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alexdok [17]
3 years ago
5

A hot air balloon starts with its temperature at 68.7°C and a pressure of 0.987 ATM and volume of 564L at what temperature in de

grees Celsius while its pressure be 0.852 ATM and its volume be 625L
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂

1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C

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What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
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The first dissociation for H2X:
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initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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