Balanced chemical equation:
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O
2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2
moles C2H2 = 84 * 2 / 5
molesC2H2 = 168 / 5 => 33.6 moles of C2H2
Answer:
4 significant digits
Explanation:
Zeroes placed between other digits are always significant
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
The new volume is 2.4 L
<u><em> calculation</em></u>
The new volume is calculated using Boyle's law formula
That is P₁V₁=P₂V₂
where,
p₁ = 802 torr
V₁ = 12.0 l
P₂ = 4010 torr
V₂=?
make V₂ the subject of the formula by diving both side of equation by P₂
V₂ = P₁V₁/P₂
V₂ is therefore = {(12.0 l × 802 torr) 4010 torr} = 2.4 L
A chemical bond is the best classification.
