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2 years ago

When water is added to a mixture Na_2O_2(s) and S(s), a redox reaction occurs, as represented by the equation below.

Chemistry

1 answer:

BARSIC [14]2 years ago

8 0

Answer:

Explanation:

Let's consider the following balanced equation.

2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂

To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).

If ΔG° < 0, the reaction will be favorable.

If ΔG° > 0, the reaction will be unfavorable.

We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T.ΔS°

As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.

ΔG° = ΔH° - T.ΔS°

ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol

The reaction is favorable and is driven by ΔH° only.

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A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$

So we solve for C2:

$C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M$

2. Now, since 111 mL of water is added, we compute the final volume, V3:

$V_3=139+111=250mL$

So, the final concentration of the 139 mL portion is:

$C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M$

Best regards!

8 0

2 years ago

I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.

Alex_Xolod [135]

Answer:

175

Explanation:

3 0

2 years ago

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 15 ∘C.

umka2103 [35]

Answer:

P(oxygen gas) = 3.75atm

P(helium) = 19.2atm

Ptotal = 22.95atm

Explanation:

8 0

2 years ago

A buffer solution is composed of 1.00 mol of acid and 2.25 mol of the conjugate base. If the p K a of the acid is 4.90 , what is

Gemiola [76]

<u>Answer:</u> The pH of the buffer is 5.25

<u>Explanation:</u>

Let the volume of buffer solution be V

We know that:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.90

$[\text{conjugate base}]=\frac{2.25}{V}$

$[acid]=\frac{1.00}{V}$

pH = ?

Putting values in above equation, we get:

$pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25$

Hence, the pH of the buffer is 5.25

4 0

2 years ago

The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate

Vedmedyk [2.9K]

Answer:

Explanation:

The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.

This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

thus

0.1 ≤ [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

Now we are equipped to answer our question:

pH range = 3.9 +/- 1 = 2.9 through 4.9

7 0

3 years ago