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3 years ago

In each pair of compounds, pick the one with the higher vapor pressure at a given temperature. Explain your reasoning.

Chemistry

1 answer:

Gnoma [55]3 years ago

8 0

Answer:

The answer is:

$PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO$

Explanation:

Parts A:

The vapor pressure is higher in $CH_4$ because it is non-polar, while $CH_3Cl$ is polar. $CH_4$ has a lower molar weight as well.

Part B:

Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of $CH_30H$ .

Part C:

$H_2CO$ does not show hydrogen due to the increased vapor pressure

$CH_3OH$ bonding.

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1 pts

soldier1979 [14.2K]

Answer:

144 g

Explanation:

Use the mole ratio of 4 mol CO2 for every 9 mol O2 to convert from mol O2 to mol CO2. Then use the molar mass of CO2 to convert from mol of CO2 to grams of CO2.

7.34 mol O2 • (4 mol CO2 / 9 mol O2) • (44.01 g CO2 / 1 mol CO2) = 144 g CO2

5 0

3 years ago

Read 2 more answers

An ion is an atom with a net electrical charge due to ______.

Serga [27]

<span>the loss of one or more electrons & the addition of one of more electrons</span>

7 0

3 years ago

Read 2 more answers

Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.15 mol of PCl3 in the flask. What is the equi

borishaifa [10]

Answer : The equilibrium concentration of $PCl_5$ is, 0.50 M

Explanation : Given,

Initial moles of $PCl_5$ = 0.65 mole

Volume of solution = 1.0 L

Moles of $PCl_3$ at equilibrium = 0.15 mole

The balanced equilibrium reaction will be,

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Initial moles 0.65 0 0

At eqm. (0.65-x) x x

Moles of $PCl_3$ at equilibrium = x = 0.15 mole

Moles of $Cl_2$ at equilibrium = x = 0.15 mole

Moles of $PCl_5$ at equilibrium = (0.65-x) = (0.65-0.15) = 0.50 mole

Now we have to calculate the concentration of $PCl_3,Cl_2\text{ and }PCl_5$ at equilibrium.

Formula used : $Concentration=\frac{Moles}{Volume}$

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M$

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.50mole}{1.0L}=0.50M$

Therefore, the equilibrium concentration of $PCl_5$ is, 0.50 M

8 0

4 years ago

Name a substance in the air which makes the surface of limestone change

Elanso [62]

A substance in the air which can change the surface of limestone would be oxygen, O2.

6 0

3 years ago

A sample from a local stream is found to have 3.0 ppm nitrates (in the form of sodium nitrate). How many grams of sodium nitrate

maw [93]

Answer:

0.01028 grams of sodium nitrate would there be in 2.5 L of the stream.

Explanation:

The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

The ppm concentration of nitrates = 3.0 ppm

Mass of nitrates = x

Mass of steam= m

Volume of steam = V = 2.5 L = 2500 ml ( 1 L = 1000 mL)

Density of steam = d = 1.0 g/mL

$M=d\times V=1.0 g/mL\times 2500 mL = 2500 g$

Putting values in above equation, we get:

$3.0=\frac{x}{2500 g}\times 10^6$

$x=0.0075 g$

Mass of nitrate = 0.0075 g

Moles of nitrate = $\frac{0.0075 g}{62 g/mol}=0.0001210 mol$

1 mole of nitrate ion is present in 1 mole of sodium nitrate.

Then 0.0001210 moles of nitrate will be present in :

$\frac{1}{1}\times 0.0001210 mol=0.0001210 mol$ of sodium nitrate;

Mass of 0.0001210 moles of sodium nitrate :

0.0001210 mol × 85 g/mol = 0.01028 g

0.01028 grams of sodium nitrate would there be in 2.5 L of the stream.

6 0

4 years ago