It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.
<u>Explanation:</u>
2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂
We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

= 337.5 g AgCl
In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.
It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.
Answer:
100g / (5.2g/cm3)
= 100g / (5.2g / 1cm3)
= 100g x 1cm3 / 5.2 g
= 19.2 cm3
Since 1 cm3 = 1 ml, your answer is 19.2 ml.
Explanation:
Density of object = gram/volume
= 100 g/ 5 cm³
= 20 g/cm³