Calculate the mass 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

The mass of calcium carbonate in this sample =  3.74 * 10 ^-3 g

Explanation:

Mass of Calcium carbonate = Density of CaCO3 * Volume of the sample

⇒ CaCO3 has a density of 0.9977 g/mL

⇒ The sample has a volume of 50 mL

Mass = 0.9977 g/mL * 50 mL = 49.885 g

Since it has a hardness of 75 ppm = 75 parts per million = 75 mg/l  = 7.5 * 10 ^-5 g /ml

⇒7.5 * 10^-5  * 48.885 g = 3.74 * 10 ^-3 g

The mass of calcium carbonate in this sample =  3.74 * 10 ^-3 g

The mass of calcium carbonate is 3,75  mg

Explanation:

Remember that when you work with ppm, you have  mass solute / volume solution.

In this case, your volume which is mL has to be in L.

50 mL = 0,05L

Ppm = mg / kg  or  mg/L

75 mg/L = mass solute / 0,05L

75 mg/L  x 0,05L = 3,75 mg

the answer is yes

it has kinetic energy

reason

an object has kinetic energy when you stop the force carrying it, the object will stay in motion.

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