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Zepler [3.9K]
3 years ago
6

40 POINTS COLLEGE CHEMISTRY

Chemistry
1 answer:
alexdok [17]3 years ago
5 0

Answer:

C2= 0.16M

Explanation:

C1= 2M, V1= 20ml, C2= ?, V2= 250ml

Applying dilution formula

C1V1= C2V2

2×20 =C2×250

C2= 0.16M

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The mass in grams of 6.0 mol hydrogen gas (contains h2) is
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3 years ago
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At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00
Levart [38]

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Option E is the right answer.

Explanation:

Data given:

Kw = 51.3 x 10^{-14}

pOH = ?

we know that pure water is neutral and will have pH pf 7.

The equation for relation between Kw and H+ and OH- ion is given by:

Kw = [H+] [OH-}

here the concentration of H+ ion and OH- ion is equal

so, [H+]= [OH-]

Putting the values in the equation of Kw

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Now, pOH is calculated by using the equation:

14 = pOH + pH

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8 0
3 years ago
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