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3 years ago

Can I get help with this, please!!

Chemistry

1 answer:

kondor19780726 [428]3 years ago

8 0

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

Explanation:

Pressure in the submarine = 108.9 kPa

Volume, V = 2.4 * 10^5 L

Pressure, P = 116k Pa

Temperature, T = 312 K

Ideal gas law: PV = nRT or n = PV / RT

So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K

= 1.073 *10^4 mol

when temperature is changed to 293K,

PV = nRT or P = nRT / V

=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L

=108.9 K Pa

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

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How many grams of H2 will be produced from 12 grams of Mg according to the following equation?

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Answer:

0.98 g of H₂

Explanation:

the balanced equation for the reaction is

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number of Mg moles reacted = 12 g/ 24.3 g/mol = 0.49 mol

according to molar ratio

when 1 mol of Mg reacts 1 mol of H₂ is formed

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Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to

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Answer:

a) kc = 0,25

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Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = $\frac{[C]^2}{[A][B]}$

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[A] = $\frac{1,60 moles}{4,0 L}$ = 0,4 M

[B] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

[C] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

kc = $\frac{[0,1]^2}{[0,4][0,1]}$ = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = $\frac{1,60-x moles}{4,0 L}$

[B] = $\frac{0,60-x moles}{4,0 L}$

[C] = $\frac{0,60+2x moles}{4,0 L}$

0,25 = $\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}$

You will obtain

3,75x² +2,95x +0,12 = 0

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x =-0,74363479081119 → No physical sense

x =-0,043031875855476

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$\frac{1,60-(-0,04 moles)}{4,0 L}$ = 0,41 M

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = $\frac{1,60 moles}{2,0 L}$ = 0,8 M

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[C] = $\frac{0,40 moles}{2,0 L}$ = 0,2M

I hope it helps!

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