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SashulF [63]
2 years ago
13

Question 1 (1 point) Saved Please choose all that describe a series circuit. (If you did not attend live session and take notes,

make sure you watch the recording before taking this assessment.) If there are many bulbs in a circuit with a battery (cell), it is very likely that the light will be brighter with each additional bulb. The electrons have multiple pathways to travel. The amount of current is the same at every point in a series circuit. A series circuit has no resistors. All of the parts of a series circuit-power source, wires, and devices-are connected along the same pathway​
Chemistry
1 answer:
diamong [38]2 years ago
8 0

Answer: All of the parts of a series circuit-power source, wires, and devices-are connected along the same pathway​,  All of the parts of a series circuit-power source, wires, and devices-are connected along the same pathway.​ These two are correct.

Explanation:

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Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
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Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

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Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

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                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

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