What is the oxidation number of s in na2s4o6?
2 answers:
Answer:
+2.5
Explanation:
So the oxidation number of S in Na2S4O6is +2.5.
Answer:
Explanation:
Oxidation number of Na = 1+ There are two of them, so it is 2*1 = 2
Oxidation number of O = 2- There are 6 of them so 6 * 2- = -12
There are 4 S's so the oxidation number is 4*S
Now set up a small equation
2 + 4s - 12 = 0
And solve.
2 - 12 + 4s =0
-10 + 4s = 0
4s = 10
4s/4 =10/4
s = 2.5
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Answer:
The number of OH group found in found in a 6 carbon alditol is 6
Explanation:
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Answer:
I hope this link helps you.
Explanation:
http://astronomy.swin.edu.au/cosmos/P/Phases
Answer:
A) pH = 2.8
B) pH = 5.5
C) pH = 8.9
D) pH = 13.72
Explanation:
a) [H⁺] = 1.6 × 10⁻³ M
pH = -log [H⁺]
pH = -log [1.6 × 10⁻³ ]
pH = 2.8
b) [H⁺] = 3 × 10⁻⁶
pH = -log [H⁺]
pH = -log [3 × 10⁻⁶ ]
pH = 5.5
c) [OH⁻] = 8.2 × 10⁻⁶
pOH = -log[OH]
pOH = -log[8.2 × 10⁻⁶]
pOH = 5.1
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 5.1
pH = 8.9
d) [OH⁻] = 0.53 M
pOH = -log[OH]
pOH = -log[0.53]
pOH = 0.28
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.28
pH = 13.72
