Answer:
It is referred to as b) premises liability
<span>M(NO</span>₃<span>)</span>₂<span> fully separates into M</span>²⁺<span> and NO</span>₃<span> </span>²⁻<span> </span><span>and M(OH)</span>₂<span> partially separates
as <span>M</span></span>²⁺<span><span> and 2OH</span></span>⁻
<span>M(NO</span>₃<span>)</span>₂<span><span> </span>→
M</span>²⁺<span> + 2NO</span>₃²⁻
<span>0.202 M 0.202 M</span>
<span> M(OH)</span>₂<span>(s) ↔ <span>M</span></span>²⁺<span><span> (aq) + 2OH</span></span>⁻<span><span>(aq)</span></span>
<span>I - -</span>
<span>C -X +X +2X</span>
<span>E X 2X</span>
<span>Ksp = [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²
4.45 * 10∧-12 = (0.202
+ X ) (2X)²
Since X is very small, (0.202 + X ) = 0.202
<span>4.45 * 10<span>-12 </span>= 0.202 *
4X</span>²
<span> X = 2.347 </span>× 10∧-6 M
Hence
the solubility of <span>M(OH)2
is 2.347 </span>× 10∧-6 M
<span>The isotopes of bromine are Bromine-79 and Bromine-81. Bromine-79 has 35 protons and 44 neutrons while bromine-81 has 35 protons and 46 neutrons. They are represented as Br-79 and Br-81.</span>
If I still remember how to solve these.. I think it’s 50? (50 m/s)
Answer: -
4.25 mol of O₂ left as excess.
3.5 mol of NO₂ formed.
Explanation: -
Number of moles of NO taken = 3.5
Number of moles of O₂ taken = 6.0
The balanced chemical equation for this reaction is
2 NO+ O₂ → 2 NO₂
From the equation we can see that
2 mol of NO react with 1 mol of O₂
3.5 mol of NO react with 
x 3.5 mol NO
= 1.75 mol O₂
So Oxygen O₂ is in excess and NO is the limiting reagent.
Moles of O₂ left over = 6 - 1.75 =4.25 mol of O₂
From the balanced chemical equation we see
2 mol of NO gives 2 mol of NO₂
3.5 mol of NO gives
x 3.5 mol NO
= 3.5 mol of NO₂
