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Ray Of Light [21]
3 years ago
7

If 9.5 × 10²⁵ molecules of CO₂ are produced in a combustion reaction, what is the mass in kg of CO₂ that is produced?

Chemistry
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

6.9428 kg of CO2

Explanation:

1) Use Avogadro's number that states 1 mole = 6.022 x 10^23 particles. Convert 9.5 x 10^25 molecules into moles.

9.5 x 10^25 CO2 moleculesx\frac{1 mole CO2}{6.022 x 10^23} = 157.75 moles CO2

2) Convert 157.75 moles of CO2 into grams. CO2's molar mass is 44.01g.

157.75moles CO2 x \frac{44.01g/mol}{1 mole CO2} = 6942.79g

3) Convert 6942.79 grams into kilograms (divide by 1000):

6.9428 kg

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Organisms breakdown _____ in order to produce____?
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0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

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Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

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T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

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V₂ =  1.92 L

4 0
3 years ago
A sealed can of your favorite soda has a carbon dioxide gas (C0) volume of 0.05 L. When it is refrigerated, the
liubo4ka [24]

Answer:

<h2>0.102 L</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{207000 \times 0.05}{101000}  =  \frac{10350}{101000}  \\  = 0.102475...

We have the final answer as

<h3>0.102 L</h3>

Hope this helps you

8 0
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TEA [102]

Explanation:

The structures of both acetone and propanal are shown below:

In the formula of propanal there is -CHO functional group at the end.

In acetone -CO- group is present in the middle that is on the second carbon.

The molecular formula is C3H6O.

Both have same molecular formula but different structural formulas.

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3 years ago
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