The vapor pressure of water at 20.0°C is . If the pressure of a gas collected ov

Question

The vapor pressure of water at 20.0°C is . If the pressure of a gas collected over water was measured to be . What is the pressure of the pure gas? A. 0.619 atm B. 0.0230 atm C. 0.596 atm D. 0.573 atm E. none of the above 6.0L of a gas having pressure of 650mmHg will have a volume of ________ when it's pressure is 1.1atm. A. 4.7L B. 0.010L C. 3.5L 250 ml of O2 is collected over water at 1.20 atm and 50oC. Calculate the volume of O2(dry) at STP. A. 247 mL B. 200 mL C. 228 mL For: 2Al(s) + 3Ca2+(aq) 2Al3+(aq) + 3Ca(s) ; [Ca2+] = 0.1 and [Al3+] = 0.010 ; calculate Ecell A. -1.22 B. -1.20 C. -1.21 Any help would be appreciated. I have posted a couple of these before with incorrect responses. I promise to rate. thank you sooo much

Explanation / Answer

Part b

1 atm = 760 Mm of hg

since P1*V1 = P2V2 (boyles law at const pressure)

hence 650*6 = 1.1 *760 V2

soo V2 = 4.66L or 4.7 L apprx

part C

using ideal gas equation relation

P1*V1/T1 = P2V2/T2

STP   means P = 1atm and T = 273K   

50 Degree means 50 + 273 = 323K

so   using above relation

1.2*250/323 = 1*V2/273

Hence V2 = 253.5mlml

part d

EoCa2+/ca = -2.87V and EoAl+/Al = -1.66VV

number of electron involve = 2*3 = 6

ca2+ beomes ca hence reduced by accepting electrons hence act as cathode and Al is anodeanode

Eo = Eocathode - Eoanode

hence Eo = -2.87-(-1.66)

Eo = -1.21V

using nernst equation

Ecell = Eocell - 0.059/nlog([Al3+]2/[ca2+]3

Ecell = -1.21 -0.059/6log([O.0001]/[0.001]

Ecell = -1.21 + 0.01

Ecell = -1.20 V

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