Answer:
4.43 g
Explanation:
The reaction between sodium chloride and flourine gas is given as;
NaCl + F2 --> NaF + Cl2
From the stochiometry of the equation;
1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2
Mass of 1 mol of F2 = 38g
Mass of 1 mol of sodium flouride, NaF = 42g
This means 38g of flourine reacted with NaCl to form 42g of NaF
xg of F2 would form 4.9g of NaF
38 = 42
x = 4.9
x = 4.9 * 38 / 42
x = 4.43 g
The Answer to your question would be A
Answer:
The answer to your question is: letter A
Explanation:
If an atom lose 3 electrons its charge will be positive, it will be +3
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
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