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3 years ago

A catalyst accelerates a reaction because:_____.

Chemistry

1 answer:

Margaret [11]3 years ago

3 0

Answer:

D. Lowering activation energy for the reaction

Explanation:

Catalysts increase the rate of a reaction without being used up. They do this by lowering the activation energy needed. With a catalyst, more collisions result in a reaction, so the rate of reaction increases.

Hope this helped :)

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Which is an aspect of the kinetic-molecular theory and can be used to explain the behavior of plasmas?

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Particle exchange energy though elastic collision , hope this helps .

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4 years ago

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

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4 years ago

Calculate the number of oxygen atoms in 3.65 mol Fe(ClO3)3

IRINA_888 [86]

Answer:

no thank you

Explanation:

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4 years ago

How many grams of calcium nitrate, Ca(NO3)2 (molecular weight 164), contains 24 grams of oxygen atoms?

Studentka2010 [4]

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3 years ago

A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of

Molodets [167]

Answer:

The volume when the conditions were altered is 0.5109 L or 510.9 mL

Explanation:

Using the general gas equation,

P1 V1 / T1 = P2 V2 / T2

where;

P1 = 756 mmHg

V1 = 475 ml = 0.475 L

T1 = 23.5°C = 23.5 + 273K = 275.5 K

P2 = 722 mm Hg

T2 = 10°C = 10 + 273 K = 283 K

V2 = ?

Rearranging to make V2 the subject of the formula, we obtain:

V2 = P1 V1 T2 / P2 T1

V2 = 756 * 0.475 * 283 / 722 * 275.5

V2 = 101, 625.3 / 198911

V2 = 0.5109 L or 510.9 mL

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4 years ago