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hjlf
3 years ago
12

For problems 2 & 3, the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. 2. Calc

ulatethemass(ingrams)ofcalciumcarbonatepresentina50.00mLsampleof an aqueous calcium carbonate standard, assuming the standard is known to have a hardness of 75.0 ppm (hardness due to CaCO3).
Chemistry
1 answer:
Alenkasestr [34]3 years ago
4 0
1) Use of density formula to calculate mass of sample, M

D = M / V => M = D* V = 0.9977 g / mL * 50.00 mL = 49.885 g of sample

Realize that the sample is the solution.

2) Use ppm concentration to calculate mass of solute (calcium carbonate)

By definition, 75.0 ppm = 75.0 g of calcium carbonate / 1,000,000 g of solution.

You use that ratio to calculate the mass of calcium carbonate in 49.885 g of solution.

Mass of solute = 49.885 g of solution * [75.0 g of calcium carbonate] / [1,000,000 g of solution] =

Mass of solute =  0.00374 g of calcium carbonate

Answer: 0.00374 g of calcium carbonate
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What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?
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Pressure = 4313.43mmHg

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What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas
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Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

NH_3+HCl\rightarrow NH_4Cl

First we have to calculate the moles of NH_3 and HCl

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

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and,

\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}

Molar mass of HCl = 36.5 g/mole

\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NH_3

So, 2.07 mole of HCl react with 2.07 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

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P = Pressure of gas = 752 mmHg = 0.989 atm     (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 14.0^oC=273+14.0=287K

Putting values in above equation, we get:

0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K

V = 56.5 L

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Now we have to calculate the mass of NH_4Cl

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