Answer:
6%
Explanation:
Hello, for this case, we assume that the volume of the solution is 1L, thus, the mass is given by using the density as follows:
Now, the mass of the ethanol:
Finally, the by mass percent is:
%
%m=6%
Best regards.
Answer:
The law is observed in the given equation.
Explanation:
CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂
In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:
- Ca ⇒ 1 on the left, 1 on the right.
- C ⇒ 1 on the left, 1 on the right.
- O ⇒ 3 on the left, 3 on the right.
- H ⇒ 2 on the left, 2 on the right.
- Cl ⇒ 2 on the left, 2 on the right.
As the numbers for all elements involved are the same, the law is observed in the given equation.
Answer:
19.4 g of alum, will be its theoretical yield
Explanation:
The reaction is:
2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O
Let's determine the amount of acid.
M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution
M = mmol /mL
M . mL = mmol
We replace: 8.3 mL . 9.9 M = 82.17 mmoles
We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles
Ratio is 4:2
4 moles of sulfuric acid can make 2 moles of alum
By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.
We convert moles to mass:
Molar mass of alum is: 473.52 g/mol.
0.041085 moles . 473.52 g/mol = 19.4 g
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
