4 mol NH₃ → 5 mol N₂
x mol NH₃ → 0.824 mol N₂
x=0.824*4/5=0.6592 mol
Answer:
The new temperature of the water bath 32.0°C.
Explanation:
Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)
Initial temperature of the water = 
Final temperature of the water = 
Specific heat capacity of water under these conditions = c = 4.18 J/gK
Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J
( 1kJ=1000 J)




The new temperature of the water bath 32.0°C.
<span>The half-life of 9 months is 0.75 years.
2.0 years is 2.0/0.75 = 2.67 half-lives.
Each half-life represents a reduction in the amount remaining by a factor of two, so:
A(t)/A(0) = 2^(-t/h)
where A(t) = amount at time t
h = half-life in some unit
t = elapsed time in the same unit
A(t)/A(0) = 2^(-2.67) = 0.157
15.7% of the original amount will remain after 2.0 years.
This is pretty easy one to solve. I was happy doing it.</span>
Answer: i think its A diatomic compound..
Explanation: hope i helped! sorry if im wrong!
Answer: A volume of 500 mL water is required to prepare 0.1 M
from 100 ml of 0.5 M solution.
Explanation:
Given:
= 0.1 M,
= ?
= 0.5 M,
= 100 mL
Formula used to calculate the volume of water is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M
from 100 ml of 0.5 M solution.