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3 years ago

What is the volume of 1.9 moles of chlorine gas (Cl2) at 298 K and 1.3 atm?

Chemistry

2 answers:

nataly862011 [7]3 years ago

6 0

Hello!

For this exercise you have to use the Ideal Gas Law Equation,  as follow:

PV= nRT

Where:

P= pressure ( in this example, atmosphere (atm) is the pressure unit)

V= volume (which is the value you want to calculate)

n= Mol (as reference to the mass of the gas Cl2)

R = Ideal Gas Constant (for this example is a number 0.0821 atm*L/mol*K)

T= Temperature in Kelvin (K)

So, for find the value of the volume, you just have to replace the values given in the equation as follow:

P= 1.3 atm

V = you don't know, is what you want to know.

n= 1.9 mol

R =0.0821 atmL/molK

T =298 K

Remember, never forget to put the units, because teacher can get down your score if you don't do it. So, back with the values to the equation:

13atm * V = 1.9mol * 0.0821 atmL/molK * 298K

Then, we have to solve the equation to find V value, so we has to send 1.3 atm to divide in the other side of the equation:

V = (1.9mol * 0.0821 atmL/molK * 298K)/ 1.3 atm

V = 35.76 L

And, that's it!

I hope this help you a lot, and good luck.

kap26 [50]3 years ago

3 0

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol

The answer is 43 L if I am correct.

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Given:

bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ $= \frac{59.7}{16} = 3.73125 \ moles$

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0

3 years ago

3(x - 2) = 5(x + 4)

Tcecarenko [31]

Answer:

Uh first of all this is algebra but I'll answer this

First distribute the three and 5 (Multiply them by both terms inside parenthesis.

3x-6=5x+20

Then add like terms

8x=14

Divide 8 by 8 and 8 by 14

x = 14/8

Explanation:

5 0

3 years ago

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A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

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3 years ago

This is the process where fossil fuels, forests, or other carbon-containing substances are burned, adding more carbon dioxide to

sattari [20]

The process where fossil fuels, forests, or other carbon-containing substances are burned, addin more carbon dioxide to the air is the combustion.

Some examples of combustion are:

Fossil fuel:

Carbon + O2

C + O2 -> CO2

Forests (wood)

Wood = cellulose = [C6H10O5]n

[C6H10O5]n + 6nO2 = 6n CO2 + 5n H2O

So, in general the combustion of organic matter produces CO2 and water.

4 0

2 years ago

If the reaction is at dynamic equilibrium at 500 K, which statement applies to the given chemical system?

Alenkinab [10]

The correct option is this: THE CONCENTRATION OF THE PRODUCTS AND THE REACTANTS DO NOT CHANGE.

A reversible chemical reaction is said to be in equilibrium if the rate of forward reaction is equal to the rate of backward reaction. At this stage, the concentrations of the products and the reactants remain constant, that is, there is no net change in the concentration even though the reacting species are moving between the forward and the backward reaction.

8 0

3 years ago

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