Calculate the mass (in grams) of calcium carbonate present in a 50.00 ml sample of an aqueous calcium carbonate standard, assuming the standard is known to have a hardness of 75.0 ppm (hardness due to caco3).

1 part per million is the same as 

1 mcg/mL (mcg = 1 micro gram = 1 * 10^-6 grams). 
125 ppm = 125 mcg/mL 

125 mcg/ml * [1 gram / 1 * 10^6 mgrm] = 1.25 * 10^-4 grams 

The density of hard water is 1.000 grams / mL 

50 mL of water contains 1.25 * 10^-4 grams. 

given mass = 1.25 * 10^-4 
Molar mass = 100.0 grams / mole 
n =  

n = 1.25 * 10^-4/100 = 1.25 * 10^-6 moles. 

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Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.

Ci think not