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:
<em><u>To determine the number of significant figures in a number use the following 3 rules:</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.A final zero or trailing zeros in the decimal portion ONLY are significant.</u></em>
I think the answer is yes
Answer:
Mass = 179.9 g
Explanation:
Given data:
Volume of solution = 450 mL
Molarity of solution = 2.00 M
Mass in gram required = ?
Solution:
Volume of solution = 450 mL× 1 L / 1000 mL = 0.45 L
Molarity = number of moles of solute/ Volume of solution in L
2.00 M = number of moles of solute / 0.45 L
Number of moles of solute = 2.00 M × 0.45 L
M = mol/L
number of moles of solute = 0.9 mol
Mass of CaBr₂ in gram:
Mass = number of moles × molar mass
Mass = 0.9 mol ×199.89 g/mol
Mass = 179.9 g