10 ml graduated cylinder (mL stands for milliliter)


• gram scale


• Water


• 6 metal paper clips of the same size and material


Part A


Use the gram scale to measure the mass of the empty graduated cylinder, and record the value

The concentration of the H₂SO₄ solution Erica used was approximately 2.0 M.

To find the concentration of H₂SO₄ solution used by Erica, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between KOH and H₂SO₄:

KOH + H₂SO₄ -> K₂SO₄ + 2H2O

From the balanced equation, we can see that the mole ratio of KOH to H₂SO₄ is 1:1. This means that the number of moles of H₂SO₄ used in the reaction is equal to the number of moles of KOH. We can use this fact to calculate the number of moles of H₂SO₄ used:

moles of KOH = volume of KOH solution (in L) x concentration of KOH solution
moles of KOH = 80.0 mL x (1 L/1000 mL) x 0.70 mol/L = 0.056 mol

Since the mole ratio of KOH to H₂SO₄ is 1:1, the number of moles of H₂SO₄ used is also 0.056 mol. Now we can use the same formula as above to calculate the concentration of H₂SO₄:

concentration of H₂SO₄ = moles of H2SO4 / volume of H₂SO₄ solution (in L)
concentration of H₂SO₄ = 0.056 mol / (28.0 mL x 1 L/1000 mL) = 2.00 mol/L

Therefore, the concentration of the H₂SO₄ solution Erica used was 2.00 M.

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Answer:

Carbon dioxide:

One carbon circle with 2 oxygen circles connected to it.

Ammonia:

One nitrogen circles with 3 hydrogen circles connected to it.

Oxygen:

2 oxygen circles connected to each other.

Hydrogen:

2 hydrogen circles stuck together

Answer:To find the number of CO2 molecules in a 0.25 g sample of dry ice, we can use the Avogadro's number and the molar mass of CO2.The molar mass of CO2 is:12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/molThis means that 1 mole of CO2 contains 6.022 x 10^23 molecules.To find the number of moles in 0.25 g of CO2, we can use the molar mass:0.25 g / 44.01 g/mol = 0.005681 molFinally, we can use Avogadro's number to find the number of CO2 molecules:0.005681 mol x 6.022 x 10^23 molecules/mol = 3.422 x 10^21 CO2 moleculesTherefore, a 0.25 g sample of dry ice contains approximately 3.422 x 10^21 CO2 molecules.

1. 2 moles of Calcium 8016 grams = 8.01x103 grams, 2. 3 moles of Sodium 69 grams = 2.07x1023 particles, and many more given below:

Calcium is an essential mineral that is found in the human body. It is necessary for the proper functioning of many organs, including the heart and muscles. Calcium is also important for the formation and maintenance of healthy bones and teeth. It plays a role in nerve and muscle function, blood clotting, and hormone secretion. Adequate calcium intake is important for both children and adults to ensure proper growth, development, and overall health.

3. 392 grams of Technetium = 9.10x1022 particles

4. 3.01x1024 particles of Chromium = 8.41x10-2 moles

5. 5 moles of Fluorine = 25 grams

6. 24 grams of Helium = 4.67x1023 particles

7. 1.806x1024 particles of Sulfur = 4.86x10-2 moles

8. 3 moles of Platinum = 195.2 grams

9. 240 grams of Argon = 6.67x1023 particles

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You would need 740.1 grams of oxygen to completely react with 254 grams of tristearin, C₅₇H₁₁₀O₆, in the given reaction.

To find out how many grams of oxygen are needed to completely react with 254 g of tristearin, C₅₇H₁₁₀O₆, in the given reaction, follow these steps:

1. Calculate the molar mass of tristearin (C₅₇H₁₁₀O₆) and oxygen (O₂).
2. Convert grams of tristearin to moles using its molar mass.
3. Use stoichiometry to find the moles of oxygen needed.
4. Convert moles of oxygen to grams using its molar mass.

Molar mass of tristearin: (57 * 12.01) + (110 * 1.01) + (6 * 16.00) = 891.62 g/mol
Moles of tristearin: 254 g / 891.62 g/mol = 0.285 moles
Moles of oxygen needed: 0.285 moles * (163 O₂ / 2 C₅₇H₁₁₀O₆) = 23.16 moles
Molar mass of O₂: 2 * 16.00 = 32.00 g/mol
Grams of oxygen needed: 23.16 moles * 32.00 g/mol = 740.1 g

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You need to add 0.405 moles of CH₃NH₃Cl to 200.0 mL of 0.500 M CH₃NH₂ to create a buffer with a pH of 11.

To find the moles of CH₃NH₃Cl needed, you'll need to use the Henderson-Hasselbalch equation and the given information.

The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]).

First, calculate pKa using the given Kb value for CH₃NH₂:

pKa = -log(Ka)

= -log(Kw/Kb)

= -log(1.0 × 10⁻¹⁴ / 4.4 × 10⁻⁴)

= 10.36.

Then, plug in the desired pH (11) and the given concentrations of CH₃NH₂ (0.500 M):

11 = 10.36 + log([CH₃NH₃Cl]/[0.500]).

Solving for [CH₃NH₃Cl], you get [CH₃NH₃Cl] = 0.405 M.

Finally, multiply this concentration by the volume of the solution in liters (0.200 L) to find the moles of CH₃NH₃Cl needed: 0.405 M × 0.200 L = 0.405 moles.

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261.52 grams of zinc are needed to fully react with 8.0 moles of silver nitrate.

To answer this question, we first need to determine the balanced chemical equation for the reaction given. The equation shows that one mole of zinc reacts with two moles of silver nitrate to produce two moles of silver and one mole of zinc nitrate. This means that the stoichiometric ratio between zinc and silver nitrate is 1:2.

Next, we can use the given amount of silver nitrate (8.0 moles) to determine how much zinc is needed to react completely with it. Since the ratio between zinc and silver nitrate is 1:2, we know that we need half as many moles of zinc as there are moles of silver nitrate.

Therefore, we can calculate the number of moles of zinc needed as follows:

Number of moles of zinc = (1/2) x Number of moles of silver nitrate
Number of moles of zinc = (1/2) x 8.0 mol
Number of moles of zinc = 4.0 mol

Finally, we can use the molar mass of zinc to convert the number of moles into grams:

Mass of zinc = Number of moles of zinc x Molar mass of zinc
Mass of zinc = 4.0 mol x 65.38 g/mol
Mass of zinc = 261.52 g

Therefore, 261.52 grams of zinc are needed to fully react with 8.0 moles of silver nitrate.

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When glass is placed in a furnace, its temperature rises in tandem with the temperature of the furnace. This is due to the fact that glass is a good conductor of heat and will absorb heat from its surroundings. The temperature of the glass, however, will not continue to rise eternally.

When the glass's temperature hits its softening point, it begins to deform and lose its shape. The glass will become less dense and its heat conductivity will decrease at this stage. As a result, the glass will absorb less furnace heat and its temperature will begin to stabilize.

Furthermore, after being heated in the furnace, modern glass manufacturing procedures frequently use a controlled cooling process to progressively reduce the temperature of the glass. This reduces heat shock and ensures that the glass is adequately annealed to avoid cracks or fractures.

In conclusion, while the temperature of the glass will initially rise in a furnace, it will eventually settle, and the glass will not absorb heat indefinitely due to its thermal qualities and manufacturing process.

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The pH of a solution with [H+] = 4.71x[tex]10^-^1^0[/tex] is approximately 9.327, as pH is a measure of the acidity or basicity of a solution as it is defined as the negative logarithm of the concentration of hydrogen ions in moles per liter (pH = -log[H+]).

The lower the pH, the more acidic the solution, while a higher pH indicates a more basic solution. In the given problem, the concentration of hydrogen ions ([H+]) is 4.71x [tex]10^-^1^0[/tex]

To calculate the pH,

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in moles per liter.

Substituting [H+] = 4.71x[tex]10^-^1^0[/tex] into the formula,

pH = -log(4.71x[tex]10^-^1^0[/tex]) = -(-9.327) = 9.327

Therefore, the pH of a solution with [H+] = 4.71x[tex]10^-^1^0[/tex] is approximately 9.327.

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The mass of KI that must be dissolved in 500 g of water to produce a 0.05 M solution of KI is approximately 8.3 grams. The answer is C) 8.3 grams.

To calculate the mass of KI required to make a 0.05 M solution, we need to use the formula for freezing point depression:

ΔT = K_f × m

where ΔT is the freezing point depression, K_f is the freezing point depression constant of water (1.86 °C/m), and m is the molality of the solution (moles of solute per kilogram of solvent).

Since we want to make a 0.05 M solution of KI, we need to find the molality of the solution. 0.05 moles of KI per liter of solution corresponds to 0.05 moles of KI per 1000 g of water, since the density of water is 1 g/mL. Therefore:

m = 0.05 moles KI / 0.5 kg water = 0.1 mol/kg

Now we can use the freezing point depression formula to find ΔT:

ΔT = K_f × m = 1.86 °C/m × 0.1 mol/kg = 0.186 °C

This means that the freezing point of the solution will be lowered by 0.186 °C compared to pure water.

To calculate the mass of KI required, we can use the formula:

moles of solute = mass of solute / molar mass of solute

Since we want 0.05 moles of KI, we can rearrange this formula to solve for the mass of KI:

mass of KI = moles of KI × molar mass of KI

The molar mass of KI is 166 g/mol. Substituting the given values, we get:

mass of KI = 0.05 mol × 166 g/mol = 8.3 g

Therefore, the mass of KI that must be dissolved in 500 g of water to produce a 0.05 M solution of KI is approximately 8.3 grams. The answer is C) 8.3 grams.

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152 grams of [tex]C2H6[/tex]would release 152 kJ of energy when it condenses at its boiling point.

Assuming you meant "[tex]C2H6[/tex]" instead of "[tex]CH Ch[/tex]", the heat of vaporization of [tex]C2H6[/tex]is 30.1 kJ/mol. The molar mass of [tex]C2H6[/tex] is 30.07 g/mol.

To calculate the heat of vaporization for 152 g of [tex]C2H6[/tex], we need to first calculate the number of moles of [tex]C2H6[/tex]:

152 g / 30.07 g/mol = 5.05 mol

Then, we can calculate the energy released using the heat of vaporization:

5.05 mol x 30.1 kJ/mol = 152 kJ

Therefore, 152 grams of [tex]C2H6[/tex]would release 152 kJ of energy when it condenses at its boiling point.

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A sample of helium gas that occupies 2.65 L at 1.20 atm would exert a pressure of 3.18 atm in a 1.50-L container at the same temperature, according to Boyle's Law.

To know the pressure exerted by a sample of helium gas that occupies 2.65 L at 1.20 atm when it's compressed to 1.50 L at the same temperature, using Boyle's Law (V₁P₁ = V₂P₂).

Here's the step-by-step explanation:
1. Identify the initial volume (V₁), initial pressure (P₁), and final volume (V₂):
  V₁ = 2.65 L
  P₁ = 1.20 atm
  V₂ = 1.50 L

2. Apply Boyle's Law to find the final pressure (P2):
  V₁P₁ = V₂P₂

3. Plug in the values and solve for P₂:
  (2.65 L)(1.20 atm) = (1.50 L)P₂

4. Calculate P₂:
  P₂= (2.65 L × 1.20 atm) / 1.50 L
  P₂= 3.18 atm
A sample of helium gas that occupies 2.65 L at 1.20 atm would exert a pressure of 3.18 atm in a 1.50-L container at the same temperature, according to Boyle's Law.

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The diagram illustrates the relationship between energy and temperature in a sample of water.

It shows that as energy is added, the temperature of the water increases until it reaches a point where the water changes state, demonstrating the importance of understanding the thermal properties of water in various scientific fields.

The diagram that shows the temperature of a sample of water as heat is added is an illustration of the thermal properties of water. As energy is added to the system, the temperature of the water increases until it reaches a point where it begins to change state.
The process of adding energy to the water is called heating, and the energy that is added is called heat. The amount of heat required to raise the temperature of water depends on its mass, specific heat capacity, and the temperature difference between the initial and final temperatures.
In the diagram, the temperature of the water increases gradually as heat is added until it reaches a point where the water begins to boil. At this point, the temperature of the water remains constant even as more heat is added, and the energy is used to break the bonds between the water molecules, resulting in the conversion of liquid water to steam.

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Answer:

[tex]\huge\boxed{\sf T_2=100.3 \ K}[/tex]

Explanation:

Volume 1 = [tex]V_1[/tex] = 3.00 L

Volume 2 = [tex]V_2[/tex] = 1.00 L

Temperature 1 = [tex]T_1[/tex] = 28 °C + 273 = 301 K

Temperature 2 = [tex]T_2[/tex] = ?

[tex]\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex] (Charles Law)

Put the given data in the above formula.

[tex]\displaystyle \frac{3.00}{301} = \frac{1.00}{T_2} \\\\Cross \ Multiply\\\\3 \times T_2=301 \times 1\\\\3T_2= 301\\\\Divide \ both \ sides \ by \ 3\\\\T_2=301/3\\\\T_2=100.3 \ K\\\\\rule[225]{225}{2}[/tex]

The complexity differences between Spectra C and D suggest that the regioselectivity of bromination of aniline versus acetanilide is different. Specifically, Spectra C shows the proton NMR spectrum of a mixture of aniline and p-bromoaniline, while Spectra D shows the proton NMR spectrum of a mixture of acetanilide and p-bromoacetanilide.

The complexity differences between Spectra C and D suggest that the regioselectivity of bromination of aniline versus acetanilide is different. Specifically, Spectra C shows the proton NMR spectrum of a mixture of aniline and p-bromoaniline, while Spectra D shows the proton NMR spectrum of a mixture of acetanilide and p-bromoacetanilide.

This indicates that the bromination of aniline is less regioselective than the bromination of acetanilide, meaning that multiple products are formed in significant amounts. In contrast, the bromination of acetanilide is more regioselective, resulting in a higher proportion of the desired product (p-bromoacetanilide) and fewer side products. The diffdifferenceerence in regioselectivity is likely due to the fact that the amino group in aniline is more strongly activating towards electrophilic aromatic substitution reactions than the amide group in acetanilide.
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The concentration of [tex]Ca(OH)_2[/tex] is 0.0217 M.

The balanced chemical equation for the reaction between the  [tex]Ca(OH)_2[/tex] and the HCl is:

[tex]Ca(OH)_2 + 2HCl[/tex] → [tex]CaCl_2 + 2H_2O[/tex]

From this equation, we can see that 1 mole of [tex]Ca(OH)_2[/tex] reacts with 2 moles of HCl.

The number of moles of HCl used can be calculated as:

moles HCl = Molarity * Volume in liters[tex]= 0.029 M\ *\ 0.0373 L = 0.0010837\ mol[/tex]

Since the stoichiometry of the reaction is 1:2 between [tex]Ca(OH)_2[/tex] and HCl, the number of moles of [tex]Ca(OH)_2[/tex] in the 25.0 mL sample can be calculated as:[tex]moles\ Ca(OH)2 = 0.0010837\ mol / 2 = 0.00054185\ mol[/tex]

The concentration of [tex]Ca(OH)_2[/tex] can then be calculated as:

[tex]Molarity = moles[/tex] ÷ [tex]Volume\ in\ liters\ = 0.00054185\ mol[/tex] ÷ 0.025 L = 0.0217M

Therefore, the concentration of [tex]Ca(OH)_2[/tex] is 0.0217 M.

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Translating the given balanced chemical equation into words :A.)Phosphorus pentachloride and water yield phosphoric acid and hydrochloric acid.

Phosphorus pentachloride and water react to yield phosphoric acid and hydrochloric acid. Balanced chemical equation shows that for every one mole of PCl₅ and four moles of H₂O that react, one mole of H₃PO₄ and five moles of HCl are produced.

Phosphorus pentachloride (PCl₅) is a chemical compound composed of one phosphorus atom and five chlorine atoms. It is yellowish-white crystalline solid that is highly reactive and can decompose violently when exposed to water or moist air.

PCl₅ is primarily used as a chlorinating agent in organic chemistry, where it is used to convert alcohols, carboxylic acids, and other functional groups into the corresponding chlorides.

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The molarity of the magnesium chloride solution is 2.38 M. This means that there are 2.38 moles of magnesium chloride per liter of solution.

The molarity is defined as the number of moles of the solute per liter of the solution. In this problem, we are given the moles of magnesium chloride (7.39 moles) and the final volume of the solution (3.10 L). We can use the formula Molarity = moles of solute / volume of solution to calculate the molarity of the magnesium chloride solution.

First, we divide the moles of magnesium chloride by the volume of the solution in liters:

[tex]Molarity = 7.39 moles / 3.10 L[/tex]

[tex]Molarity = 2.38 M[/tex]

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Answer: To determine the pressure change of a gas when it is heated at constant volume, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

Since the volume of the gas is constant, we can simplify the equation to:

P/T = nR/V

The quantity nR/V is a constant, which means that P/T is also a constant at constant volume. Therefore, we can use the following equation to calculate the pressure at a new temperature:

P2/T2 = P1/T1

where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

We can convert the temperatures to Kelvin by adding 273.15:

T1 = 30.0 °C + 273.15 = 303.15 K

T2 = 40.0 °C + 273.15 = 313.15 K

We can plug in the given values and solve for P2:

P2/313.15 K = 2.50 atm/303.15 K

P2 = (2.50 atm)(313.15 K)/(303.15 K)

P2 = 2.58 atm

Therefore, the pressure of the gas increases from 2.50 atm to 2.58 atm when it is heated from 30.0 °C to 40.0 °C at constant volume.

Explanation:

1) The balanced reaction is [tex]2CO + O_{2} ---- > 2CO_{2}[/tex]

2) 313.6 mL is required so that 630 mL of carbon monoxide gas is completely converted to carbon dioxide.

3) 3.136 L are produced if the catalytic converter processes 6.25 L of carbon monoxide.

4) The volume of oxygen is 1.23L.

If 1  mole of CO occupies 22400 mL

x moles of CO occupies 630 mL

x = 0.028 moles

If 2 moles of CO reacts with 1 mole of oxygen

0.028 moles of CO reacts with 0.028 moles * 1/2

= 0.014 moles

Volume of oxygen required = 0.014 moles * 22400 mL

= 313.6 mL

If 1 mole of CO occupies 22.4 L

x moles of CO occupies 6.25 L

x = 0.28 moles

If 2 moles of CO produces 1 mole of carbon dioxide

0.28 moles of CO produces  0.28 * 1/2

= 0.14 moles

Volume of the carbon dioxide = 0.14 moles * 22.4 L

= 3.136 L

If 1 mole of carbon dioxide occupies 22.4 L

x moles of carbon dioxide occupies 2.5 L

x = 2.5 L * 1/22.4 L

x = 0.11 moles

If 1 mole of oxygen produces 2 moles of carbon dioxide

x moles of oxygen produces 0.11 moles of carbon dioxide

x = 0.055 moles

Volume of oxygen =  0.055 moles * 22.4 L

= 1.23L

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Aspartic acid (Asp) and glutamic acid (Glu) are the two amino acids that are regarded as acidic amino acids. These amino acids are acidic due to the Carboxylic acid group (-COOH) in their side chains, which has the ability to donate a hydrogen ion (H+) to the environment.

The three amino acids lysine (Lys), arginine (Arg), and histidine (His), on the other hand, are regarded as basic amino acids. These amino acids are classified as basic because they include basic amine groups (-NH2) in their side chains that can accept a hydrogen ion (H+) from the environment. It is significant to remember that amino acids can become more basic or acidic depending on the pH of the surroundings.

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Aspartic acid (Asp) and glutamic acid (Glu) are acidic amino acids. Lysine (Lys), arginine (Arg), and histidine (His) are basic amino acids under physiological conditions.

The two amino acids commonly referred to as acidic amino acids are aspartic acid (Asp) and glutamic acid (Glu). They are called acidic amino acids because their side chains can ionize and release a proton, resulting in a negatively charged carboxylate group. The ionization occurs under physiological conditions when the pH is higher than the pKa value of the side chain.

The three amino acids commonly referred to as basic amino acids are lysine (Lys), arginine (Arg), and histidine (His). They are called basic amino acids because their side chains can accept a proton, resulting in a positively charged amino group. The ionization occurs under physiological conditions when the pH is lower than the pKa value of the amino group.

It's important to note that the ionization and charges of amino acids depend on the specific pH and pKa values of their side chains. The mentioned ionization states are commonly observed under physiological conditions, where the pH is around 7. However, at different pH values, the ionization states may vary.

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The heat energy associated with the reaction of 35g of aluminum with excess hydrochloric acid is -5,380 kJ. This is calculated by multiplying the number of moles of aluminum (0.2 mol) by the enthalpy change of the reaction (-152 kJ/mol) to give -30.4 kJ.

This is then multiplied by the mass of aluminum (35g) to give -5,380 kJ.

In this reaction, heat energy is released as a result of the formation of bonds between the aluminum and the hydrochloric acid.

This means that the enthalpy change is negative, indicating that the reaction is exothermic. The reaction can be represented by the equation 2Al + 6HCl → 2AlCl3 + 3H2, with an enthalpy change of -152 kJ/mol.

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The total number of moles of solute (HCl) in 0.50 L of 3.0 M HCl is 1.5 moles.

To determine the total number of moles of solute in a solution, we can use the formula:

moles of solute = concentration of solution x volume of solution

In this case, we are given that the volume of the solution is 0.50 L and the concentration of the solution is 3.0 M HCl.

Using the formula above, we can calculate the number of moles of HCl in the solution:

moles of HCl = 3.0 M x 0.50 L

moles of HCl = 1.5 moles

This result can be explained by the fact that the concentration of a solution is defined as the amount of solute (in moles) per unit volume of the solution (in liters).

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The types of rocks being described in this process are sedimentary and metamorphic. (A)

Dead plant material, which is organic, is compressed into coal, which is a type of sedimentary rock. The coal is then buried within the earth and subjected to heat and pressure from the overlying material, which turns it into anthracite coal, a type of metamorphic rock.

Sedimentary rocks are formed from the accumulation and cementation of sediment, such as dead plant material. Metamorphic rocks are formed from the transformation of existing rocks under intense heat and pressure. In this case, the coal is transformed into anthracite coal through the process of metamorphism.

Igneous rocks are not mentioned in this process, as they are formed from the cooling and solidification of magma or lava, and do not play a role in the formation of coal. Therefore, the correct answer is A: sedimentary and metamorphic.

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The primary alcohol required for the synthesis of heptanoic acid is heptanol, which has the chemical formula C₇H₁₆O. The aldehyde required for this synthesis is heptanal, which has the chemical formula C₇H₁₄O.

Heptanoic acid is a carboxylic acid with seven carbon atoms. It can be synthesized from primary alcohol and an aldehyde via oxidation.

To synthesize heptanoic acid, heptanol, and heptanal are reacted in the presence of an oxidizing agent, such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). The oxidation of heptanol produces heptanal, which is further oxidized to heptanoic acid. The chemical equation for the synthesis of heptanoic acid is as follows:

C₇H₁₆O + O → C₇H₁₄O + H₂O

C₇H₁₄O + O → C₇H₁₂O₂ + H₂O

The resulting product, heptanoic acid, is a colorless liquid with a pungent odor and is commonly used as a flavoring agent and in the production of esters for fragrances and plastics.

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An atom is an indivisible particle of the matter and it is the fundamental building blocks of the matter. Some examples of atoms are sodium atom, fluorine atom, etc. It is the smallest unit of matter.

The elements are defined as the substance which is made up of same kind of atoms and that cannot be broken down into simpler form by any physical or chemical methods. Carbon is an element.

Carbon - C = 1 C atom

Oxygen molecule - O₂ = 2 'O' atoms

Methane - CH₄ = 1 'C' and 4 'H' atoms

Iron - Fe = 1 'Fe' atom

Glucose - C₆H₁₂O₆ = 6 'C', 12 'H' and 6 'O' atoms

Hydrogen chloride - HCl

Sulfur dioxide - SO₂ = 1 'S' and 2 'O' atoms

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The molality of a solution formed by mixing 104 g. Of silver nitrate(AgNO₃) with 1. 75 kg of water is 0.350 mol/kg.

The molality of a solution formed by mixing 104 g of silver nitrate (AgNO₃) with 1.75 kg of water can be calculated as follows:

1. First, convert the mass of silver nitrate to moles:

104 g AgNO₃ * (1 mol AgNO₃/169.87 g AgNO₃) = 0.6122 mol AgNO₃

2. Then, calculate the mass of water in kilograms:

1.75 kg water = 1750 g water

3. Finally, divide the moles of AgNO₃ by the mass of water in kilograms to get the molality:

molality = 0.6122 mol AgNO₃ / 1.75 kg water = 0.350 mol/kg

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The number average molar mass for the sample of poly(ethylene oxide) is 13000 g/mol.

The number of moles of acetic anhydride used in the reaction can be calculated as follows:

Moles of acetic anhydride = (mass of acetic anhydride) / (molar mass of acetic anhydride)

Molar mass of acetic anhydride = (2 x molar mass of carbon) + (3 x molar mass of oxygen) = (2 x 12.011) + (3 x 15.999) = 102.09 g/mol

Moles of acetic anhydride = (2.5 × 10⁻³) / 102.09 = 2.45 × 10⁻⁵ mol

Since the hydroxyl end groups of each molecule of poly(ethylene oxide) react with one molecule of acetic anhydride, the number of moles of poly(ethylene oxide) can be calculated as follows:

Moles of poly(ethylene oxide) = moles of acetic anhydride / 2 = 1.23 x 10⁻⁵ mol

The mass of the sample of poly(ethylene oxide) is given as 2.00 g, so the number average molar mass can be calculated as follows:

Number average molar mass = (mass of sample) / (moles of sample)

Number average molar mass = 2.00 / 1.23 x 10⁻⁵ = 1.626 x 10⁸ g/mol

However, each molecule of poly(ethylene oxide) has two hydroxyl end groups, so the actual number average molar mass is half of this value:

Number average molar mass = 1.626 x 10⁸ / 2 = 13000 g/mol.

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