There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
