Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Within the options, we have four organic compounds. Let's see what the skeletal structure of the compounds is in order to identify them better:
The first compound CH3OCH3 has two methyl groups linked by a carbon atom, this type of compound is called an Ether
The second compound has a double bond, it is badly written but it seems that is an alkene.
The third compound has two methyl groups linked by nitrogen atoms, therefore will be an amine.
The last compound has a hydroxyl group, therefore it is an alcohol
Answer:
CH3OCH3 Ether
CH2CH2CHCH2CH3 Alkene
CH3NHCH3 Amine
CH3CH(OH)CH₂CH3 Alcohol
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
