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Ratling [72]
3 years ago
14

Consider 55 ml of water (h2o) in a beaker and 55 ml of acetone [(ch3)2co] in an identical beaker under identical conditions. com

plete the sentences to explain the relationship between the vapor pressure of water and acetone. match the words in the left column to the appropriate blanks in the sentences on the right.
Chemistry
1 answer:
alisha [4.7K]3 years ago
4 0
I cannot answer this question exactly if the choices are not given. There is obviously more to this question. However, I can still give an idea about the vapor pressure of water and acetone. Acetone is a liquid ketone. It is more volatile compared to water. Therefore, you will expect acetone to have a lower vapor pressure than water. With the same conditions and same amount of initial volume, at a certain amount of time, the amount of liquid left in the container for water than acetone. This is because most of the acetone has already vaporized.
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Read 2 more answers
What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
6 0
3 years ago
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