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2 years ago

For which compound does 0.256 mole weigh 12.8 g?

1 answer:

6 0

Number of moles = $\frac{mass (g)}{Molar mass}$
so to calculate molar mass (Molecular weight of compound):
Molecular weight = $\frac{mass (g)}{number of moles}$
= $\frac{12.8 g}{0.256 mole}$ = 50 g / mole
Now we calculate molecular weight of each compound in choices:
a) C₂H₄O = 44
b) CO₂ = 44
c) CH₃Cl = 50.4 ALMOST 50 so this is the correct answer
d) C₂H₆ = 30

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. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy

gayaneshka [121]

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

$density=\frac{mass}{volume}$

Density is a characteristic property of every body or substance.

The most commonly used units of density are $\frac{kg}{m^{3} }$ or $\frac{g}{cm^{3} }$ for solids, and $\frac{kg}{L}$ or $\frac{g}{mL}$ for liquids and gases.

In this case, you know:

density= 0.87 $\frac{g}{mL}$
mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)
volume= ?

Replacing:

$0.87\frac{g}{mL} =\frac{3,600 g}{volume}$

Solving:

$volume =\frac{3,600 g}{0.87\frac{g}{mL}}$

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

<u><em>The volume is 4.13793 L</em></u>

5 0

3 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

2 years ago

Help me Plz, I am really confused

avanturin [10]

O is a blood donor
AB receiver

5 0

2 years ago

Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c

kvv77 [185]

Mercury naturally exists in Liquid state.

On Condensing it can exist in Solid state as well.

Hope it helps...

Regards;

Leukonov/Olegion.

3 0

3 years ago

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ANSWER FAST PLZ!!!!!!!!!

valina [46]

I think the answer is B

7 0

2 years ago

Read 2 more answers