Is it asking for multiple answers?
If not the nucleus is one organelle found in the <span>eukaryotic.
Hope This Helps :)</span>
Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
Answer:- C. H
Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.
As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.
The oxidation number in elemental form is zero.
In 
, the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in 
is -1. On product side, the oxidation number of hydrogen in 
is zero and in 
the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in 
is 0.
From above data, Oxidation number of O is -2 on both sides so it is not reduced.
Oxidation number of Cl is changing from -1 to 0 which is oxidation.
Oxidation number of H is changing from +1 to 0 which is reduction.
So, the right choice is C.H
Answer:
I believe it is B,fluorine to complete the octet rule
Explanation:
