The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M
<u><em>calculation</em></u>
This is calculated usind M₁V₁=M₂V₂ formula
where,
M₁( molarity ₁) = ?
V₁( volume ₁) = 2.00 L
M₁ (molarity ₂) = 0.25M
V₂( volume₂) = 4.5 L
make M₁ the subject of the formula by diving both side of the formula by V₁
M₁ is therefore = M₂V₂/V₁
M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M
B. 1, 1, 1, 2
Explanation:
You only need to balance the NaNO3 on the right. Since there is 2 NO3 on the left, you need to put a 2 in front of the NaNO3 on the right. Everything else is already balanced so the only coefficient needed is 2 in front of the NaNO3.
The scientist's results is that at a temperature of 35<span>°C, the solubility of the substance in water is 146.2 grams in 200 grams of water. There isn't really a different method to determine the solubility of a substance in water. Another procedure could be that a lesser amount of the substance is used and the water required to dissolve it is determined. The solubility of the substance based on the two procedures can then be compared.</span>
Answer:
Salt is the pure substance out of them all.
The equilibrium constant (Kp) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at 530 degree C is 2.9 x 10 to the minus eleventh
N2(g) + O2 (g) = 2 NO (g)
