6840 is the answer hope this help
At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D
The concentration of hydroxide ion is 5 
10^
−
14 M.
<u>Explanation:</u>
Consider the equilibrium of this acid's dissociation,
H
C
l
O
4 ⇌ H
+ + C
l
O 4
-
Moreover, let's assume that H
C
l
O
4 is a strong acid and will fully dissociate.
Hence,
[
H
+
] = 0.20 M
Now, recall,
K
w = [
H
+
]
[
O
H
−
] = 1.0 10
^−
14
Hence,
⇒
[
O
H
−
] = K
w / [
H
+
] = 5 10^
−
14 M.
Answer: 75%
Explanation:
The following information can be gotten from the question:
Waste = 70kg
Theoretical yield = 280kg
Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:
= 280kg - 70kg = 210kg
The percent yield will now be:
= Actual yield / Theoretical yield × 100
= 210/280 × 100
= 3/4 × 100
= 75%
