Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
I don’t know what exactly is but see at the picture.
Answer:
I think that the answer is A.
M = n x Mr
Mr of H20 - 18.006
M = 1.15mol x 18.006g/mol
= 20.7069g
= 20.7g (3sfg)
