Answer:
-1
Explanation:
According to this question, the oxidation state/number of H and O in C2H4O is +1 and -2 respectively.
The oxidation state of carbon in the compound can be calculated thus:
Where;
x represents the oxidation number of C
C2H4O = 0 (net charge)
x(2) + 1(4) - 2 = 0
2x + 4 - 2 = 0
2x + 2 = 0
2x = -2
Divide both sides by 2
x = -1
The oxidation number of C in C2H4O is -1.
The answer to your question would be C. Also you put two "B"'s XD
This can be done with some simple dimensional analysis. First you convert your gram of gold into moles of gold There are 196.97 grams of gold for every one mole of gold so you divide 4.27 by 196.67 to get 0.0217 moles of gold. You then multiply the moles of gold by Avogadro's number, 6.02 X 10^23. Avogadro's number indicates the number of atoms in one mole. Therefore, there are 1.307 X 10^22 atoms in 4.27 grams of gold.
Answer:
Answer choice A
Explanation:
When an electron is transferred to another atom, both atoms involved become ions.
[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
