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Llana
17 days ago
14

If kc = 7.04 × 10-2 for the reaction: 2 hbr(g) ⇌h2(g) + br2(g), what is the value of kc for the reaction: 1/2 h2(g) + 1/2 br2 ⇌h

br(g)
Chemistry
1 answer:
Anarel [2.6K]17 days ago
7 0
For the reaction 2HBr(g) ⇄ H2(g) + Br2(g), the equilibrium constant Kc is expressed as [H2] [Br2] / [HBr]^2. Hence, Kc equals 7.04X10^-2 = [H2][Br2] / [HBr]^2. In the subsequent reaction, where 1/2 H2(g) + 1/2 Br2(g) ⇄ HBr, its Kc is given as [HBr] / ([H2]^(1/2)*[Br2]^(1/2)). To manipulate the first equation, we reverse it: 1/7.04X10^-2 equals [HBr]^2 / ([H2][Br2]). By taking the square root, we find that √(1/7.04X10^-2) equals [HBr] / ([H2]^(1/2)*[Br2]^(1/2)). Thus, Kc for the second reaction equals √(1/7.04X10^-2) = 3.769.
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Is this a multiple-choice question? Is there an educational issue that needs to be resolved?

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7 0
14 days ago
Compounds A and B are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 g of sulf
lorasvet [2515]

Answer:

1.5

Explanation:

It is given that:

Compound A and B originate from Sulfur + Oxygen.

Compound A:

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

By comparing the ratios:

Compound A:

S: O = 6.00: 5.99

S/0 = 6.0g S / 5.99g O

Compound B:

S: O = 8.60: 12.88

S / O = 8.60g S / 12.88g O

The mass ratio of A and that of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

= 77.28 / 51.514

= 1.50017

= 1.5

4 0
10 days ago
How many molecules are in 13.5g of sulfur dioxide, so2?
alisha [2704]
Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.
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1 month ago
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In a chemical reaction that takes place at a fixed pressure and volume, the enthalpy change (ΔH) is –585 kJ/mol. Will this react
VMariaS [2690]

According to the sign convention, a negative ΔH indicates that the reaction is exothermic, resulting in a loss of heat and a reduction in temperature.

5 0
1 month ago
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