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Monica [59]
4 years ago
5

If 12 mol of lithium were reacted with excess nitrogen gas, how many moles of lithium nitride would be produced

Chemistry
2 answers:
ludmilkaskok [199]4 years ago
7 0

4 is the number of mols


Lyrx [107]4 years ago
5 0
The balanced reaction that describes the reaction between lithium  and nitrogen gas to produce lithium nitride is expressed 6Li + N2 = 2Li3N.Hence for every 6 moles of lithium used, there are 2 moles of lithium nitride produced. For this problem, 12 moles of lithium yields 4 moles of lithium nitride.
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Gre4nikov [31]

Answer: 114.91L

Explanation:

You need to know the conversion factor first in order to solve this. Any gas occupies 22.4L per mol.

5.13mol(\frac{22.4L}{1mol})= 114.91L of nitrogen gas.

7 0
3 years ago
For each (P, V) pair, type the pressure in the x-
lisabon 2012 [21]

Answer:

using three significant figures, to match the data

v = 51.4

p = -0.999

6 0
3 years ago
A container holds 0.490 m3 of oxygen at an absolute pressure of 5.00 atm. A valve is opened, allowing the gas to drive a piston,
mash [69]

<u>Answer:</u> The new volume will be 2.04m^3

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2            (at constant temperature)

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=5.00atm\\V_1=0.490m^3\\P_2=1.20atm\\V_2=?m^3

Putting values in above equation, we get:

5atm\times 0.490m^3=1.20\times V_2\\\\V_2=2.04m^3

Hence, the new volume will be 2.04m^3

5 0
3 years ago
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Read 2 more answers
. How much energy is lost by a 30.0g sample of water that decreases in temperature from 56.7C to 25.0C?
lbvjy [14]

Answer:

Q = -3980.9 j

Explanation:

Given data:

Mass of sample = 30 g

Initial temperature = 56.7 °C

Final temperature = 25 °C

Specific heat of water = 4.186 j/g.°C

Amount of heat released = ?

Formula:

Q = m.c.ΔT

Q = heat released

m = mass of sample

c = specific heat of given sample

ΔT = change in temperature

Solution:

ΔT = T2 -T1

ΔT = 25 °C - 56.7 °C = - 31.7°C

Q = m.c.ΔT

Q = 30 g × 4.186 j/g.°C ×  - 31.7°C

Q = -3980.9 j

8 0
3 years ago
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