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alexandr1967 [171]
3 years ago
14

The following reaction mixture contains 0.22 M CH4, 0.67 M CO2 and 1.3 M H2O. Which of the following statements is TRUE concerni

ng this system?
4 CuO(s) + CH4(g) ? CO2(g) + 4 Cu(s) + 2 H2O(g)
Kc = 1.10
a)the reaction is running fastest toward the reactants
b)the reaction is running fastest toward the products
c)the value of the equilibrium constant will increase
d)the value of the reaction quotient will increase
e)the reaction is at equilibrium
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

Reaction is shifting left toward reactant side

Explanation:

4CuO(s) + CH₄(g) <=?=> CO₂(g) + Cu(s) + 2H₂O(g)

---------      0.22M             0.67M     -------    1.30M

Qc = [O₂(g)][H₂O(l)]²/[CH₄(g)] = (0.67)(1.30)²/(0.22) =5.12

Kc = 1.1 < Qc = 5.12 => Reaction is shifting LEFT and reactant concentrations are increasing with product concentrations decreasing. Qc will be decreasing and final Kc value will be lower than original Qc upon reaching equilibrium stability.  

_____________________

FYI Note => For reaction mixtures the following is a good guide to reaction response to applied system stress. Note the conclusion follows the direction the inequality symbol is pointing.  

Kc  <  Qc  => Rxn system is shifting Left

Kc =   Qc  => Rxn system is at equilibrium

Kc >  Qc  => Rxn system is shifting Right

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Which formula equation represents the burning of sulfur to produce sulfur dioxide?
Dmitriy789 [7]

Answer:

option A = S(s) + O₂(g)   →   SO₂ (s)

Explanation:

Chemical equation:

S(s) + O₂(g)   →   SO₂ (s)

when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.

Uses of sulfur dioxde:

It is used as a solvent and reagent in laboratory.

Sulfur dioxide is used to produce sulfuric acid.

It is used as a disinfectant

It is also used as a reducing agent.

It is used to preserve the dry food.

7 0
4 years ago
How many moles are 8.8 grams of CO2?
Lemur [1.5K]
The answer is .2grams
8 0
4 years ago
How many grams of NaCl are required to make 150.0 mL of a 5.000 m solution
Oksi-84 [34.3K]

The  grams   of NaCl  that are required  to  make  150.0 ml of  a  5.000 M  solution is  43.875 g


calculation

Step 1:calculate  the  number of moles

moles =  molarity  x volume  in L

volume  = 150 ml / 1000 = 0.15 L

= 0.15 L  x 5.000  M  = 0.75  moles

Step 2:  calculate mass

mass =  moles x  molar mass

molar mass  of NaCl = 23 + 35.5 = 58.5 mol /L


mass is therefore =0.75  moles  x 58.5  mol /l =43.875 g

4 0
3 years ago
Read 2 more answers
Type the correct answer in each box.
Arturiano [62]

Answer:

5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2

Explanation:

balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.

8 0
3 years ago
26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to
frozen [14]

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

7 0
3 years ago
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