Answer:
First one is 5.0 M ammonia and the Second one ?
Explanation:
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
When the amount of heat gained = the amount of heat loss
so, M*C*ΔTloses = M*C* ΔT gained
when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.
∴( M * C * ΔT )W = (M*C*ΔT) Al
when Mw is the mass of water = 100 g
and C the specific heat capacity of water = 4.18
and ΔT the change in temperature for water= 28-25 = 3 ° C
and ΔT the change in temperature for Al = 100-28= 72°C
and M Al is the mass of Al block
C is the specific heat capacity of the block = 0.9
so by substitution:
100 g * 4.18*3 = M Al * 0.9*72
∴ the mass of Al block is = 100 g *4.18 / 0.9*72
= 19.35 g
It will form a tetraamminezinc sulphate and a blue solution will be formed...the reaction should be carried out in a fume chamber because a poisonous substance will be released to the atmosphere
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
