Answer: The concentration of excess ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of 
give = 1 mole of 
Thus 0.019 moles of give = 0.019 mole of
2. moles of 
According to stoichiometry:
1 mole of 
gives = 2 moles of 
Thus 0.012 moles of give = 
moles of
As 1 mole of neutralize 1 mole of
0.019 mole of will neutralize 0.019 mole of
Thus (0.024-0.019)= 0.005 moles of will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of in solution is 0.017 M.
Letter d, because they are both alkali metals (group one)
Oxygen 4. The fatty acid chain of the lipids is often referred to as a hydrocarbon chain.
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
