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Flura [38]
3 years ago
15

(2pts) Post-lab Questions (1pts) 1. Do you expect the solubility of Borax to increase or decrease as temperature increases? Sele

ct the option that best explains why. Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger. Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get smaller. Solubility will decrease, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get smaller. Solubility will decrease, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger. Choose... (1pts) 2. Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why. To make sure no more sodium borate would dissolve in solution. To ensure the dissolution process was at equilibrium. To make sure the solution was saturated with sodium and borate ions. All of the above
Chemistry
1 answer:
saw5 [17]3 years ago
6 0

Answer:

1.  Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.

2. To ensure the dissolution process was at equilibrium.

Explanation:

Given that;

ΔG°= -RTlnK

and

ΔG° = ΔH° - TΔS°

So;

-RTlnK = ΔH° - TΔS°

lnK = ΔH°/-RT - TΔS°/-RT

lnK = -(ΔH°/RT) + ΔS°/R

K = e^-(ΔH°/RT) + ΔS°/R

Hence, Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.

2.

Since solubility is an equilibrium process, it means that some undissolved solute must be present in order to determine the solubility product correctly.

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Over time pollution generally leads to which of the following
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A decrease in population which means animals and plants.

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Q10.An aqueous solution contains 5.00x10-2 mol/L of Ca2+ and 7.00x10-3 mol/L of SO4
romanna [79]

Answer:

The precipitate will form.

Explanation:

Let's write the equilibrium expression for the solubility product of calcium sulfate:

CaSO_4(s) ⇄ Ca^{2+}(aq)+SO_4^{2-}(aq)

The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:

K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}

Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:

Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}

Now given the K value of:

K_{sp}=10^{-4.5}=3.2\cdot10^{-5}

Notice that:

Q_{sp}>K_{sp}

This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.

6 0
4 years ago
Sodium carbonate can be made by heating sodium bicarbonate: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) At 25°C, for this reaction,
insens350 [35]

Answer:

401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.

Explanation:

The expression for the standard change in free energy is:

\Delta G=\Delta H-T\times \Delta S

Where,  

\Delta G is the change in the Gibbs free energy.

T is the absolute temperature. (T in kelvins)

\Delta H is the enthalpy change of the reaction.

\Delta S is the change in entropy.

Given at:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25.0 + 273.15) K = 298.15 K

\Delta H = 128.9 kJ/mol

\Delta G = 33.1 kJ/mol

Applying in the above equation, we get as:-

33.1=128.9-298.15\times \Delta S

-29815\Delta S=-9580

\Delta S=-9580 = 0.32131 kJ/Kmol

So, For reaction to be spontaneous, \Delta G

Thus, For minimum temperature:-

\Delta H-T\times \Delta S=0

128.9-T\times 0.32131=0

T=\frac{128.9}{0.32131}=401.17\ K

<u>Hence, 401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.</u>

6 0
3 years ago
explain observation made when anhydrous calcium chloride and anhydrous copper (ii) sulphate are separately exposed to the atmosp
Burka [1]

Answer:

Anhydrous calcium chloride  dissolves and becomes liquid

Anhydrous copper (ii) sulphate will produce crystal particles

Explanation:

Anhydrous calcium chloride is deliquescent and hence when it is exposed to air, it absorbs water from air. After absorbing water, it dissolves and after some time a pool of clear liquid appears.

Anhydrous copper (ii) sulphate will form crystal structures  and the following reaction will takes place

CuSO4 + 5 H20 --> CuSO4.5H2O

6 0
3 years ago
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