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maksim [4K]
3 years ago
12

Calculate the volume of 5.00 mol of helium at 120C and 1520 mm Hg

Chemistry
1 answer:
levacccp [35]3 years ago
4 0

Answer:

80.7 L

Step-by-step explanation:

This looks like a case where we can use the Ideal Gas Law to calculate the volume.

pV = nRT       Divide both sides by p

V = (nRT)/p

=====

Data:

n = 5.00 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = (120 +273.15) K = 393.15K

p = 1520 mmHg × 1 atm/760 mmHg = 2.00 atm

=====

Calculation:

V = (5.00 × 0.082 06 × 393.15)/2.00

V = 161.3/2.00

V = 80.7 L  

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Why are there 2 bonds holding the oxygen atoms together?​
tino4ka555 [31]

Answer:

Because Oxygen shares 2 electrons with mutual bond interaction forming covalent bond . thus it is diatomic due to K shell 2 electrons mutual sharing .

Explanation:

8 0
3 years ago
HELP ME PLEASE!!!
777dan777 [17]

Answer:

A bonding that occurs between high electronegative atoms such are N, F, O and H atoms, is called a hydrogen bond. Hydrogen bond is a very strong bond. (C)

If hydrogen bonds are not formed between H atoms and N, F, O atom, then the atoms interact through dispersion forces (also known as london dispersion forces). Dispersion forces are weak and they are temporary forces formed by overlapping of orbitals. (B)

8 0
3 years ago
How much heat is released when 15.7g of methane (c2h6) is combusted if the enthalpy of the reaction is - 1560.7 kj
lyudmila [28]

- 407.4 kJ of heat is released.

<u>Explanation:</u>

We have to write the balanced equation as,

2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O

Here 2 moles of ethane reacts in this reaction.

Now we have to find out the amount of ethane reacted using its given mass and molar mass as,

2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆

Heat released = ΔH × given mass / 60.14

                        = - 1560. 7 kj ×15.7 g / 60. 14 g  = -407. 4 kJ

5 0
2 years ago
How are estuaries connected to oceanic organisms?
Viktor [21]

Answer:

A: Estuaries provide a nursery grounds for oceanic organisms

Explanation:

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5 0
3 years ago
How do I do this? What are the answers to the 5 questions shown?
frozen [14]

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

6 0
3 years ago
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