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umka2103 [35]
3 years ago
11

[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.1

5 _____[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.15 _____

Chemistry
1 answer:
ankoles [38]3 years ago
7 0

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

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1) 1.52 atm.

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5) 254.22 K = -18.77 °C.

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1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

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P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

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2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.


3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

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4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

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P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

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P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

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