Find the volume of hydrogen gas that reacts with 200 cm of nitrogen gas
measured at the same temperature and pressure.
N2 + 3H2(g) 2NH3(g)

Answers

Answer 1

Answer:

15 L

Explanation:

The important thing to remember about reactions that involve gases kept under the same conditions for pressure and temperature is that the mole ratios that exist between the gases are equivalent to volume ratios.

This can be proven using the ideal gas law equation for two gases kept at a temperature

T

and a pressure

P

.

P

V

1

=

n

1

R

T

the ideal gas law equation for the first gas

P

V

2

=

n

2

R

T

the ideal gas law equation for the second gas

Divided these two equations to get

P

V

1

P

V

2

=

n

1

R

T

n

2

R

T

This means that you have

n

1

n

2

=

V

1

V

2

the mole ratio is equivalent to the volume ratio

In your case, the balanced chemical equation looks like this

N

2(g]

+

3

H

2(g]

2

NH

3(g]

The problem tells you that this reaction takes place at constant temperature and pressure. You don't need to know the exact values, you just need to know that they are constant for all three chemical species.

You can thus say that since you have a

1

:

3

mole ratio between nitrogen gas and oxygen gas, you will also have a

1

:

3

the volume ratio between these two gases

n

N

2

n

H

2

=

V

N

2

V

H

2

=

1

3

The volume of hydrogen gas needed to react with that much nitrogen gas will thus be

5.0

L N

2

3

L H

2

1

L N

2

=

15 L H

2


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Answer:

A) antigens present on the invaders

Explanation:

Antigens are substances (usually proteins) on the surface of cells, viruses, fungi, or bacteria. Nonliving substances such as toxins, chemicals, drugs, and foreign particles (such as a splinter) can also be antigens. The immune system recognizes and destroys, or tries to destroy, substances that contain antigens

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Answers

The mass percentage of sugar in the sweet tea is 10%

From the question,

We are to determine the mass percentage of sugar in the sweet tea

[tex]Mass\ percentage\ of\ sugar\ in\ the\ sweet\ tea = \frac{Mass\ of\ sugar\ in\ the\ solution}{Mass\ of\ the\ solution} \times 100\%[/tex]

Mass of sugar in the solution = 20 g

Mass of the solution = 20g + 180g = 200g

[tex]Mass\ percentage\ of\ sugar\ in\ the\ sweet\ tea = \frac{20}{200} \times 100\%[/tex]

[tex]Mass\ percentage\ of\ sugar\ in\ the\ sweet\ tea = \frac{20}{2} \%[/tex]

∴ Mass percentage of sugar in the sweet tea = 10%

Hence, the mass percentage of sugar in the sweet tea is 10%

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Answers

Answer:

The weather caused by a warm from and a cold front compare by:

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Answers

Answer:

Explanation:

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Answers

Answer:

The presence of drugs or chemicals in an organisms environment can also influence gene expression in the orgains.

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CH4+4S-CS2+2H2S
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Answers

Let's check the equation

[tex]\\ \sf\longmapsto {CH_4\atop 1mol}+{4S\atop 4moles}\longrightarrow {CS_2\atop 1mol}+{2H_2S\atop 2moles}[/tex]

4moles of S gives 1 mol of CS21mol of S gives 1/4=0.25mol of CS210.6mol of S gives:-

[tex]\\ \sf\longmapsto 10.6(0.25)=2.65moles\:of\:CS_2[/tex]

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Answer:

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Answer:

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Answers

Answer:

True

Explanation:

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Answers

Answer:

Chemistry is the branch of science that deals with the properties, composition, and structure of elements and compounds, how they can change, and the energy that is released or absorbed when they change.

Answer:

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Explanation:

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Answers

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8.1
-1.09
-8.1
1.09

Answers

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Fill in the blanks:
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Answers

87 picograms are equivalent to [tex]8.7\times 10^{-11}[/tex] grams.

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[tex]1\,pg = 1\times 10^{-12}\,g[/tex]

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[tex]87\,pg[/tex]

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[tex]8.7\times 10^{-11}\,g[/tex]

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Answers

Answer:

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Explanation:

Because I've checked


A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81
g Cl. What is this compound's percent composition?

Answers

Explanation:

getting the empirical formula

For carbon

3.01/12 = 0.2508

For oxygen

4/16 = 0.25

For Cl

17.81/35.5= 0.501

0.2508 ÷ 0.25 = 1.0032

0.25 ÷ 0.25 = 1

0.501 ÷ 0.25 = 2.004

Empirical formula of the compound = CCl20

Molar mass = (1×12) + (2×35.5) + (1×16)

= 12 + 71 + 16

= 99g/mol

percentage composition= 24.81/99 × 100

= 25.06%

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. the compound's percent composition is  25.06%.

What are the application of carbon ?

The impure form of carbon is Amorphous which is used for the manufacture of car batteries and dry cells, ink and paints; So the majority element of carbon is present in charcoal, so these are used as fuel.

Graphite is derivative form of carbon used as a lubricant in heavy industrial machinery, making pencil leads, these carbon are used in the production of steel.

The empirical formula, For carbon 3.01/12 = 0.2508

For oxygen 4/16 = 0.25, For Cl 17.81/35.5= 0.501, 0.2508 ÷ 0.25 = 1.0032

0.25 ÷ 0.25 = 1

0.501 ÷ 0.25 = 2.004

Empirical formula of the compound = CCl20

Then the Molar mass = (1×12) + (2×35.5) + (1×16)

= 12 + 71 + 16

= 99g/mol

percentage composition= 24.81/99 × 100

= 25.06%

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Explanation:

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Answers

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Answers

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Answers

Answer:

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Answers

Answer:

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Answer Choices:

A. The electron moved down to an energy level and has an energy of 10.92 x 10−19 J.

B. The electron moved down to an energy level and has an energy of 21.72 x 10−19 J.

C. The electron moved up to an energy level and has an energy of 10.92 x 10−19 J.

D. The electron moved up to an energy level and has an energy of 21.72 x 10−19 J.

Answers

Answer:

D. The electron moved up to an energy level and has an energy of 21.72 x 10−19 J.

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Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen gas (O2) according to the following equation. 2H2O Right arrow. 2H2 O2 The molar mass of H2O is 18. 01 g/mol. The molar mass of O2 is 32. 00 g/mol. What mass of H2O, in grams, must react to produce 50. 00 g of O2? 14. 07 23. 05 28. 14 56. 28.

Answers

The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.

The reaction for the decomposition of water has been:

[tex]\rm 2\;H_2O\;\rightarrow\;H_2\;+\;O_2[/tex]

From the balanced equation, 2 moles of water decomposes to form 1 moles of hydrogen and 1 mole of oxygen.

The mass of oxygen produced has been 50 g. The moles of oxygen has been given by:

[tex]\rm Moles=\dfrac{mass}{molar\;mass}[/tex]

The moles of oxygen has been:

[tex]\rm Moles_O_2=\dfrac{50}{32}\;mol\\Moles_O_2=1.5625\;mol[/tex]

The moles of oxygen produced has been 1.5625 mol.

The moles of hydrogen decomposed has been given from the balanced chemical equation as:

[tex]\rm 1 \;mole\;O_2=2\;mole\;H_2O\\1.5625\;mol\;O_2=1.5625\;\times\;2\;mol\;H_2O\\1.5625\;mol\;O_2=3.125\;mol\;H_2O[/tex]

The moles of hydrogen decomposes has been 3.125 mol.

The mass of hydrogen decomposed has been given by:

[tex]\rm Mass=moles\;times\;molar\;mass\\Mass_{H_2O}=3.125\;\times\;18.01\;g\\Mass_{H_2O}=56.28\;g[/tex]

The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.

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