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1 year ago

You collect 552 mL of argon gas at 23.0 C. What volume will the gas occupy at 46.0C if the pressure remains constant?

Chemistry

1 answer:

taurus [48]1 year ago

8 0

Answer:

1027.9 mL

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

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What is the maximum amount of ammonia that can be produced when 0.35 mol Nz reacts with 0.90 mol Hz?

professor190 [17]

Answer:

0.6 moles NH₃

Explanation:

The reaction that takes place is:

N₂ + 3H₂ → 2NH₃

First we determine the limiting reactant:

0.35 mol N₂ would react completely with (3*0.35) 1.05 moles of H₂. There are not as many H₂ moles, so H₂ is the limiting reactant.

Then we convert H₂ moles (the limiting reactant) to NH₃ moles, keeping in mind the stoichiometry of the reaction:

0.90 mol H₂ * $\frac{2molNH_3}{3molH_2}$ = 0.6 moles NH₃

5 0

2 years ago

A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir

Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H: 0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

5 0

2 years ago

Ethene is the first member of the

DanielleElmas [232]

It will be:

b. Alkyne* series

* You have done a spelling mistake here. It will be "Alkene" not "Alkyne".

5 0

2 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

2 years ago

describe the trends in the atomic size, location energy and electrongavity from left to right across a period in periodic table

Elenna [48]

Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.

<h3>Describe the trends in the atomic size, ionization energy and electronegativity?</h3>

Atomic radius decreases across a period because of nuclear charge increases whereas atomic radius of atoms generally increases from top to bottom within a group because there is again an increase in the positive nuclear charge.

Ionization energy increases when we move from left to right across an period and decreases from top to bottom.

Electronegativity also increases from left to right across a period and decreases from top to bottom.

So we can conclude that atomic size decreases in a period but the ionization energy and electronegativity increases across a period.

Learn more about Electronegativity here: brainly.com/question/24977425

#SPJ1

3 0

1 year ago