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lutik1710 [3]
3 years ago
10

Consider the reaction, 2 D(g) + 3 E(g) + F(g) => 2 G(g) + H(g) When D is decreasing at 0.15 mol/Ls, how quickly is E decreasi

ng? Give your answer to 3 decimal places.
Chemistry
1 answer:
muminat3 years ago
5 0

Answer:

0.023 mol/(L*s).

Explanation:

Equation of the reaction:

2 D(g) + 3 E(g) + F(g) => 2 G(g) + H(g)

Using stoichiometry, 2 moles of D reacts with 3 moles of E and 1 mole of F to give the above products.

1 mole of D will react with 3/2 moles of E.

Delta[D]/delta t = 0.15 mol/L.s

Delta[E]/delta t = (3/2) * 0.15 =

= 0.023 mol/(L*s).

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A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
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The correct answer is protein
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A student is given 2.19 g of an unknown acid, which can be either oxalic acid, H2C2O4, or citric acid, H3C6H5O7. To determine wh
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Answer:

The unknown acid is citric acid.

There is 0.0342 moles of NaOH consumed.

Explanation:

Step 1: Data given

Mass of the unknown acid = 2.19 gram

Titrating with 0.560 M of NaOH

The equivalence point is reached when 61.0 mL are added

Molar mass of oxalic acid = 90.03 g/mol

Molar mass of citric acid = 192.12 g/mol

<u>Step 2:</u> The balanced equations for both acids

The reaction between oxalic acid and NaOH is:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

The reaction between citric acid and NaOH is:

H3C6H5O7 +3NaOH → Na3C6H5O7 + 3 H2O

<u>Step 3:</u> Calculate the number of moles of the acid

Moles = mass / Molar mass

In case of oxalic acid: 2.19 grams / 90.03 g/mol = 0.0243 moles

In case of citric acid: 2.19 grams /192.12 g/mol = 0.0114 moles

Step 4: Calculate number of moles of NaOH

The mole of NaOH required for titration;

number of moles  = Molar mass * volume = (0.560 M * 0.061 L) = 0.03416 mol

<u>Step 5:</u> Calculate which acid

For each mole of oxalic 2 moles of NaOH is required, for 0.0243 mol citric acid 0.0243 *2= 0.0486 mol NaOH is required. This is more than the number of moles consumed.

For each mole of citric acid 3 moles of NaOH is required, for 0.0114 mol citric acid 0.0114 * 3= 0.0342 mol NaOH is required. This is the number of moles NaOH used for the titration.

The unknown acid is citric acid. There is 0.0342 moles of NaOH consumed.

8 0
3 years ago
A chemist is studying the rate of the Haber synthesis: N2 + 3H2 2NH3
Slav-nsk [51]

Answer:

1.17 M

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇒ 2 NH₃

Step 2: Calculate the rate of disappearance of H₂

We will use the following expression.

rH₂ = - Δ[H₂]/t = - (0.25 M - 0.50 M)/44 s = 0.0057 M/s

Step 3: Calculate the rate of disappearance of N₂

The molar ratio of N₂ to H₂ is 1:3.

0.0057 mol H₂/L.s × 1 mol N₂/3 mol H₂ = 0.0019 mol N₂/L.s

Step 4: Calculate the final concentration of N₂

We will use the following expression.

[N₂] = [N₂]₀ - rN₂ × t

[N₂] = 1.25 mol/L - 0.0019 mol/L.s × 44 s

[N₂] = 1.17 M

6 0
3 years ago
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