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Harman [31]
1 year ago
15

Is each of these statements true? If not, explain why.(m) For most reactions, ΔH°rnx is lowered by a catalyst.

Chemistry
1 answer:
barxatty [35]1 year ago
3 0

A  catalyst reduces H°rnx in most reactions. The answer is false

<h3>Do catalysts reduce delta H?</h3>

By reducing the activation energy required for the reaction to occur, a catalyst just modifies the route used to go from reactants to products. However, because it doesn't alter the state of the products or reactants, delta H is unaffected.

A catalyst reduces a reaction's activation energy, enabling a chemical reaction to occur. The number of reactant particles with energy above the activation energy increases as the temperature of a reaction rises.

learn more about catalyst refer

brainly.com/question/12507566

#SPJ4

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An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the
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Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

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