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xxTIMURxx [149]
3 years ago
7

How many grams of chlorine gas are in a 17.8 liter sample at 1.1 atmospheres and 29 degrees Celsius

Chemistry
1 answer:
jasenka [17]3 years ago
5 0
For this problem, we can use the Ideal Gas Law to solve for the number of moles of chlorine gas present in this container, and then we can convert this molar measurement into grams.

The Ideal Gas Law is presented as:

PV=nRT

where P is pressure, V is volume, in liters, n is the number of moles of substance, R is the gas constant associated with the units for pressure, and T is temperature, in Kelvins.

Starting with the IGL, we can substitute the information given in the problem, and then we can simplify to a number of moles of Chlorine gas.

PV=nRT
(1.1 atm)(17.8 L) = n(0.0821 L•atm/mol•K)(302.15 K)
19.58=24.807n
n=0.7893 moles

Chlorine is a gas in this problem, which means it is the di-Chlorine form most commonly found as a gas (Cl2). This molecule weighs twice as much as elemental chlorine (2 mol x 35.453 g/mol). So, we can multiply this new weight by the number of moles of chlorine gas we calculated:

70.9 g/mol • 0.7893 mol = 55.96 g Cl2

55.96 grams of diChlorine gas is present in the sample.
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A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal
Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent:  0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

                |   masss  |  volume  |  mol

solute       |                |                |          

solvent     |                |                |  

solution    |                |                |

Let's complete, what we have.

                 |   masss  |  volume  |  mol

solute       |  10.8g     |                |          

solvent     |                |  133 mL   |  

solution    |                |  137 mL    |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

                 |   masss  |  volume  |  mol

solute       |   10.8g     |                |   0.110 mol      

solvent     |   133g      |  133 mL   |  7.39 mol

solution    |   143.8g   |  137 mL   | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles  or  solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

5 0
3 years ago
What is the volume, in liters, of 2.00 moles of hydrogen at STP?
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The volume of one mole of any gas at STP is 22.4 L. So, at STP, the volume of 2.00 moles of hydrogen gas would be (22.4 L/mol)(2 mol H2) = 44.8 L.
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What is the correct name for the cyclic alkane shown on the right? 1,3-dimethyl, 3-ethylcyclohexane 1-ethyl, 1,4-dimethylcyclohexane 1-ethyl-1,3-dimethylcyclohexane 1-ethyl-1,3-dimethylcyclopentane 1,3 dimethyl, 3- ethylcyclopentane A
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zlopas [31]

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salantis [7]

Answer: A

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