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amid [387]
3 years ago
12

A buffer contains the weak acid HA and its conjugate base A-. the weak acid has a p Ka of 4.82 and the buffer has a pH of 4.25.

Which statement is true of the relative concentration of the weak acid and its conjugate base in the buffer?
a) [HA} > [A-]
b) [Ha} < [A-]
c) [HA]=[A-]
Chemistry
1 answer:
klasskru [66]3 years ago
3 0
<span>I think the correct answer is A. A buffer is a substance that resists small change in the acidity of a solution when an acid or base is added to the solution. Usually, a buffer involves a weak acid or a weak alkali and one of its salt.</span>
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Solve this problem using the appropriate law. (Remember that ) What is the pressure of 1.9 mols of nitrogen gas in a 9.45 L tank
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PV=nRT

You are funding pressure and you have V = 9.45L, n = 1.9 moles, R = gas constant, and T = 228 K

P(9.45L) = (1.9moles)(0.0821)(228K)

Find P

Multiply

9.45P = 35.57

Divide

P = 3.76 L of N

Answer should be 3.76 liters of nitrogen
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Chemical energy to heat and light

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The element chlorine has two stable isotopes, chlorine-35 with a mass of 34.97 amu and chlorine-37 with a mass of 36.95 amu. Fro
sergejj [24]

<u>Answer: </u>One isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

<u>Explanation:</u>

We are given the two stable isotopes of chlorine with their respective masses. The average atomic mass of chlorine is also given.

Average atomic mass of chlorine = 35.45 amu.

Let us assume the fractional abundance of one isotope be 'x' and the fractional abundance for another isotope will be (1 - x) because the total fractional abundance is always equal to 1.

  • For Chlorine-35 isotope:

Fractional abundance = x

Mass = 34.97 amu

  • For Chlorine-37 isotope:

Fractional abundance = 1 - x

Mass = 36.95 amu

The formula for the calculation of average atomic mass is given by:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Putting values in above equation, we get:

35.45=34.97(x)+36.95(1-x)\\\\35.45=34.97x+36.95-36.95\\\\-1.5=-1.98x\\\\x=\frac{-1.5}{-1.98}=0.7575

1-x=1-0.7575=0.2424

Converting these two fractional abundances into percentage abundances by multiplying it with 100.

x=0.7575\times 100=75.75\%\\\\1-x=0.2424\times 100=24.24\%

Hence, one isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

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