1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
= 32.026804 calories
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
First you calculate the concentration of [OH⁻] in <span>solution :
POH = - log [ OH</span>⁻]
POH = - log [ 0.027 ]
POH = 1.56
PH + POH = 14
PH + 1.56 = 14
PH = 14 - 1.56
PH = 12.44
hope this helps!
