Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = ![\frac{[H+][IO-]}{[HIO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%2B%5D%5BIO-%5D%7D%7B%5BHIO%5D%7D)
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
Answer:
Mass = 112 g
Explanation:
Given data:
Mass of CO₂ produced = 90.6 g
Mass of oxygen needed = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂+ 4H₂O
Number of moles of CO₂:
Number of moles = 90.6 g/ 44 g/mol
Number of moles = 2.1 mol
Now we will compare the moles of CO₂ and oxygen:
CO₂ : O₂
3 : 5
2.1 : 5/3×2.1 = 3.5
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 3.5 mol × 32 g/mol
Mass = 112 g
Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal
Explanation:
To calculate the moles :
The balanced chemical equuation is:
According to stoichiometry :
4 moles of 
produce == 2 moles of 
Thus 0.556 moles of will produce=
of
Mass of 
Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.
Reactives
-> Products
CuO
and water are products.
I
found this reaction which has CuO and water as products: decomposition of
Cu(OH)2.
Cu(OH)2
-> CuO + H2O
Stoichiometry calculus involve the mole
proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of
CuO and 1 mole of H2O are formed.
Considering
the molar masses:
Cu(OH)2
= 83.56 g/mol
CuO
= 79.545 g/mol
H2O
= 18.015 g/mol
Then:
When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.
You
should use that numbers in the rule of three:
79.545
g CuO __________18.015 g water
3.327
g CuO__________ x =3.327*18.015 /79.545 g water
x= 0.7535 g water
