O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V
Answer:
39 mol AgNO3
Explanation:
We have the equation 4HNO3 + 3Ag -----> 3AgNO3 + NO + 2H2O
We want to calculate the number of silver nitrate (AgNO3) moles that would be produced from 52 moles of nitric acid ( HNO3 )
We can calculate this by using mole ratio as well as dimensional analysis.
The mole ratio of Silver nitrate to nitric acid based on the balanced equation is 3AgNO3:4HNO3.
Using this we can create a table: The table is attached.
Breakdown of the table.
The moles of nitric acid cancel out and we multiply 52 by 3/4 to get 39 moles of Silver nitrate.
Cumulus, stratus, and cirrus, there's many more but these are the main ones ^^
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L