles of Pb(NO3)2 = M × V = 0.120 mol/L × 0.180 L = 0.0216 mol

Since we need 2 moles of potassium iodide for every mole of lead (II) nitrate, we need:

moles of KI = 2 × moles of Pb(NO3)2 = 2 × 0.0216 mol = 0.0432 mol

Finally, we can calculate the volume of 0.259 M potassium iodide solution needed to provide this many moles of KI:

V = moles of KI / M = 0.0432 mol / 0.259 mol/L = 0.167 L = 167 ml

Therefore, the minimum volume of 0.259 M potassium iodide solution required to completely precipitate all of the lead in 180.0 ml of a 0.120 M lead (II) nitrate solution is 167 ml.

The closest answer from the given options is 167 mL. So, the minimum volume of 0.259 M potassium iodide solution required to completely precipitate all of the lead in 180.0 mL of a 0.120 M lead(II) nitrate solution is approximately 167 mL.

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please help me answer this asap

Answers

Answer: the answer is A

Explanation: its alot to put down

which substance is the least soluble in water? ammonia uric acid urea sodium chloride

Answers

Sodium chloride is the least soluble in water among the given substances. This is the correct option.

Sodium chloride (NaCl) is an ionic compound composed of positively charged sodium ions (Na+) and negatively charged chloride ions (Cl-).

When NaCl is added to water, the polar water molecules interact with the ions and pull them apart from each other, resulting in the dissociation of NaCl into Na+ and Cl- ions. These ions are then surrounded by water molecules, and the compound dissolves in water.

However, NaCl is relatively insoluble in nonpolar solvents like benzene and hexane, which do not interact with the ions and are unable to dissociate the compound.

This property is due to the strong ionic bond between Na+ and Cl- ions, which requires a significant amount of energy to break apart.

In contrast, ammonia, uric acid, and urea are organic compounds that contain polar functional groups that allow them to form hydrogen bonds with water molecules, making them more soluble in water than NaCl.

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The ground-state electron configuration of a Cr²⁺ ion is 1s²2s²2p⁶3s²3p⁶3d⁴. Therefore Cr²⁺ is

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The ground-state electron configuration of a Cr2+ ion is 1s2 2s2 2p6 3s2 3p6 3d4. Therefore, Cr2+ is a chromium ion that has lost two electrons from its neutral atom configuration of 1s2 2s2 2p6 3s2 3p6 3d5, resulting in a 3d4 electron configuration.

The ground-state electron configuration of a Cr2+ ion is 1s2 2s2 2p6 3s2 3p6 3d4. Therefore, Cr2+ is an ion with 20 electrons. 1. Chromium (Cr) has an atomic number of 24, meaning it has 24 electrons in its neutral state. 2. Cr2+ indicates that the chromium atom has lost 2 electrons, leaving it with 22 electrons.

3. The given electron configuration (1s2 2s2 2p6 3s2 3p6 3d4) accounts for 20 electrons, meaning there's an error in the configuration. 4. The correct electron configuration for Cr2+ should be 1s2 2s2 2p6 3s2 3p6 3d4 4s2, which accounts for all 22 electrons in the ion.

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Determine the pressure of the collected hydrogen gas,
P H 2


gas, in
mmHg
. The atmospheric pressure acting upon the surface of the water in the beaker will be equal to the sum of the partial pressures in the buret: the pressure of the liquid water column (hydrostatic pressure), the pressure of water vapor above the liquid water column, and the pressure of the collected hydrogen gas. Applying Dalton's Law of Partial Pressures:
P atm

=P H 2

O liquid ​
+P H 2

O vapor ​
+P H 2

gas ​
Where,
P atm

= atmospheric pressure P H 2

O liquid ​
= Height difference of solutions (mm)× 13.60 mm water 1mmHg

P H 2

O vapor ​
= vapor pressure of water at recorded temperature ​
Use the following table of vapor pressure of water at various temperatures to determine
P H 2

O

vapor. You may need to round your recorded temperaure to the nearest degree. Submit your answers to this question to the appropriate number of significant figures, unless otherwise specified in the table, but carry forward all the digits of your calculated answers for use in subsequent calculations. 3. Carry out the necessary conversions, then use the ideal gas law to determine moles of hydorgen gas that were collected. Note: Remember that the buret was inverted. Make sure to report a positive volume of
H 2

gas. Submit your answers to this question to the appropriate number of significant figures, but carry forward all the digits of your calculated answers for use in subsequent calculations. Report Table VCE.5: Moles of hydrogen gas collected Table view List view Report Table VCE.5: Unit conversions and calculation of moles of hydrogen gas using the ideal gas law Trial 1 Trial 2 Collected
H 2

pressure,
P H 2


(atm) Temperature
( ∘
C)
21.9 21.9 Temperature (K) Initial buret reading
(mL)
Final buret reading
(mL)
12.80 12.99 Volume of
H 2

gas (mL) Volume of
H 2

gas (L) Moles of collected
H 2

( molH 2

)

Answers

The pressure of the collected hydrogen gas is 738.7 mmHg , and the moles of hydrogen gas collected is 9.73 x 10⁻⁶ mol.

To determine the pressure of the collected hydrogen gas, we need to apply Dalton's Law of Partial Pressures, which states that the total pressure ([tex]P_{total[/tex]) of a gas mixture is equal to the sum of the partial pressures of each gas:

[tex]P_{total[/tex] = [tex]P_{H{2} O[/tex] liquid + [tex]P_{H2O[/tex] vapor + [tex]P_{H2[/tex] gas

where [tex]P_{H{2} O[/tex] liquid is the hydrostatic pressure due to the water column in the burette, [tex]P_{H{2} O[/tex] vapor is the vapor pressure of water at the temperature of the experiment, and [tex]P_{H2[/tex] gas is the pressure of the collected hydrogen gas.

To calculate [tex]P_{H{2} O[/tex] liquid, we need to determine the height difference of the solutions in the buret, which is given by the difference between the initial and final buret readings. The volume of the liquid in the buret can be converted to millimeters of water using the conversion factor 1 mL = 13.60 mm H₂O. Thus, we have:

Height difference of solutions = (final buret reading - initial buret reading) x 13.60 mm H₂O/mL

To determine [tex]P_{H2O[/tex] vapor, we need to look up the vapor pressure of water at the recorded temperature in the provided table. We then convert the vapor pressure to millimeters of mercury (mmHg) using the conversion factor 1 atm = 760 mmHg.

Once we have calculated [tex]P_{H2O[/tex] liquid, [tex]P_{H2O[/tex] vapor, and the atmospheric pressure (which we assume is equal to 1 atm), we can solve for [tex]P_{H2[/tex] gas. We then use the ideal gas law to determine the moles of hydrogen gas collected.

Let's apply this process to the given data:

Trial 1:

Collected H₂ pressure, [tex]P_{H2[/tex] = 21.9 atm

Temperature = 21.9°C = 295.05 K

Initial buret reading = 12.80 mL

Final buret reading = 12.99 mL

Volume of H₂ gas = final buret reading - initial buret reading = 0.19 mL

Height difference of solutions = (12.99 - 12.80) x 13.60 mm H2O/mL = 2.572 mmHg

Vapor pressure of water at 21.9°C = 18.7 mmHg (from the provided table)

Atmospheric pressure = 1 atm = 760 mmHg

[tex]P_{total[/tex] = PH₂O liquid + PH₂O vapor + PH₂ gas

PH₂ gas = [tex]P_{total[/tex] - PH₂O liquid - PH₂O vapor

PH₂ gas = 760 - 2.572 - 18.7 = 738.728 mmHg

Converting the volume of H₂ gas to liters and the pressure of H₂ gas to atmospheres:

Volume of H₂ gas = 0.19 mL = 0.00019 L

PH₂ gas = 738.728 mmHg / 760 mmHg/atm = 0.9712 atm

Using the ideal gas law:

PV = nRT

n = PV/RT = (0.9712 atm)(0.00019 L)/(0.08206 L·atm/mol·K)(295.05 K) = 9.73 x 10⁻⁶ mol

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URGENT
What is the pH of 6.00 M H2CO3 if it has 7% dissociation? SHOW YOUR WORK!!!

Answers

pH is a numerical indicator of how acidic or basic aqueous or other liquid solutions are.

Thus, The pH, which is frequently used in chemistry, biology, and agronomy, converts the hydrogen ion concentration, which typically ranges between 1 and 1014 gram-equivalents per liter, into numbers between 0 and 14.

The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral (neither acidic nor alkaline).

A solution with a pH below 7 is referred to as acidic, and one with a pH over 7 is referred to as basic, or alkaline.

Thus, pH is a numerical indicator of how acidic or basic aqueous or other liquid solutions are.

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For a drug following first order degradation kinetics, if t1/2 is 1 hour and the conc. at 4 pm is 0.1M what is the conc. at 7 pm?

Answers

The concentration of the drug at 7 pm is approximately 0.031 M.

If a drug follows first-order degradation kinetics, the rate of degradation is proportional to the concentration of the drug.

The half-life (t1/2) is the time it takes for the concentration of the drug to decrease by half, and it is related to the rate constant (k) as follows:

t1/2 = ln(2)/k

Rearranging this equation, we get:

k = ln(2)/t1/2

Using the given half-life of 1 hour, we can calculate the rate constant as:

k = ln(2)/1 hour = 0.693/hour

Now we can use the first-order degradation equation to calculate the concentration of the drug at 7 pm.

Let's assume that the concentration of the drug at 4 pm is 0.1 M, and we want to find the concentration at 7 pm, which is 3 hours later.

The first-order degradation equation is:

ln(Ct/C0) = -kt

where Ct is the concentration at time t, C0 is the initial concentration, k is the rate constant, and t is the time elapsed.

Substituting the given values, we get:

ln(Ct/0.1 M) = -(0.693/hour) x 3 hours

ln(Ct/0.1 M) = -2.079

Ct/0.1 M = e^-2.079

Ct = 0.031 M

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Help with last three

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In the first  two  chemical equations, double displacement reactions occur as the ions are exchanged between the 2 reactants while in last one no reaction takes place as both ammonium and potassium salts are soluble.

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .

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draw the neutral organic product that results from the following reaction. draw all hydrogen atoms.

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The neutral compound formed is 2-butanone, it is also known as methyl ethyl ketone, with a chemical formula [tex]CH_3C0CH_2CH_3[/tex].

When 2-butanol reacts with [tex]Cr_2O_7^{2-}[/tex] which is also known as a dichromate ion, in the presence of [tex]H^+[/tex] (protons), the 2-butanol undergoes an oxidation reaction and forms an aldehyde. The resultant aldehyde which is formed will depend on external conditions like temperature, concentration of the reactants, etc.

Oxidation reaction is the process in which in order to form the ions and chemical bonds, the elements lose the electrons. The electron loss will result in the element attaining a stable complete electronic configuration.

The overall reaction that can be drawn out from the equation is:

[tex]CH_3CH_2CHO + Cr^3+ H_2O[/tex]

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The complete question is: draw all the neutral organic product that is formed when [tex]CH_3CH0HCH_2CH_3[/tex] (l) is reacted with [tex]Cr_2O_7^{2-}[/tex] (aq) in the presence of [tex]H^+[/tex](aq). draw all the hydrogen atom

The image shows a large bar magnet with the poles labeled.

How would the poles of the magnet change if it were divided into four equal parts?

Answers

Bar magnets, cube magnets and block magnets are the most common magnet shape for every day mounting and holding applications.

Thus, They have completely flat surfaces with right angles (90°).  Square, cube or rectangular in shape, these magnets are widely used for holding and mounting applications, and can be combined with other hardware (such as channels) to increase their holding force.

These magnets, which can be square, cube, or rectangular in shape, are frequently used for holding and mounting purposes and can be paired with other hardware (such channels) to strengthen their gripping power.

Due to the fact that they are magnetized down their length, bar magnets operate well at both ends.

Thus, Bar magnets, cube magnets and block magnets are the most common magnet shape for every day mounting and holding applications.

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what is the difference between a battery and a fuel cell? take advantage of spontanous redox reactions

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Both batteries and fuel cells are devices that convert chemical energy into electrical energy through spontaneous redox reactions, but the main difference is that batteries are self-contained and have a finite amount of stored energy, while fuel cells require a continuous supply of fuel to sustain the reaction.

Batteries and fuel cells are both electrochemical devices that convert chemical energy into electrical energy through spontaneous redox reactions.

In batteries, this energy is stored chemically within the cell and can be released as electrical energy as needed. However, batteries have a finite amount of stored energy and need to be recharged or replaced when depleted.

On the other hand, fuel cells require a continuous supply of fuel, such as hydrogen or methanol, and an oxidizing agent, such as oxygen, to sustain the redox reaction that produces electricity.

Fuel cells can operate continuously as long as fuel and an oxidizing agent are supplied, making them useful for applications such as electric vehicles and stationary power generation. Additionally, fuel cells produce only water and heat as byproducts, making them a cleaner alternative to traditional combustion engines.

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TRUE/FALSEThe rate of enzymatic reaction depends on conditions in the immediate environment

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TRUE. The rate of enzymatic reaction is influenced by various factors such as temperature, pH, substrate concentration, and presence of inhibitors or activators in the immediate environment.

Changes in these conditions can affect the activity and efficiency of enzymes, leading to alterations in the rate of the enzymatic reaction. Enzymes are highly specific and their activity can be modulated by altering these factors in their immediate environment.

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Calculate the formula mass for each of the following halogen containing compounds to five or more significant figures.
(a) {eq}BrN_3 {/eq} (explosive)
(b) {eq}CCl_2F_2 {/eq} (former refrigerant)

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The formula mass for BrN₃ is 121.925.

The formula mass for CCl₂F₂ is 120.913.

(a) To calculate the formula mass for BrN₃, follow these steps:
1. Determine the atomic mass of each element: Br = 79.904, N = 14.007
2. Multiply the atomic mass by the number of atoms in the compound: (1 × 79.904) + (3 × 14.007)
3. Add the masses together: 79.904 + 42.021 = 121.925

The formula mass for BrN₃ is 121.925.

(b) To calculate the formula mass for CCl₂F₂, follow these steps:
1. Determine the atomic mass of each element: C = 12.011, Cl = 35.453, F = 18.998
2. Multiply the atomic mass by the number of atoms in the compound: (1 × 12.011) + (2 × 35.453) + (2 × 18.998)
3. Add the masses together: 12.011 + 70.906 + 37.996 = 120.913

The formula mass for CCl₂F₂ is 120.913.

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At 520 nm, if 60% of light transmit through a cuvetteand 40% of light is absorbed by the solution in the cuvette, please calculate transmittance T% and absorbance A.

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The transmittance (T%) at 520 nm is 60%, and the absorbance (A) of the solution in the cuvette at 520 nm is approximately 0.22.


Transmittance and absorbance are important parameters used to measure the amount of light passing through a sample. Transmittance is the ratio of the amount of light transmitted through the sample to the amount of light incident on the sample. Absorbance is a measure of how much light is absorbed by the sample.

Transmittance is the percentage of light that passes through a medium, such as a cuvette.
Step 1: Given that 60% of light is transmitted, the transmittance (T%) is already provided.
T% = 60%
Absorbance is a measure of how much light is absorbed by a solution.
Step 1: Calculate the fraction of light transmitted by dividing T% by 100.
Fraction transmitted = T% / 100 = 60 / 100 = 0.6

Step 2: Use the Beer-Lambert Law formula to calculate absorbance (A), which is A = -log10(I/I₀), where I is the transmitted light intensity, and I₀ is the incident light intensity.
Since the fraction transmitted is I/I₀, we have A = -log10(0.6).

Step 3: Calculate the absorbance (A).
A = -log10(0.6) ≈ 0.22

The transmittance (T%) at 520 nm is 60%, and the absorbance (A) of the solution in the cuvette at 520 nm is approximately 0.22.

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Cual es la concentracionmaltilde estaba realizando una investigación sobre la solubilidad de una sal en agua y encontré que esta era de 0. 5g/ml su maestra ñe pidió que preparar una solución utilizando 50g de sal en 100ml de agua

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The concentration of the solution prepared by Maltilde is calculated to be the same as the solubility of the salt in water, which is equal to0.5 g/mL.

In the given scenario, the solubility of the salt in water is mentioned as 0.5 g/mL. This means that at a given temperature and pressure, one milliliter of water can dissolve 0.5 grams of the salt.

The teacher has asked Maltilde to prepare a solution using 50g of salt in 100ml of water. To find the concentration of this solution, we can use the formula:

Concentration (in g/mL) = Mass of solute (in g) / Volume of solution (in mL)

Substituting the values, we get:

Concentration = 50 g / 100 mL

Concentration = 0.5 g/mL

Therefore, the concentration of the solution prepared by Maltilde is the same as the solubility of the salt in water, which is 0.5 g/mL.

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The complete question is :

Maltilde was doing an investigation on the solubility of a salt in water and found that it was 0.5g/ml her teacher asked her to prepare a solution using 50g of salt in 100ml of water. What is the concentration?

Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water H2O and carbon dioxide CO2 chemically react to form the simple sugar glucose C6H12O6 and oxygen gas O2.

What mass of simple sugar glucose is produced by the reaction of 8.87g of carbon dioxide?

Round your answer to 3 significant digits.

Answers

The stoichiometric coefficient is used here to determine the mass of water produced. Here we use the ratios from the balanced equation to calculate the amount of a substance. The mass of glucose is 5.94 g.

Stoichiometry is an important concept which use the balanced equation to determine the amounts of reactants and products.

Here the balanced equation is:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Molar mass of glucose = 180.156

Moles of CO₂ = 8.87 / 44.01 = 0.2015 mol

Moles of glucose = 0.2015 mol × 1 mol glucose / 6CO₂ = 0.033

Mass of glucose = 0.033 × 180.156 = 5.94 g

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