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3 years ago

Consider the reaction and the rate law below.

Chemistry

1 answer:

ruslelena [56]3 years ago

3 0

The overall order of the reactants in this rate law : 2

<h3>Further explanation</h3>

Given

Reaction

2O3 → 302

The rate law :

R= k[03]²

Required

The overall order

Solution

The overall order : the sum of orders for each reactant

For A + B ---> C + D

r = k [A]ᵃ[B]ᵇ

The overall reaction order is m+n

So for the rate law of R= k[03]², the overall order = 2

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A reaction has a theoretical yield of 124.3 g SF6, but only 113.7 g SF6 are obtained in the lab, what is the percent yield of SF

ratelena [41]

Answer:

Percent yield = 91%

Explanation:

Given data:

Theoretical yield of SF₆ = 124.3 G

Actual yield of SF₆ = 113.7 g

Percent yield of SF₆ = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield)× 100

By putting values,

Percent yield = (113.7 g/ 124.3 g) × 100

Percent yield = 0.91 × 100

Percent yield = 91%

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3 years ago

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3 years ago

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xxMikexx [17]

Answer:

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Explanation:

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3 years ago

Calculate the pressure of the CO₂ (g) in the container at 425 K.

Elanso [62]

The pressure of the CO₂ = 0.995 atm

<h3>Further explanation</h3>

The complete question

<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>

<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>

<em />

Gay Lussac's Law

When the volume is not changed, the gas pressure is proportional to its absolute temperature

$\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

P₁=0.7 atm

T₁=299 K

T₂=425 K

$\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{0.7\times 425}{299}=0.995 `atm$

<em />

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