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Anastaziya [24]
3 years ago
5

15. What volume of CCI, (d = 1.6 g/cc) contain

Chemistry
1 answer:
anastassius [24]3 years ago
4 0

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

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During region C of the graph above the energy being absorbed is a. increasing the velocity of the water particles. b. increasing
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7 0
2 years ago
What mass (in grams) of silver contains the same number of atoms as 5.59 grams of sulfur?
Amanda [17]

Answer:

18.84 g of silver.

Explanation:

We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole of sulphur contains 6.02×10²³ atoms.

1 mole of sulphur = 32 g

Thus,

32 g of sulphur contains 6.02×10²³ atoms.

Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.

From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.

Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.

This is illustrated below:

1 mole of silver = 6.02×10²³ atoms.

1 mole of silver = 108 g

108 g of silver contains 6.02×10²³ atoms.

Therefore, Xg of silver will contain 1.05×10²³ atoms i.e

Xg of silver = (108 × 1.05×10²³)/6.02×10²³

Xg of silver = 18.84 g

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7 0
2 years ago
3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O
AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cu: 3 moles
  • HNO₃: 8 moles
  • Cu(NO₃)₂: 3 mole
  • NO: 2 moles
  • H₂O: 4 moles

The molar mass of the compounds is:

  • Cu: 63.55 g/mole
  • HNO₃: 63 g/mole
  • Cu(NO₃)₂: 187.55 g/mole
  • NO: 30 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Cu: 3 moles ×63.55 g/mole= 190.65 grams
  • HNO₃: 8 moles ×63 g/mole= 504 grams
  • Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
  • NO: 2 moles ×30 g/mole= 60 grams
  • H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3}  x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }

<u><em>mass of Cu(NO₃)₂=  5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

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