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)

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

if you're asking what comes out, it's steam.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.