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mr_godi [17]
3 years ago
12

How many molecules are in 165 g of carbon dioxide?

Chemistry
2 answers:
devlian [24]3 years ago
3 0

The molar mass of CO2 is 44 grams per mole.

165 grams / 44 grams per mole of CO2 = 3.75 moles CO2

Using Avogadro’s law where 1 mole of substance equals 6.023 x 10^23 molecules

3.75 moles CO2 (6.023 x 10^23 molecules /mole) = 2.26 x 10^24 molecules CO2

Alchen [17]3 years ago
3 0

Answer: 22.6 \times 10^{23} molecules

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

For CO_2

Given mass= 165 g

Molar mass of  CO_2 = 44 g/mol

Putting values in above equation, we get:

\text{Moles of}CO_2 =\frac{165g}{44g/mol}=3.75mol

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of CO_2 contains =6.023\times 10^{23} molecules of   CO_2

Thus 3.75 moles CO_2 contains =\frac{6.023\times 10^{23}}{1}\times 3.75=22.6 \times 10^{23} molecules of CO_2

Thus there will be 22.6 \times 10^{23} molecules of CO_2

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
NeTakaya

Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

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Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

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The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

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8 0
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