Answer:
Magnesium reacts in strong and weak acids.
Explanation:
Acid solutions also, have pH values, less than 7. This is the correct answer, to this question.
Hope this helps!!!
Kyle.
Answer:
We'll have 1.00 mol H2
Explanation:
Step 1: Data given
Number of moles of Mg = 1.00 moles
Step 2: The balanced equation
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
Step 3: Calculate moles of H2
For 1 mol Mg we need 1 mol H2SO4 to produce 1 mol MgSO4 and 1 mol H2
We'll have 1.00 mol H2
It's called ”wire gauze”, a metal mesh used with a ring clamp to support glassware over a Bunsen burner. Spreads flame out for more even heating rates.
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
