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stellarik [79]
3 years ago
12

For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.

Chemistry
1 answer:
HACTEHA [7]3 years ago
3 0

Answer :

The concentration of H^+ before any titrant added to our starting material is 0.200 M.

The pH based on this H^+ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of H^+ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion H^+ and bromide ion Br^-.

As, 1 M of HBr dissociates to give 1 M of H^+

So, 0.200 M of HBr dissociates to give 0.200 M of H^+

Thus, the concentration of H^+ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this H^+ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (0.200)

pH=0.698

Thus, the pH based on this H^+ ion concentration is 0.698

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Answer:

The statements that correctly describes pyruvate dehydrogenase includes:

- Several copies each of E 1 and E 3 surround E 2.

-A regulatory kinase and phosphatase are part of the mammalian PDH complex.

-E 2 contains three domains.

Explanation:

Pyruvate dehydrogenase is a hydrolase key enzyme in glucose metabolism which converts pyruvate to acetyl- ChoA. It also forms a complex that catalyzes an irreversible reaction that is the entry point of pyruvate into the TCA cycle. Pyruvate dehydrogenase complex contains E1, E2 and E3 enzymes that transform pyruvate, NAD+, coenzyme A into acetyl-CoA, CO2, and NADH. Also, A regulatory kinase and phosphatase are part of the mammalian PDH complex and E 2 contains three domains.

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3 years ago
Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p
likoan [24]

Answer: 3) 39.96 amu

Explanation:

Mass of isotope Ar- 36 = 35.97 amu

% abundance of isotope Ar- 36= 0.337% = \frac{0.337}{100}=3.37\times 10^{-3}

Mass of isotope Ar- 38 = 37.96 amu

% abundance of isotope 2 = 0.063 % = \frac{0.063}{100}=6.3\times 10^{-4}

Mass of isotope Ar- 40 = 39.96 amu

% abundance of isotope 2 = 99.600 % = \frac{99.600}{100}=0.996

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]

A=39.96amu

Therefore, the average atomic mass of argon is 39.96 amu

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