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ArbitrLikvidat [17]
3 years ago
13

Apply the Law of Conservation of Mass to the following problem: During a combustion reaction, 12.2 grams of methane reacts with

14 g of oxygen. The reaction produces carbon dioxide and water. If 20 grams of water are produced, how many grams of carbon dioxide are produced?
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Answer:

5.8 g of carbon dioxide are produced

Explanation:

The Law of Conservation of Mass states that the mass of the reactants must equal the mass of the products in all chemical reactions.

This is the chemical reaction (combustion)

 CH₄     +   2O₂    →    CO₂   +    2H₂O

12.2 g        14 g             x             20g

Mass in reactants = 12.2 g + 14 g = 26.2 g

Mass in products =  x + 20 g

26.2 g = x + 20g

26.2 g - 20g = x

5.8 g = x

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Vlad [161]
A is the answer I know bc I take biology
7 0
3 years ago
Kc= [3.1*10^-2]^2. (mol/L^2) divide by [8.5*10^-1] [3.1*10^-3]^3 (mol/L)^4​
jok3333 [9.3K]

Answer:

Kc = 12.58

Explanation:

Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3

Kc = (0.052441)(0.10513)/(0.002385)(0.18373)

Kc = 0.0005513/0.000438

Kc = 12.58

Hope that helps!!

5 0
3 years ago
PLEASE HELP ME
ki77a [65]

Answer:

1.15 atm

Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide

We were given the following:

Total pressure = 2.45 atm

Pressure of oxygen = 0.65 atm

Pressure of carbon monoxide = x

Pressure of carbon dioxide = 0.65 atm

Therefore:

2.45 = x + 0.65 + 0.65

2.45 = x + 1.3

x = 2.45 - 1.3

x = 1.15 atm

6 0
3 years ago
\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of
atroni [7]

<u>Answer: </u>The amount of heat released is 84 calories.

<u>Explanation: </u>

The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

where,

Q = heat gained  or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

8 0
3 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0
3 years ago
Read 2 more answers
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