Gives away hydrogen ion Is an acid...while the one that accepts is a base
<span>Molten barium
chloride is separetes:</span><span>
BaCl</span>₂(l) →
Ba(l) + Cl₂(g), <span>
but first ionic bonds in this salt are separeted
because of heat:
BaCl</span>₂(l) →
Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction
at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation
at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive
and the cathode is negative.
We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound
C H O
mass 77.87 g 11.76 g <span>10.37 g
number of moles 77.87/12 11.76/1 10.37/16
moles = 6.48 = 11.76 =0.648
divide by least number of moles
6.48/0.648 11.76/0.648 0.648/0.648
= 10 =18.1 = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C</span>₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154
number of empirical units = molecular mass / mass of one empirical unit
= 154.25 / 154 = 1.00
Therefore molecular formula = C₁₀H₁₈O
Answer:
C
Explanation:
The oxidation number of Sulphur in SO4^2- is;
x + 4(-2) = -2
x - 8 = -2
x = -2 + 8
x = 6
Now,
the oxidation number of sulphur in H2SO3 is
2 (1) + x + 3(-2) = 0
2 + x -6 = 0
-4 + x = 0
x = 4
Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.
I think it’s a, it has to be
