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:
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Explanation:
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Answer:
70 mL of 5% HCl and 30 mL of 15% HCl
Explanation:
We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:
5x + 15y = 8
Since x and y are fractions of a total, they must equal one:
x + y = 1
This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:
y = 1 - x
This expression is substituted into the first equation and we solve for x.
5x + 15(1 - x) = 8
5x+ 15 - 15x = 8
-10x = -7
x = 7/10 = 0.7
We then calculate the value of y:
y = 1 - x = 1 - 0.7 = 0.3
Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:
(100 mL)(0.7) = 70 mL
Similarly, the volume of 15% HCl we need is:
(100 mL)(0.3) = 30 mL
Answer:
The molar mass and molecular weight of Al(CH3CO2)3 is 204.1136.
Explanation:
