Answer:
Sodium reacts with the oxygen in air. It reacts vigorously with oxygen and the moisture that is already present in the air and thus catches fire.
Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
The molar mass of carbon is 12, hydrogen is 1, and
nitrogen is 14, hence the ratio are:
C = 38.65 / 12 = 3.22
H = 16.25 / 1 = 16.25
N = 45.09 / 14 = 3.22
Divide the three by the lowest ratio which is 3.22:
C = 3.22 / 3.22 = 1
H = 16.25 / 3.22 = 5
N = 3.22 / 3.22 = 1
So the empirical formula is:
CHN
2NH4ClO4 --------> N2 + Cl2 + 2O2 + 4H2O
from reaction 2 mol 1 mol
given x mol 0.10 mol
Proportion:
<u>2 mol NH4ClO4 </u>= <u>1 mol Cl2</u>
x mol NH4ClO4 0.10 mol Cl2
x= (2*0.10)/1 = 0.20 mol NH4ClO4
