Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
Answer:
We need 92.3 grams of sodium azide
Explanation:
Step 1: Data given
Mass of nitrogen gas = 59.6 grams
Molar mass of nitrogen gas = 28.0 g/mol
Molar mass of sodium azide = 65.0 g/mol
Step 2: The balanced equation
2NaN3 → 2Na + 3N2
Step 3: Calculate moles nitrogen gas
Moles N2 = mass N2 / molar mass N2
Moles N2 = 59.6 grams/ 28.0 g/mol
Moles N2 = 2.13 moles
Step 4: Calculate moles NaN3
for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2
For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3
Step 5: Calculate mass NaN3
Mass NaN3 = Moles NaN3 * molar mass NaN3
Mass NaN3 = 1.42 moles * 65.0 g/mol
Mass NaN3 = 92.3 grams
We need 92.3 grams of sodium azide
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The average mass of a single chlorine atom is 35.453 grams