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melomori [17]
3 years ago
11

I need help I can't chose between A or D

Chemistry
2 answers:
djyliett [7]3 years ago
7 0
That’s a hard one, but honestly i would go with a.
maria [59]3 years ago
7 0

Answer:

Increasing the temperature  

Step-by-step explanation:

Increasing the temperature gives the reacting molecules more kinetic energy.

More of the molecular collisions will have enough energy to get over the activation energy barrier and the reaction rate will increase.

B) is wrong. If you halve the amounts in the recipe for the experiment, the concentrations stay the same. There is no change in the rate of reaction.

C) is wrong. If you decrease the surface area of a solid, there will be fewer of its molecules at the surface to react. The rate of reaction will decrease.

D) is wrong. If you decrease the concentrations of the reactants, their molecules will be further apart. There will be fewer collisions per second, so the reaction rate will decrease.

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How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
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<h2><u>Answer:</u></h2>

n = 0.0989 moles

<h2><u>Explanation:</u></h2>

n = PV / RT

P = 2.09atm

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R = 0.08206

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Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

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iren2701 [21]

Answer : The correct option is, (B) CO_2

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According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

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Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

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The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

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