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astra-53 [7]
3 years ago
9

What mass of sulfur would have precisely 4.7x10^22 atoms of sulfur?

Chemistry
2 answers:
Svetllana [295]3 years ago
7 0

Answer:

2.498g

Explanation:

First off it's important to know Avogadro's number relationship with number of moles and molar mass.

It is given as;

1 mole = 6.02 * 10^23 atoms.

A mole's weight is equal to it's molar mass. Hence the equation can be changed to;

1 mole = molar mass = 6.02 * 10^23 atoms

In this problem, molar mass of Sulphur is 32g/mol. This means 1 mole of sulphur weighs 32 g and contains 6.02 * 10^23 atoms. What mass of sulphur would then contain 4.7x10^22 of atoms?

This leads us to;

32 g =  6.02 * 10^23

x g =  4.7x10^22

Upon cross multiplication, we have;

x =  (4.7x10^22 * 32) / 6.02 * 10^23

x = 24.98 * 10^-1

x = 2.498g

butalik [34]3 years ago
7 0

Answer: It would be 2.5

Explanation:

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An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture
earnstyle [38]

Answer:

The final volume will be 24.7 cm³

Explanation:

<u>Step 1:</u> Data given:

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Pressure and amount = constant

<u>Step 2: </u>Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

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C.

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The answer is Hydrogen bonds
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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper
finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0
3 years ago
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ludmilkaskok [199]

Answer:

C) 2 H₂ + O₂  →  2 H₂O

Explanation:

4 atoms of hydrogen on reactant side

2 atoms of oxygen on reactant side

4 atoms of hydrogen on product side

2 atoms of oxygen on product side

6 0
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