The $ -(Br) at that instant is calculated as follows
BrO3 + 5Br ---> 3Br2 + 3 H2o
by use of reacting ratio BrO3 to Br2 which is 1:3
therefore $ of Br = 3 x (1.5 x10^-2)= 4.5 x10^-2
Answer:
51.3% Ca 48.7% F
Explanation:
MM CaF2 = 40.1+2x19 = 78.1 g/mol
MM Ca = 40.1 g/mol
MM F = 19 g/mol
% Ca = (40.1/78.1) x 100 = 51.3% Ca
% F = 100-51.3 = 48.7% F
(a)
Write balanced half-reactions for the process:
Oxidation: Se^2- (aq) → Se
(s) + 2e-
Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- →
S2O3^2- + 6OH- (aq)
(b)
If E sulfite is 0.57 V, calculate E selenium:
E anode = E cathode – E cell
= -0.57 – 0.35
=
-.092
Answer:
D
Explanation:
I think it is d sorry if wrong hoped I helped