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%
Answer:
19.12 L
Explanation:
At STP(i.e. Standard temperature and pressure).
The volume occupied by one mole of gas = 22.4 L
The pressure = 1 atm
The temperature = 273 K
Thus, since 1 mole of gas = 22.4 L;
Then 0.853 moles of N2 gas will occupy:
= (0.853 moles of N2 gas × 22.4 L)/ 1 mole of N2 gas
= 19.12 L
Im pretty sure the answer is <span> 0.01218859659g
not 100% sure tho so please consult someone else b4 answering
i hope this helps!!</span>
Water can be split into O2 and H2 gas
