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Ann [662]
3 years ago
11

• Find the Percentage Composition for CO

Chemistry
2 answers:
gulaghasi [49]3 years ago
8 0

Answer:

C = 42.86%

O = 57.14%

Explanation:

Percentage composition = (mass/molar mass) x 100

CO = Molar Mass 28g/mol

C = 12g/mol

O = 16g/mol

C = 12/28 x 100 = 42.86%

O = 16/28 x 100 = 57.14%

Aleks04 [339]3 years ago
4 0

Answer:

Carbon or the symbol C has a mass percentage of 42.880% while Oxygen or symbol O has a mass percentage of 57.120%

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Which type of organic reaction produces both water and carbon dioxide?
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the correct answer is combustion

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an example for combustion reaction is as follows

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CH₄ + 2O₂ --> CO₂ + 2H₂O

correct answer is

2. combustion


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The component of solution present in smaller quantity is called
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A solute

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A physical change is a change (4 points) in the properties of matter that does not change the identity of the substance in the p
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Why is the air in a jet aircraft flying at high altitudes pressurized? *
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Answer:

Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

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4 years ago
Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

7 0
3 years ago
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