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A solution is made by mixing 0.82 grams of sodium acetate, 1 mL of 12 M acetic acid (pKa=4.76) and water to give a final volume

kompoz [17]

Answer:

This solution acts as an efficient buffer

Explanation:

the pH of a buffer solution can be described like this: $pH=pKa+log\frac{[base]}{[acid]}$

[acid]=[acetic acid]= $1mL.\frac{12mol}{1000mL} . \frac{1}{1L} = 0.012M$

[base]=[sodium acetate]= $0.82g.\frac{1mol}{82g} .\frac{1}{1l} = 0.01M$

replacing, $pH=4.76+log\frac{0.01M}{0.012M} =4.84$

If we add an acid, pH will decrease a little bit and if we add a base, pH wil increase a little bit.

lets supose that we change the rate by increasing [base] to 0.1, then

$pH=4,76+log\frac{0.1}{0.012} = 5.68$

and now lets supose that we increase [acid] to 0.1 $pH=4.76+log\frac{0.01}{0.1} = 3.76$

Big changes in concentration of base or acid doesn´t produce big changes in pH, in that way the mix of sodium acetate with acetic acid is a good buffer solution.

4 0

2 years ago

A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the

Gnesinka [82]

Answer:

CaAsHO₄

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

Step 1. Calculate the amount in moles of each element.

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

Step 3. Write the empirical formula with the rounded numbers obtained

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

CaAsHO₄

3 0

3 years ago

BaO2 is a peroxide. Why?

melamori03 [73]

Answer:

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7 0

2 years ago

Read 2 more answers

Compare the life cycles of a flowering and a conifer non-flowering plant.

damaskus [11]

Answer:

The main difference between flowering and nonflowering plants is their method of reproduction. Flowering plants rely on pollination for reproduction, where as nonflowering plants rely on dispersion to continue their life cycle.

3 0

2 years ago

the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107

oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0

3 years ago