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tia_tia [17]
3 years ago
5

A mixture of CO (g) and excess O2(g) is placed in a 1.0 L reaction vessel at 100.0C and a total pressure of 1.50 atm. The CO is

ignited and it burns according to the balanced equation below. After reaction, the vessel is cooled to 100.0C and the pressure of the mixture of product gases (O2(g) , CO2(g)) is 1.40 atm. What is the partial pressure of of CO2 in the product mixture
Chemistry
1 answer:
umka2103 [35]3 years ago
7 0

Answer:

Partial pressure of of CO₂ in the product mixture is 0,20atm

Explanation:

The balance equation is:

2CO(g) + O₂(g) → 2CO₂(g)

Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:

X₁ + Y₁  = 1,50atm <em>(1)</em>

Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂

After reaction partial pressures are:

X₂ = X₁ - 2n = 0; <em>2n = X₁</em>

Y₂ = Y₁ - n

Z₂ = 2n

Where Z₂ is final partial pressure of CO₂

After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:

1,40 atm = (Y₂ + Z₂)

1,40 atm = Y₁ - n + 2n

1,40atm = Y₁ + n

1,40 atm = Y₁ + X₁/2 <em>(2)</em>

Replacing (1) in (2)

1,40 atm = 1,50atm - X₁ + X₁/2

-0,10 atm = - X₁/2

<em>0,20 atm = X₁</em>.

As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>

I hope it helps!

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Answer:

The addition of sulfate ions shifts equilibrium to the left.

Explanation:

Hello!

In this case, according to the following ionization of strontium sulfate:

SrSO_3(s)\rightleftharpoons Sr^{2+}+SO_4^{2-}

It is evidenced that when sodium sulfate is added, sulfate, SO_4^{2+} is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:

The addition of sulfate ions shifts equilibrium to the left.

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What stress will shift the following equilibrium system to the left?
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<u>Answer:</u> Increasing temperature

<u>Explanation:</u>

The Principle of Le Chatelier states that <u>if a system in equilibrium is subjected to a change of conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium. </u>

The variation of one or several of the following factors can alter the equilibrium condition in a chemical reaction:

  • Temperature
  • The pressure
  • The volume
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In the case of the reaction in the question, <u>the change that moves the balance to the left will be the one that moves it towards the reagents</u>, that is, that favors the production of reagents instead of products.

  • Decreasing the concentration of SO3 and increasing the concentration of SO2 <u>will favor the production of SO3</u>, which is the product of the reaction.
  • Decreasing the volume increases the pressure of the system and the balance will move to where there is less number of moles. In the case of the reaction in question, we have 3 moles of molecules in the reactants (1 mole of O2 + 2 moles of SO2) while in the products there are 2 moles of SO3 only, therefore, <u>decreasing the volume will displace the balance to the right</u>, which corresponds to the sense in which there is less number of moles.

The reaction of the question is an exothermic since ΔH <0, therefore in the reaction heat is produced and it can be written in the following way,

2SO2(g) + O2(g) ⇌ 2SO3(g) + heat

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What is the molar mass of water in the hydrate? FeSO4 • 7H2O
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Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alco
natka813 [3]

Answer:

The approximate molar mass of lauryl alcohol is 174.08 g/m

Explanation:

An excersise to apply the colligative property of Freezing-point depression.

This is the formula: ΔT = Kf . m

First of all, think the T° of fusion of benzene → 5.5°C

ΔT = T° pure solvent - T° fusion solution

Kf for benzene: 5.12 °C/m

5.5°C - 4.5°C = 5.12 °C /m  . m

1°C /  5.12 m /°C = m

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This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.

I have to find out in 0.2 kg.

1 kg sv ____ 0.195 moles solute

0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute

The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles

6.80 g/ 0.039 moles = 174.08 g/m

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