The answer for your question is <span>No. This is because in given conditions, it is not the most stable form of oxygen's element. It will not equate into zero because there will be charge remained after balancing the equation.
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<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:

where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:

Hence, the experimental van't Hoff factor is 1.21
30.58 grans of Calcium ...
Answer:
Explanation:
a )
3NO₂(g) + H₂O(l) — -→ 2HNO₃(aq) + NO(g)
3 x 46 g 18 g 2 x 63 g 30 g
138 g of NO₂ requires 18 g of H₂O
28 g of NO₂ requires ( 18 / 138) x 28
= 3.65 g of H₂O.
b )
18 g of H₂O produces 30 g of NO gas
15.8 g of H₂O produces ( 30/18 ) x 15.8
= 26.33 g of NO gas .
c )
138 g of NO₂ produces 126 g of HNO₃
8.25 g of NO₂ produces (126 / 138 ) x 8.25
= 7.53 g of HNO₃