Part A 0.59 g/(100 mL) The Henry\'s law equation, Ch/P-lP, allows you to calcula

Question

Part A 0.59 g/(100 mL) The Henry's law equation, Ch/P-lP, allows you to calculate the solubility when you are given values for throe of the four known variables, Ch. F GM-G/BC i te sohtleyofco, i water, Gis the solblity ofco, i asofdrink Riste pre re of CO2 water, andisthe pre sure of the O,inas think The known quantities are Cs,P. and P First restate the Henry's law equation to isolate the unknown variable, C You are given C=0.15 g/(100 mL). n=700 mmHg and-3.9atm Since n andare i, to dleent units, use the equality 760 mrmHg =iatm to convert them G-Kaisd/(100 mil 19 Ar-(0.50 g)/(100 mL) Part B Part c

Explanation / Answer

We have henry's equation as:

C1/P1 = C2/P2

C1/P1 = 0.59g/100mL/3.9atm = C2/P2 = C2/0.00038 atm

C2 =  0.59g/100mL/3.9atm x 0.00038 atm = 0.0008744g/100 mL

% of C1 = C2/C1 x 100 = (0.0008744g/100 mL) / (0.59g/100mL ) x 100 = 0.1482 %

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