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Vika [28.1K]
4 years ago
15

How does increasing temperature increase the number of reactions

Chemistry
2 answers:
r-ruslan [8.4K]4 years ago
7 0

Answer:

I think the answer is option A.

Explanation:

Because when the temperature increases, the kinetic enegry increases which creates the effective collisions which lets the chemical to react.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

Nitella [24]4 years ago
4 0

Answer:

A

Explanation:

the increase in the temperature let's the particles to collide more so it increases the rate of reaction.

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3 years ago
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Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, g
Nostrana [21]

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_4 will be,

C(s)+2H_2(g)\rightarrow CH_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.4kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.7kJ/mole

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1) CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)     \Delta H_1=890kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.7kJ)=-571.4kJ/mole

The expression for enthalpy of formation of CH_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)

\Delta H=-74.8kJ/mole

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole

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Answer:

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What is the empirical formula for propene (C3H6)?<br> C2H4<br> O C4H8<br> O C3H6<br> O CH2
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Answer: C2H4 C4H8 C3H6.

Explanation:

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In a science lab, Cash mixes two clear liquids together in a beaker. Bubbles are produced, and a white solid forms and settles t
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Answer:

A chemical change occurred, so even if Cash heats it, the white solid will remain.

Explanation:

  • A physical change occurred, so if Cash heats it, the white solid will dissolve to form the original mixture of clear liquids.
  • A physical change occurred, so even if Cash heats it, the white solid will remain.
  • A chemical change occurred, so if Cash heats it, the white solid will dissolve to form the original mixture of clear liquids.
  • A chemical change occurred, so even if Cash heats it, the white solid will remain.

<em>The correct answer would be that a chemical change occurred, so even if Cash heats it, the white solid will remain.</em>

Chemical changes occur when atoms in a substance or mixture of substances are rearranged, leading to the formation of new substances.

<u>In this case, the mixture of the two clear liquids led to the formation of gas in form of bubbles and a white solid forms that settle at the bottom. The formation of these 2 substances from the mixture of the two clear liquid is an indication of chemical reaction/changes. Hence, even if cash heats the resulting mixture, the white solids will likely remain.</u>

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