Answer:
Shifts the equilibrium to the left. reduces solubility.
Explanation:
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S S 2S
∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²
⇒ 4S² * S = 6.4 E-9
⇒ 4S³ = 6.4 E-9
⇒ S³ = 1.6 E-9
⇒ S = 1.1696 E-3 M
- NaF(s) → Na+(aq) + F-(aq)
0.10M 0.10M 0.10M
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S' S' 2S' + 0.10
⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²
If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:
⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'
⇒ S' = 6.4 E-7 M
∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption
We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).
Answer:
Molecular formula → PbSO₄ → Lead sulfate
Option c.
Explanation:
The % percent composition indicates that in 100 g of compound we have:
68.3 g of Pb, 10.6 g of S and (100 - 68.3 - 10.6) = 21.1 g of O
We divide each element by the molar mass:
68.3 g Pb / 207.2 g/mol = 0.329 moles Pb
10.6 g S / 32.06 g/mol = 0.331 moles S
21.1 g O / 16 g/mol = 1.32 moles O
We divide each mol by the lowest value to determine, the molecular formula
0.329 / 0.329 = 1 Pb
0.331 / 0.329 = 1 S
1.32 / 0.329 = 4 O
Molecular formula → PbSO₄ → Lead sulfate
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
Answer:
1 M
Explanation:
Magnesium chloride will furnish chloride ions as:
Given :
Moles of magnesium chloride = 0.20 mol
Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:
Moles of chloride ions by magnesium chloride = 0.40 moles
Potassium chloride will furnish chloride ions as:
Given :
Moles of potassium chloride = 0.10 moles
Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:
Moles of chloride ions by potassium chloride = 0.10 moles
Total moles = 0.40 + 0.10 moles = 0.50 moles
Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)
Concentration of chloride ions is:
<u>
The final concentration of chloride anion = 1 M</u>
Answer:
35 KJ.
Explanation:
The activation energy is the minimum energy that must be overcome for a reaction to take place.
In the diagram given above, the activation energy lies between the energy of the reactants and that at the peak.
Thus we can calculate the activation energy as follow:
Energy of reactants = 30 KJ
Energy at the peak = 65 KJ
Activation energy =..?
Activation energy = Energy at the peak – Energy of reactants
Activation energy = 65 – 30
Activation energy = 35 KJ
Therefore, the activation energy of th reaction is 35 KJ