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attashe74 [19]
3 years ago
7

If i have 4 moles of a gas ata pressure of 5.2 atm and a volume of 15 liters, what is the temperature?

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

Answer:

237.5 K.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT. </em>

where, P is the pressure of the gas in atm (P = 5.2 atm).  

V is the volume of the gas in L (V = 15.0 L).  

n is the no. of moles of the gas in mol (n = 4.0 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),  

T is the temperature of the gas in K (T = ??? K).

∴ T = PV/nR = (5.2 atm)(15.0 L)/(4.0 mol)(0.0821 L.atm/mol.K) = 237.5 K.

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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
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
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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