Explanation:
Molar mass of
= 39.1 + 35.5 + 3(16.0) = 122.6 g
Molar mass of KCl = 39.1 + 35.5 = 74.6 g
Molar mass of
= 32.0 g
According to the equation, 2 moles of
reacts to give 3 moles of oxygen.
Therefore, 2 (122.6) = 245.2 g of
will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of
gives 96.0 g of oxygen.
(a) Calculate the amount of oxygen given by 2.72 g of
as follows.
of
(b) Calculate the amount of oxygen given by 0.361 g of
as follows.
of
c) Calculate the amount of oxygen given by 83.6 kg
as follows.
of 
Convert kg into grams as follows.
= 32731 g of 
(d) Calculate the amount of oxygen given by 22.5 mg of
as follows.

Convert mg into grams as follows.
of 
<u><em>Answer:</em></u>
- The correct option is C.
- Formation of a precipitate
<u><em>Explanation:</em></u>
During a chemical reaction, new substances are formed known as a products, mostly reaction occur and their product is obtained as precipitates.
<u><em>Example</em></u>
Arylidene-2-thiobarbituric acid is obtained as precipitates when aldehyde and thiobarbituric acid react to each other.
melting of a substance
It is just indication of physical changes, like melting of ice, composition remained same as before.
boiling of a substance
It is just indication of physical changes, like boiling of water into vapors, composition remained same as before.
freezing of a substance
It is just indication of physical changes, like freezing of water into ice, composition remained same as before
<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:



Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:

<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:

Hence, the value of equilibrium constant for the given reaction is 56.61
Precision is obtained by getting values that are very close together. If you mess around with the protocol, you'll end up with crazy values that probably are neither accurate or precise.
Water is formed when this happens