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2 years ago

10. i really really need help plz don't waste time, thx :)

Chemistry

2 answers:

egoroff_w [7]2 years ago

6 0

Lollll lollll lolll lolll

Serjik [45]2 years ago

3 0

Dude use a calculator app

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The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th

Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $T_2$

$K_1$ = rate of reaction at $T_1$

$Ea$ = activation energy of the reaction

R = gas constant = 8.314 J/K mol

$E_a=300 kJ/mol=300,000 J/mol$

$T_2=625^oC=898.15 K,T_1=525^oC=798.15 K$

$\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]$

$\log (\frac{K_2}{K_1})=2.185666$

$K_2=153.344\times K_1$

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0

3 years ago

Determine the equilibrium constant, K, for the equilibrium below given the

meriva

Answer:

or reactions that are not at equilibrium, we can write a similar expression called the reaction quotient QQQ, which is equal to K_\text cK

K, start subscript, start text, c, end text, end subscript at equilibrium.

Explanation:

5 0

2 years ago

If the H3O is 4.950 x 10-12 what is the ph?

romanna [79]

Answer:

pH = 11.3

Explanation:

From the question given above, the following data were obtained:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

The pH of a solution is defined by the following equation:

pH = –Log [H₃O⁺]

Thus, with the above formula, we can obtain the pH of the solution as follow:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

pH = –Log [H₃O⁺]

pH = –Log 4.950×10¯¹²

pH = 11.3

8 0

3 years ago

Please help! Chemistry 113 smartworks question. Thanks !!!

Dominik [7]

Kc' =Kc^1/3
=3√0.0061
=0.182716013

3 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago