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Finger [1]
3 years ago
15

In the first 22.0 s of this reaction, the concentration of HBr dropped from 0.590 M to 0.465 M . Calculate the average rate of t

he reaction in this time interval.
Chemistry
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

The average rate is 2.84 X 10⁻³ Ms⁻¹

Explanation:

Average rate = -0.5*Δ[HBr]/Δt

given;

[HBr]₁ = 0.590 M

[HBr]₂ = 0.465 M

Δ[HBr] = [HBr]₂  - [HBr]₁ = 0.465 M - 0.590 M = -0.125 M

Δt Change in time = 22.0 s

Average rate = -0.5*Δ[HBr]/Δt

Average rate = - 0.5(-0.125)/22

Average rate = 0.00284 Ms⁻¹ = 2.84 X 10⁻³ Ms⁻¹

Therefore, the average rate is 2.84 X 10⁻³ Ms⁻¹

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Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how
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Given equation : 5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O

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Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

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Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}

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Summary : The above three steps can also be done in a singe setup as shown below.

2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

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