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3 years ago

3.9 X 1022 molecules of CO2 = _ moles of CO2

Chemistry

1 answer:

agasfer [191]3 years ago

7 0

Answer:

moles CO2 = 0.065 moles

Explanation:

mole = 6.022 E23 molecules.....Avogadro

∴ molecules CO2 = 3.9 E22 molecules

⇒ moles CO2 = (3.9 E22 molecules CO2)×( moles CO2/ 6.022 E23 molecules CO2)

⇒ moles CO2 = 0.065 moles

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A sample of TNT, C7H5N3O6 , has 8.94 × 1021 nitrogen atoms. How many hydrogen atoms are there in this sample of TNT?

Bess [88]

Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

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3 years ago

In the alveoli and lung capillaries, co2 and o2 are exchanged by means of _____.

-Dominant- [34]

Through hemoglobin transport in blood, material exchange is carried out on the alveolar membrane.

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3 years ago

Which of the following statements is true?

liubo4ka [24]

Answer:

a is the anwser

Explanation:

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3 years ago

Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A

UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

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As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

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