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Sever21 [200]
3 years ago
14

A salt crystal has a mass of 0.11 mg. How many NaCl formula units does it contain? Express your answer using two significant fig

ures.
Chemistry
1 answer:
Ann [662]3 years ago
6 0

<u>Answer:</u> The number of formula units in the given amount of NaCl are 1.1\times 10^{18}

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of NaCl = 0.11 mg = 1.1\times 10^{-4}g      (Conversion factor:  1 g = 1000 mg)

Molar mass of NaCl = 58.5 g/mol

Putting values in above equation, we get:

\text{Moles of NaCl}=\frac{1.1\times 10^{-4}g}{58.5g/mol}=1.88\times 10^{-6}mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of formula units

So, 1.88\times 10^{-6}mol moles of NaCl will contain = (1.88\times 10^{-6}\times 6.022\times 10^{23})=1.1\times 10^{18} number of formula units

Hence, the number of formula units in the given amount of NaCl are 1.1\times 10^{18}

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Which of the following statements about gamma rays are TRUE? I. Gamma rays have low energy II. Gamma rays are a form of light. I
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A standard solution contained 0.8 mg/mL. A student took 2 mL of the standard solution and added 10 mL of water. What is the new/
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therefore

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It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
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Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L

1 m^3=1000 l

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

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F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%

Total mass of hydrogen hexafluorosilicate = m'

79.16\%=\frac{25,434 g}{m'}\times 100

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

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