Answer:
CuCl2-Ion-dipole forces
CuSO4-Ion-dipole forces
NH3-Dipole-dipole forces
CH3OH-Dipole-dipole forces
Explanation:
Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.
Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.
Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles
Answer:

Explanation:
Given that,
Force applied to a soccer player, F = 56.6 N
The mass of the ball, m = 0.46 kg
We need to find the acceleration of the soccer ball. The force acting on the ball is given by :
F = ma
Where
a is the acceleration

So, the required solution is
.
Answer:
Boron and Aluminium
Explanation:
Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
- CH3COOH + NaOH ↔ CH3COONa + H2O
- CH3COONa + NaOH ↔ CH3COONa
∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ <em>C</em> CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ <em>C</em> CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7