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2 years ago

How many grams of NH3 are needed to provide the same number of molecules as in 0.85 grams of SF6?

Chemistry

2 answers:

Akimi4 [234]2 years ago

7 0

Answer:

0.099 grams of ammonia are needed to provide the same number of molecules as in 0.85 grams of sulfur hexafluoride.

Explanation:

Mass of sulfur hexafluoride = 0.86 g

Moles of sulfur hexafluoride = $\frac{ 0.86 g}{146 g/mol}=0.005822 mol$

1 mole = $N_A=6.022\times 10^{23} molecules/ atom$

This means 0.005822 moles of ammonia will have same number of molecules as that of the molecules in 0.005822 moles of sulfur hexafluoride .

Moles of ammonia = 0.005822 mole

Mass of 0.005822 moles of ammonia = $0.005822 mol\times 17 g/mol=0.099 g$

0.099 grams of ammonia are needed to provide the same number of molecules as in 0.85 grams of sulfur hexafluoride.

solniwko [45]2 years ago

3 0

Above it says the molecular weights are

NH3- 17g/mol and SF6-146 g/mol

Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.

= 1mole / 146.048g * 0.85g

so we would need 0.00582000438 moles of NH3 to have the same number of molecules.

One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:

= 17.030519999989988 g / mole * 0.00582000438moles

that equals 0.09911770099 grams.

so 0.09911770099 grams is the answer if you round that you get about 0.1 grams

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1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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Calculate the molarity of the solution formed when 0.72 moles of CaBr2 is dissolved in 1.50 L water.

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Answer:

The molarity of the formed CaBr2 solution is 0.48 M

Explanation:

Step 1: Data given

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Volume of water = 1.50 L

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Molarity of CaBr2 solution = moles CaBr2 / volume water

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The molarity of the formed CaBr2 solution is 0.48 M

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