When sodium amide i.e.
reacts with water i.e.
results in the formation of sodium hydroxide i.e.
and ammonia
.
The chemical reaction is given by:

Now, when ammonia i.e.
reacts with water results in the formation of ammonium hydroxide i.e. 
The chemical reaction is given by:

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).
The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.
Answer:
During cellular respiration animal cells combine oxygen with food molecules to release energy to live and function. Remember that cellular respiration produces carbon dioxide as a waste product. Animals use energy to grow, reproduce, and to function. They release the carbon dioxide into the air as a waste product
Explanation:
Answer:
The entropy change in the environment is 3.62x10²⁶.
Explanation:
The entropy change can be calculated using the following equation:

Where:
Q: is the energy transferred = 5.0 MJ
: is the Boltzmann constant = 1.38x10⁻²³ J/K
: is the initial temperature = 1000 K
: is the final temperature = 500 K
Hence, the entropy change is:
Therefore, the entropy change in the environment is 3.62x10²⁶.
I hope it helps you!
Your answer is going to be A, because it was shoved harder, it will go faster
The balanced chemical reaction is expressed as follows:
<span>CuCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + CuNO32 (aq)
To determine the </span><span>concentration of copper(II) chloride contaminant in the original groundwater sample, we use the final amount of silver chloride that was produced from the reaction and the relation of the substances from the chemical reaction. We calculate as follows:
mmol AgCl = 6.1 mg AgCl ( 1 mmol / 143.35 mg ) = 0.0426 mmol
mmol CuCl2 = </span>0.0426 mmol AgCl ( 1 mmol CuCl2 / 2 mmol AgCl ) = 0.0213 mmol CuCl2
concentration of CuCl2 in the original water sample = 0.0213 mmol CuCl2 / 200.0 mL = 1.0638 x 10^-4 mmol / mL or 1.0638 x 10^-4 mol/L