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NISA [10]
4 years ago
11

What substance would you add to KF(aq) to form a buffer solution? View Available Hint(s) What substance would you add to to form

a buffer solution? HClO NaF NH3 KOH HF KCl
Chemistry
1 answer:
AnnyKZ [126]4 years ago
4 0
<h3>Answer</h3>

HF

<h3>Explanation</h3>

A buffer solution contains <em>a weak acid</em> and<em> its conjugate base</em>. The two species shall have a similar concentration in the solution. It's also possible for <em>a weak base</em> and <em>its conjugate acid</em> to form a buffer solution.

The KF solution already contains large number of \text{F}^{-} ions. The objective is to thus find its conjugate acid or base.

\text{F}^{-} contains no proton \text{H}^{+} and is unlikely to be a conjugate acid. Assuming that \text{F}^{-} is a conjugate base. Adding one proton to \text{F}^{-} would produce its conjugate acid.

\text{H}^{+} \; (aq) + \text{F}^{-} \; (aq) \rightleftharpoons \text{HF}\; (aq)

Therefore \text{HF} is the conjugate acid of \text{F}^{-}. \text{HF} happens to be a weak acid. As a result, combining \text{F}^{-} with \text{HF} would produce a solution with large number of both the weak acid and its conjugate base, which is a buffer solution by definition.

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6 0
3 years ago
What is the answers and pls show work if possible!!
taurus [48]

D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

Diameter = 348 pm


46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold


4 0
4 years ago
Please help me please ​
sweet-ann [11.9K]

Explanation:

1 sugar solution =Distillation

2 Iron powder and sand=magnetic separation

3 petrol and diesel= Fractional distillation

4 Camphor and glass powder = Sublimation

7 0
3 years ago
What is the molar mad of CO2
stich3 [128]
<u>Molar mass of CO₂:
</u>

mC + 2×mO = 12g + 2×16g = 12g + 32g =<u> 44 g/mol</u>
4 0
3 years ago
[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.1
ankoles [38]

Answer:

See explanation below

Explanation:

I'm assuming this is a table you need to complete, so, you are not putting this in order, but I already found it in another place. let's do a little summary of the expressions we need to use in order to complete the chart.

To calculate pH we need the following expression:

<em>pH = -log[H3O+] (1)</em>

From this expression we can solve for [H3O+] in case we need it:

<em>[H3O+] = 10^(-pH)   (2)</em>

When we have [OH-] we calculate the pOH and from there, the pH:

<em>pOH = -log[OH-]  (3)</em>

and [OH-]:

<em>[OH-] = 10^(-pOH) (4)</em>

Finally to get the pH from pOH:

14 = pH + pOH

<em>pH = 14 - pOH (5)</em>

With these 5 expressions we can complete the chart. In picture 1, you have the actual chart.

To know if it's acidic or basic, that depends on the value of pH.

If pH <7 it's acidic

If pH >7 it's basic

If pH = 7 it's neutral

<u><em>First case:</em></u>

[H3O+] = 3.5x10^-3

pH = -log(3.5x10^-3) = 2.46  (acidic)

pOH = 14 - 2.46 = 11.54

[OH-] = 10^(-11.54) = 2.88x10^-12 M

<u><em>Second case:</em></u>

pOH = -log(3.8x10^-7) = 6.42

pH = 14 - 6.42 = 7.58 (it's basic)

[H3O+] = 10^(-7.58) = 2.63x10^-8 M

<u><em>Third case:</em></u>

pH = -log(1.8x10^-9) = 8.74 (it's basic)

pOH = 14 - 8.74 = 5.26

[OH-] = 10^(-5.26) = 5.5x10^-6 M

<u><em>Fourth case:</em></u>

[H3O+] = 10^(-7.15) =7.08x10^-8 M   (Basic)

pOH = 14 - 7.15 = 6.85

[OH-] = 10^(-6.85) = 1.41x10^-7 M

Hope this can help you

7 0
3 years ago
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