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3 years ago

What is the pressure if 3 mol of gas has a volume of 2.3 L at 15°C?

Chemistry

1 answer:

Lana71 [14]3 years ago

8 0

Answer:

30.84atm

Explanation:

Using the ideal gas formula:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821Latm/molK)

T = temperature (K)

According to the provided information in this question,

P = ?

V = 2.3 L

n = 3 mol

T = 15°C = 15 + 273 = 288K

Using PV= nRT

P = nRT/V

P = (3 × 0.0821 × 288) ÷ 2.3

P = 70.9344 ÷ 2.3

P = 30.84atm

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Suppose that 3.33 g of acetone at 25.0 °C condenses on the surface of a 44.0-g block of aluminum that is initially at 25 °C. If

jeyben [28]

Answer:

68.6 °C

Explanation:

From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'

Q = Q'

Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g

Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown

So, mL = m'c(θ₂ - θ₁)

θ₂ - θ₁ = mL/m'c

θ₂ = mL/m'c + θ₁

substituting the values of the variables into the equation, we have

θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C

θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C

θ₂ = 43.56 °C + 25 °C

θ₂ = 68.56 °C

θ₂ ≅ 68.6 °C

So, the final temperature (in °C) of the metal block is 68.6 °C.

5 0

3 years ago

Explain why a spinning top will not remain in motion forever

AlexFokin [52]

Friction of the air, and the surface it is on

7 0

4 years ago

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Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. Use molecular orbital theory to determine whet

julia-pushkina [17]

Answer:

The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).

Explanation:

Molecular orbital of He₂⁺

$1\sigma_{1s}^21\sigma(star)_{1s}^1$

There are two electrons in bonding and 1 electron in antibonding orbital

Bond order = $\frac{(2-1)}{2}$

= $\frac{1}{2}$

Molecular orbital of He₂⁺²

$1\sigma_{1s}^21\sigma(star)_{1s}^0$

There are two electrons in bonding and 0 electron in antibonding orbital

Bond order = $\frac{(2-0)}{2}$

= 1

So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is $\frac{1}{2}$ .

7 0

3 years ago

TRUE or FALSE. The particles of a gas are packed closely together and don't have as much movement as the particles of a solid or

ddd [48]

Gas particles spread out to fill a container evenly, unlike solids and liquids. Gas is a state of matter that has no fixed shape and no fixed volume. Gases have lower density than other states of matter, such as solids and liquids. Hence above statement is FALSE.

3 0

3 years ago

I need to know what is oxidized and what is reduced.

pishuonlain [190]

Answer:

1. Reduction: O₂ ; Oxidation: S

2. Reduction: F ; Oxidation: Al

3. Reduction: Zn ; Oxidation: H

Explanation:

Determine the oxidation state of each element:

2 CoS → Cobalt acts with +2, S with -2

O₂ is at ground state. In ground state the elements has oxidation state of 0.

In CoO, the oxygen acts with -2

The oxidation state has decreased; oxygen was reduced.

In SO₂, sulfur acts with +4; oxidation state has increased, so S has been oxidized.

2. Both reactants are at ground state.

In AlF₃ Al has oxidation state of +3 and F, -1

Al increased the oxidation state → oxidation

F decreased the oxidation state → reduction

3. Hydrogen has 0 as oxidation state. In water, H has +1

It was increased → oxidation

Zn acts with +2, in the zinc oxide. As we have elemental Zn, the oxidation state is 0; it has decreased → reduction

4 0

3 years ago