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Aleksandr [31]
3 years ago
12

At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.

Chemistry
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

D) a four-fold increase

Explanation:

According to Gay-Lussac's law, which states that the pressure of a given amount of gas is directly proportional to the temperature at a constant volume, the pressure increases with an increase in temperature.

According to this question, at a fixed volume, a four-fold increase in the temperature of a gas will lead to a four-fold increase in the pressure as well.

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kari74 [83]

It is a base due to the color change!

Hope this helps!

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Are the particles in a gas farther apart or closer together than the particles in a liquid? Use your answer to explain each of t
Mademuasel [1]

farther apart just look up states of matter :)

6 0
2 years ago
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

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 C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

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

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

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

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
A potassium atom can achieve a stable octet by
frez [133]
I think the answer is D, gaining one electron.
4 0
4 years ago
3. Perform each calculation to the correct number of significant figures.
Gwar [14]

Answer:

0.14

Explanation:

6 0
3 years ago
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