I would answer this but i have to go to bed i have a reading test tommorow
Dear friend,
Because the water remains water throughout this process, it is not undergoing any chemical change. There is no new substance or element added but only a change of form.
1) Calculate the number of moles in 1.15 liter of 0.100 M HNO3 solution.
M = n / V => n = M*V = 0.100M * 1.15 l = 0.115 moles
2) Calculate the mass of 0.115 moles of HNO3
mass = number of moles * molar mass
molar mass of HNO3 = 1.00 g/mol + 14.0 g/mol + 3*16.0g/mol = 63.0 g/mol
mass = 0.115 mol * 63.0g/mol = 7.245 g
3) Calculate the mass of 70.3% HNO3 solution that contains 7.245 grams of HNO3
% = (mass of solute / mass of solution) * 100
=> mass of solution = mass of solute * 100 / % = 7.245 g * 100 / 70.3%
mass of soltuion = 10.3 g.
4) Convert 10.3 grams of HNO3 solution into volume, using density, D
D =mass / Volume => Volume = mass / D
=. Volume = 10.3 g / 1.41 g/cm^3 = 7.30 cm^3
Answer: 7.30 cm^3
(a) <u>7.32 moles of N₂</u>
(b) = 5.72 moles of NO
(c) 0.460 moles of O₂
Explanation:
We begin by balancing the chemical reaction involved;
N₂ + 0₂ → 2NO
The number in front of the elements or compounds indicate the mole ration involved in the reaction. One (1) mole of N2 and one (1) mole of O2 react to form two (2) moles of NO.
(a) Because the mole ratio between O₂ and N₂ is 1 : 1 then <u>7.32 moles of N₂</u> are needed to react with 7.32 mol of O₂
(b) Because the ration of N₂ and NO in the chemical equation is 1 : 2, then (2.86 * 2) moles of NO can be formed when 2.86 mol of N₂ reacts.
= 5.72 moles of NO
(c) (b) Because the ration of O₂ and NO in the chemical equation is 1 : 2, then (0.230 * 2) moles of O₂ must react to produce 0.230 mol of NO.
= 0.460 moles of O₂