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kherson [118]
3 years ago
11

The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear

ance of Br2 is ________ M s-1. The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appearance of Br2 is ________ M s-1.
a. 0.0700
b. 1.28
c. 0.0196
d. 0.280
e. 0.374
Chemistry
1 answer:
djverab [1.8K]3 years ago
6 0

Answer: The rate of appearance of Br_2 is 0.0700Ms^{-1}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr = -\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2 = \frac{1d[H_2]}{dt}

Rate in terms of appearance of Br_2 = \frac{1d[Br_2]}{dt}

-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}

Given :

-\frac{1d[HBr]}{dt}=0.140Ms^{-1}

The rate of appearance of Br_2;

\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}

Thus rate of appearance of Br_2 is 0.0700Ms^{-1}

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The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

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6 0
3 years ago
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The total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.

HOW TO CALCULATE NUMBER OF ATOMS:

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  • no. of atoms = no. of moles × 6.02 × 10²³ mol-¹

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no. of atoms = 1.204 × 10²³atoms.

  • Therefore, the total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.

Learn more: brainly.com/question/15488332?referrer=searchResults

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