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ki77a [65]
2 years ago
4

Be sure to answer all parts. Predict the effect of increasing the container volume on the amounts of each reactant and product i

n the following reactions: (a) CH3OH(l) ⇌ CH3OH(g) There will be more CH3OH(l) and less CH3OH(g). There will be no change in the amounts. There will be less CH3OH(l) and more CH3OH(g). (b) CH4(g) + NH3(g) ⇌ HCN(g) + 3H2(g) There will be no change in the amounts There will be more CH4 and NH3 and less HCN and H2. There will be less CH4 and NH3 and more HCN and H2.
Chemistry
1 answer:
Marina CMI [18]2 years ago
3 0

Answer:

a) There will be less CH3OH(l) and more CH3OH(g).

b) There will be less CH₄ and NH₃ and more HCN and H₂.

Explanation:

Based on LeChatelier's principle, the increasing of volume in a container with a gas reaction will make the reaction produce more gas to counteract the change in pressure because of the increasing of volume. Thus:

a) CH₃OH(l) ⇌ CH₃OH(g). The increasing in volume will produce more CH₃OH(g) to counteract the decreasing in pressure due the increasing of volume. In other words, <em>There will be less CH3OH(l) and more CH3OH(g).</em>

<em />

b) CH₄(g) + NH₃(g) ⇌ HCN(g) + 3H₂(g). In this reaction, you have in the left two moles of gas and in the right 4 moles of gas. Thus, the increasing of volume in the system will produce more HCN and H₂ trying to counteract the change in pressure decreasing concentration of CH₄ and NH₃. That is: <em>There will be less CH₄ and NH₃ and more HCN and H₂.</em>

<em></em>

<em>There will be no changes in the amounts of substances if in the right and the left you have the same number of gas moles.</em>

<em></em>

I hope it helps!

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Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
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ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
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b = 0.463 mol ÷ 0.07918 kg.
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2 years ago
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Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

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The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

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Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

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<u>Explanation:</u>

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This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of calcium carbonate follows:

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