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3 years ago

A. 14.8

Chemistry

1 answer:

Scilla [17]3 years ago

6 0

Answer:

Explanation:

Given data:

Initial volume = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

Given data:

Initial volume = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1 mL

Given data:

Initial volume = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75 L

T₂ = 720.78 L.K / 2.75 L

T₂ = 262.1 K

Given data:

Initial volume = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89 L

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8 L

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79 mL

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Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri

Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______ 46.0684 g

x ______ 50.0 g

x = 1.085 mol CH₃CH₂OH

1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g

y ______ 35.9 g

y = 0.48 mol CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w _____ 0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

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Which is a characteristic of all ions?

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Answer:

I guess B They have a charge

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Carbon dioxide is produced by the human body through

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Answer:

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What is the correct sequence of coefficients when this equation is balanced? ___cs2(l) + ___o2(g) → ___co2(g) + ___so2(g)?

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Answer: the sequence is 1, 3, 1, 2.

Explanation:

1) Given equation: CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)

2) To balance:

i) initially C is balanced: 1 atom in the left side, and 1 atom in the right sides

ii) Add coefficient 2 in front of SO₂ to have 2 atoms in each side.

That leads to: CS₂(l) + O₂(g) → CO₂(g) + 2SO₂(g)

iii) Count the O atoms; there are 2 atoms in the left, and 6 atoms in the right side. So, add a coefficient 3 in front of O₂ (in the left) and you get:

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

4) You can verigy that now the equation is balanced:

C: 1 in the left, and 1 in the right

S: 2 in the left, and 2 in the right

O: 6 in the left, and 6 in the right.

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