<u>Given:</u>
Initial volume of He, V1 = 19.2 L
Initial mass of He, m1 = 0.0860 g
Mass of He removed = 0.205 g
<u>To determine:</u>
The new volume of He i.e V2
<u>Explanation:</u>
Based on Avogadro's law:
Volume of a gas is directly proportional to the # moles of the gas
Volume (V) α moles (n) -----(1)
Atomic mass of He = 4 g/mol
Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles
Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles
Based on eq(1) we have:
V1/V2 = n1/n2
V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L
Ans: New volume is 14.6 L
The equation is already balanced so the mole ratio is just 1/1 and from the question you could conclude that the equation is a single displacement
<span>When atoms lose or gain electrons in chemical reactions they form?
</span>Ions
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.
