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AleksandrR [38]
3 years ago
6

Which of the following is a true statement?

Chemistry
1 answer:
maria [59]3 years ago
3 0

Answer:

the second one seems right :)

Explanation:

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The atomic number of potassium is 19. How many electrons does this element have?
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A buffer contains equal amounts of a weak acid and its conjugate base and has a ph of 5.25. part a which would be a reasonable v
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Hello!

After the addition of a small amount of acid, a reasonable value of buffer pH would be 5,15.

If initially there are equal amounts of a weak acid and its conjugate base, the pH would be equal to the pKa, according to the Henderson-Hasselbach equation:

pH=pKa+log( \frac{[A^{-}] }{[HA]} ) \\ \\ if [A^{-}]=[HA] \\ \\ pH=pKa + log (1) \\ \\ pH=pKa + 0=5,25

So, when adding a little amount of acid the pH should be only a little lower than the pKa. The value from the list that is a little lower than the pKa is 5.15

Have a nice day!
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Read 2 more answers
The activation energy (E*) for 2N2O ---&gt; 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be
Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature T_1 = 1001 K

Final temperature T_2 = 298 K

Applying the equation of Arrhenius theory.

In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})

where ;

R gas constant = 8.314  J/K/mol

In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})

In \dfrac{k_2}{0.380 }= -70.8655

\dfrac{k_2}{0.380 }= e^{-70.8655}

\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}

{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }

{k_2}= 6.3575 \times 10^{-32}  /M .sec

Half life:

At 1001 K.

t_{1/2} = \dfrac{In_2}{k_1}

t_{1/2} = \dfrac{0.693}{0.38}

t_{1/2} = 1.82368 secc

At 298 K:

t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}

t_{1/2} =1.09 \times 10^{31} \ sec

3 0
3 years ago
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