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weeeeeb [17]
3 years ago
8

At a given temperature, Kc=1.3x10-2 for reaction 1. What is the value of Kc for reaction 2? Reaction 1: begin mathsize 12px styl

e N subscript 2 open parentheses g close parentheses plus 3 H subscript 2 open parentheses g close parentheses rightwards harpoon over leftwards harpoon 2 N H subscript 3 open parentheses g close parentheses end style Reaction 2: begin mathsize 12px style 4 N H subscript 3 open parentheses g close parentheses rightwards harpoon over leftwards harpoon space 2 N subscript 2 open parentheses g close parentheses plus 6 H subscript 2 open parentheses g close parentheses end style

Chemistry
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

See explaination

Explanation:

For a reversible reaction, the equilibrium constant for a backward reaction is reciprocal.

If the coefficients in a balanced equation are multiplied by a factor, n, the equilibrium expression is raised to the nth power.

K' = (K)^n

In the second reaction the value kf n is 2. The first reaction is multiplied by 2.

See attachment for further solution

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The student's test average is : 0.92

Total score obtained = 369

Number of tests = 4

Since Each test is exactly 100 points ;

The total score obtainable is (100 × 4) = 400

Average = score obtained / total score obtainable

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5 0
3 years ago
What are the reactants in the following equation: hcl(aq) nahco₃(aq)→ co₂(g) h₂o(l) nacl(aq)
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3 years ago
32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac
Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

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The difference in temperature of the two glasses of water is 30°C. What is their difference in temperature on the Kelvin scale?
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Answer:

D

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