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3 years ago

11

Can someone help with this please? greatly appreciated!

1 answer:

maw [93]3 years ago

7 0

Answer:

Explanation:

a) When the iron rod is heated, it expands (gets larger). That's the key to this whole procedure. It doesn't expand much -- depending on what temperature the iron rod is heated up to.

When the rod looks like it has expanded as much as it can, the rod is tightened as much as it can be tightened before damage is done. When the rod cools, it contracts (gets smaller). That brings the walls closer together or it stops the walls from bulging.

Very interesting question. I've never seen this. Thanks for asking.

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3 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

2 years ago