Answer:
0.535 g
Explanation:
The reaction that takes place is:
- NaCl + AgNO₃ → AgCl + NaNO₃
First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- moles = Molarity * liters
<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L
- moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃
Then we <u>convert AgNO₃ moles into NaCl moles</u>:
- 0.00915 mol AgNO₃ *
= 0.00915 mol NaCl
Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:
- 0.00915 mol NaCl * 58.44 g/mol = 0.535 g
The concentration as % by mass is calculated as below
mass of solute/mass of solvent x100
mass of solute(sugar) = 4g
mass of solvent(water) =46 g
= 4g/ 46 g x100 = 8.7%
The answer is: 197 grams
Why?
All 20 grams of aluminum was used to form aluminum bromide. Only 177 grams of bromide were used to make aluminum bromide, since 23 grams were left over.
20 grams + 177 grams = 197 grams, so the answer is 197 grams.
I think A would be the correct answer to this question.
Answer:
The balance of chemical equation is given below.
3Ca + N2------ Ca3N2