Answer:
For the production of 77.4 L water 90.3 L oxygen is required.
Explanation:
Given data:
Volume of oxygen required = ?
Volume of water produced = 77.4 L
Solution:
Chemical reaction equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
1 mole = 22.414 L
There are 6 moles of water = 6×22.414 = 134.5 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
Now we will compare the litters of water and oxygen:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
5 types of air pollution would include: Carbon Monoxide, Lead, Nitrogen oxides, sulfur oxides. I only know those 4 actually… hope it helped
Answer:
It becomes insoluble again.
Explanation:
The chemical equation showing the reaction between benzoic acid, and sodium hydroxide,NaOH is given below;
C6H5COOH + NaOH ------------------------> C6H5COO^- Na^+ + H2O.
The Benzoic acid, C6H5COOH is insoluble but when it reacted with 1M Sodium hydroxide, NaOH it changes to sodium benzoate, C6H5COO^- Na^+ which is more soluble than the benzoic acid.
C6H5COO^- Na^+ + HCl -------------------> C6H5COOH + NaCl.
When the sodium benzoate, C6H5COO^- Na^+ reacts with 6M HCl, it is converted back to benzoic acid which is insoluble. Hence, precipitation is observed.
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:
pH=pKa+log[base]/[acid]
When the the concentrations of both buffer components (the weak acid and its conjugate base) are equal:
[base] = [acid]
So, pH=pKa+log1 = pKa
<u>pH is equal to pKa of weak acid of buffer system
.</u>
When buffer contains more of weak acid than conjugate base:
[base] < [acid]
log [base]/[acid] = Negative,
So,
<u>When more of acid component is present, the pH is more acidic. (It decreases)</u>
When buffer contains more of conjugate base than weak acid:
[base] > [acid]
log [base]/[acid] = Positive,
So,
<u>When more of acid component is present, the pH is more acidic. (It increases)</u>
