To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ =
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃
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Answer:
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Explanation:
Let volume of the 40% acid solution be x.
Let volume of the 60% acid solution be y.
Volume of solution formed after mixing both solution = 40 L
x + y = 40 L..[1]
Volume of acid 40% solution = 40% of x= 0.4x
Volume of acid 60% solution = 60% of y= 0.6y
Volume of acid formed = 45% of 40 L = 
..[2]
Solving [1] and [2]
x = 30 L , y = 10 L
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
