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3 years ago

Of the following solutions, which has the greatest buffering capacity?

Chemistry

1 answer:

Goshia [24]3 years ago

4 0

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[Base]}{[Acid]} )$

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. $\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36$

b. $\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417$

c. $\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868$

d. $\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959$

Therefore, the d. solution has the best buffering capacity.

Regards.

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The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.145 M solution of dichromate to reach the equivalence p

Svetlanka [38]

Answer:

0.64 M

Explanation:

Given:

Volume of iron(II) solution (V₁) = 25.0 mL = 0.025 L

Molarity of iron(II) solution (M₁) = ?

Number of moles of iron(II) solution (n₁) = ?

Volume of dichromate solution (V₂) = 18.0 mL = 0.018 L

Molarity of dichromate solution (M₂) = 0.145 M

Number of moles of dichromate solution (n₂) = ?

Molarity is equal to the ratio of moles and volume.

So, molarity of dichromate solution is given as:

$M_2=\frac{n_2}{V_2}\\\\n_2=M_2\times V_2=0.145\times 0.018 = 2.61\times 10^{-3}\ mol$

Now, let us write the complete balanced reaction for the given situation.

So, the complete balanced equation is given below.

$6Fe^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq)\to 6Fe^{3+}(aq)+2Cr^{3+}(aq)+7H_2O$

From the equation, it is clear that, 1 mole of dichromate is required for 6 moles of iron(II) solution.

So, using unitary method, we find the number of moles of iron(II) solution.

1 mole of dichromate = 6 moles of iron(II)

∴ n₂ moles of dichromate = 6n₂ moles of iron(II)

= $6\times 2.61\times 10^{-3}=0.016\ mol\ Fe^{2+}$

So, 0.016 moles of iron(II) is needed. Therefore, $n_1=0.016\ mol$

Now, molarity of iron(II) solution is given as:

Molarity = Moles ÷ Volume

$M_1=\frac{n_1}{V_1}\\\\M_1=\frac{0.016\ mol}{0.025\ L}=0.64\ M$

Therefore, the molarity of the iron(II) solution is 0.64 M.

4 0

3 years ago

Please show all of your work! :)

Paladinen [302]

Answer:

Explanation:

To answer this, we need to use Gay-Lussac's law, which states that:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$ , where P is pressure and T is temperature

The initial pressure we're given is 4.5 atm (so P1 = 4.5) and the temperature is 45.0°C; however, we need to change Celsius to Kelvins, so add 273 to 45.0: 45.0 + 273 = 318 K (so T1 = 318).

The final pressure is what we want to find, but we do know the final temperature is 3.1°C. Converting this to Kelvins, we get: 3.1 + 273 = 276.1 K, which means T2 = 276.1.

Plug these values in:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$

$\frac{4.5}{318}= \frac{P_2}{276.1}$

Multiply both sides by 276.1:

$P_2$ ≈ 3.9 atm

The answer is thus A.

3 0

3 years ago

Please Help, will give 30 points! The following data was collected when a reaction was performed experimentally in the laborator

Sedaia [141]

Answer:

9 moles of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaCl —> 3NaNO3 + AlCl3

Next, we shall determine the number of mole of Al(NO3)3 and NaCl that reacted and the number of mole of NaNO3 produced from the balanced equation.

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl to produce 3 moles of NaNO3.

Next, we shall determine the limiting reactant.

The limiting reactant can be obtained as follow:

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl.

Therefore, 4 moles of Al(NO3)3 will react with = (4 x 3)/1 = 12 moles of NaCl.

From the calculations made above, we can see that it will take a higher amount i.e 12 moles than what was given i.e 9 moles of NaCl to react completely with 4 moles of Al(NO3)3.

Therefore, NaCl is the limiting reactant and Al(NO3)3 is the excess reactant.

Finally, we shall determine the maximum amount of NaNO3 produced from the reaction.

In this case, the limiting reactant will be used as it will produce the maximum amount of NaNO3 since all of it is consumed by the reaction.

The limit reactant is NaCl and the maximum amount of NaNO3 produced can be obtained as follow:

From the balanced equation above,

3 moles of NaCl reacted to produce 3 moles of NaNO3.

Therefore, 9 moles of NaCl will also react to produce 9 moles of NaNO3.

From the calculations made above, the maximum amount of NaNO3 produced is 9 moles

6 0

3 years ago

Which of the following molecules is nonpolar?

Rudik [331]

Answer:

1. BF3 This is a trigonal planar molecule; the electron density is drawn into a cloud that circles the Boron, this is made nonpolar by the geometrically equivalent structure of the surrounding electronegative Fluorines.

2. H2O The 2 lone pairs of e- of Oxygen makes the O partially negative, the H’s, partially positive. Polar.

3. NF3 Lone pair on Nitrogen overwhelmed by the 3 incredibly electronegative Fluorines. Polar

4. CH3Br The “Soft Ion” of Bromine is negative; it is electronegative. Polar.

5. SO2 the lone pairs of Oxygen, at approximately 119°-120° angles to one another will form a reasonance structure; there will be more lone pairs about the Oxygen than the Sulfur; the Sulfur will be partially positive compared to the oxygens. Polar.

7 0

4 years ago

Can someone help me!

Art [367]

Answer:

Explanation:

5 0

3 years ago

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