Answer:
D) Oxygen is oxidized and hydrogen is reduced.
Explanation:
In the electrolysis of water, an electric current passes through an electrolytic solution (e.g. aqueous NaCl), leading to the following redox reaction.
H₂O(l) → H₂(g) + 1/2 O₂(g)
The corresponding half-reactions are:
Reduction: 2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻
Oxidation: 2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻
As we can see, H in water is reduced (its oxidation number decreases from 1 to 0), while O in water is oxidized (its oxidation number increases from -2 to 0).
Answer:
Explanation:
First we look generally at what makes K2SO4.
In one mole of K2SO4, there are 2 moles of the potassium ion (K+) and 1 mole of sulfate ion (SO4 2-).
Knowing that; in 1.75 moles of K2SO4, there must be 2 x 1.75moles of potassium ion (K+) and 1 x 1.75moles of Sulfate ion (SO4 2-)
This gives us 3.5moles of K+ and 1.75moles of SO4 2-
Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
Answer:d
Explanation:I’m not 100% sure how to explain it but I’m almost for sure it’s d.
