a 9.84 ounces ingot of unknown metal is heated from 73.2 degrees fahrenheit - 191.2 degrees fahrenheit this requires 3.912 calor

To determine the specific heat capacity of the metal and assist in its identification, the heat absorbed by the calorimeter can be computed using: Energy = mass * specific heat capacity * temperature change Q = 250 * 1.035 * (11.08 - 10) Q = 279.45 cal/g. Next, we employ the same formula for the metal because the heat taken in by the calorimeter should equal the heat expelled by the metal. -279.45 = 50 * c * (11.08 - 45) [the minus sign indicates energy release] solving for c gives us 0.165. Therefore, the specific heat capacity of the metal amounts to 0.165 cal/g°C.

Response:

22.9 moles (Option 5)

Clarification:

A substance being 35.7% carbon by mass signifies that in every 100 g of the substance, there is 35.7 g of carbon.

Using a rule of three, we get:

100 g of substance ___ has __ 35.7 g of C

769 g of substance ___ has ___ ( 769.35.7) / 100 = 274.5 g of C

1 mol of C = 12 g/m

Mass / Molar mass = Moles

274.5 g / 12 g/m = 22.9 moles

Response:

a. To purify a gypsum sample, you will need the following equipment: Bunsen burner, beaker, filter funnel, stirring rod, and filter paper.

b. Gypsum, a sulfate mineral consisting of calcium sulfate dihydrate, can be purified by following these steps:

1. Add water to the gypsum in a beaker.

2. Stir the mixture thoroughly with the stirring rod.

3. Use the filter paper and filter funnel to remove excess solids from the mixture.

4. Heat the filtered mixture on the Bunsen burner to evaporate the remaining water.

5. After cooling, filter again through the filter paper to obtain pure gypsum.

1) The ionic compound present in solution b is K₂CrO₄ (potassium chromate). This compound contains two potassiums (oxidation state +1), a single chromium (oxidation state +6), and four oxygen atoms. The oxidation state of oxygen is -2, resulting in a neutral compound: 2 · (+1) + 6 + x · (-2) = 0. Hence, x = 4, denoting the count of oxygen atoms. 2) The ionic compound in solution a is AgNO₃ (silver nitrate). ω(N) = 8.246% ÷ 100%. Thus, ω(N) = 0.08246, indicating the mass percentage of nitrogen. M(MNO₃) = M(N) ÷ ω(N). It follows that M(MNO₃) = 14 g/mol ÷ 0.08246, leading to M(MNO₃) = 169.8 g/mol; the molar mass of the metal nitrate. M(M) = M(MNO₃) - M(N) - 3 · M(O). Consequently, M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol, resulting in M(M) = 107.8 g/mol which is the atomic mass of silver (Ag). 3) The balanced chemical equation is: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq). In ionic form: 2Ag⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq). The net ionic equation is: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). Thus, the red precipitate is identified as silver chromate (Ag₂CrO₄). 4) The mass of solid silver chromate created is m(Ag₂CrO₄) = 331.8 g. The amount is determined by n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄). Therefore, n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol yields n(Ag₂CrO₄) = 1 mol. From the balanced equation, n(Ag₂CrO₄): n(AgNO₃) = 1: 2, it follows n(AgNO₃) = 2 · 1 mol, which means n(AgNO₃) = 2 mol. Then, the mass of silver nitrate is computed as m(AgNO₃) = n(AgNO₃) · M(AgNO₃). Hence, m(AgNO₃) = 2 mol · 169.8 g/mol gives m(AgNO₃) = 339.6 g; thus, m(AgNO₄) equals m(K₂CrO₄). Therefore, m(K₂CrO₄) = 339.6 g; amount of potassium chromate is n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄). Thus, n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol thus arrives at n(K₂CrO₄) = 1.75 mol. 5) The dissociation of silver nitrate in water is expressed as: AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq). Volume of solution a = 500 mL ÷ 1000 mL/L results in V(solution a) = 0.5 L. Concentration equation c(AgNO₃) = n(AgNO₃) ÷ V(solution a), thus c(AgNO₃) = 2 mol ÷ 0.5 L, yielding c(AgNO₃) = 4 mol/L = 4 M. As a result: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻). Thus, c(Ag⁺) = 4 M; the concentration of silver ions in the initial solution a. 6) The dissociation of potassium chromate in water is represented as: K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq). Volume of solution b = 500 mL ÷ 1000 mL/L results in V(solution b) = 0.5 L. Following, c(K₂CrO₄) is calculated as n(K₂CrO₄) ÷ V(solution b). So c(AgNO₃) = 1.75 mol ÷ 0.5 L gives c(AgNO₃) = 3.5 mol/L = 3.5 M. Consequently: c(K⁺) = 7 M; the concentration of potassium ions in solution b. Therefore, c(CrO₄²⁻) = 3.5 M; the concentration of chromium ions in the same solution. 7) The total final volume is V(final solution) = V(solution a) + V(solution b). Thus, V(final solution) = 500.0 mL + 500.0 mL leads to V(final solution) = 1000 mL ÷ 1000 mL/L results in V(final solution) = 1 L. Then n(NO₃⁻) = 2 mol. Therefore, c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution) finds c(NO₃⁻) = 2 mol ÷ 1 L and results in c(NO₃⁻) = 2 M; the concentration of nitrate anions in the final solution. 8) In solution b, there are 3.5 mol of potassium cations while part of that combines with 2 moles of nitrate anions: K⁺(aq) + NO₃⁻(aq) → KNO₃(aq). From the reaction: n(K⁺): n(NO₃⁻) = 1: 1. Thus, Δn(K⁺) = 3.5 mol - 2 mol results in Δn(K⁺) = 1.5 mol, signifying the remaining potassium anions in the final solution. Thus, c(K⁺) = Δn(K⁺) ÷ V(final solution) yields c(K⁺) = 1.5 mol ÷ 1 L, leading to c(K⁺) = 1.5 M; the final concentration of potassium cations.

Answer: The percentage abundance for isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

  .....(1)

From the information provided:

Let the fractional abundance for isotope be 'x'

• For isotope:

Mass of isotope = 27.9769 amu

Percentage abundance of isotope = 92.22 %

Fractional abundance for isotope = 0.9222

• For isotope:

Mass of isotope = 28.9764 amu

Percentage abundance for isotope = 4.68%

Fractional abundance of isotope = 0.0468

• For isotope:

Mass of isotope = 29.9737 amu

Fractional abundance for isotope = x

• The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

To convert this fractional abundance into a percentage, multiply by 100:

This shows that the percentage abundance for isotope is 3.09 %.