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ale4655 [162]
3 years ago
9

An ideal gas is allowed to expand from 5.40 L to 35.1 L at constant temperature. By what factor does the volume increase?

Chemistry
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

a) 6.5

b) pressure decreases

c)22 atm

Explanation:

Since the initial volume V1 is 5.40 L and the final volume V2 is 35.1 L, the ratio of V2:V1= 35.1/5.40= 6.5 hence the volume increases by a factor of 6.5.

When the volume increases, the pressure decreases accordingly in accordance with Boyle's law. Boyle's law states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.

c) From Boyle's law

Initial volume V1= 5.40 L

Final volume V2= 35.1 L

Initial pressure P1= 143 ATM

Final pressure P2 = the unknown

P1V1= P2V2

P2= P1V1/V2

P2 = 143 × 5.40/ 35.1

P2= 22 atm

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A 7.74 L balloon is filled with water at 3.88 atm. If the balloon is squeezed into a 0.23 L beaker and does NOT burst, what is t
Vladimir [108]

Answer:

131 atm

Explanation:

To find the new pressure, you need to use Boyle's Law:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.

P₁ = 3.88 atm                       P₂ = ? atm

V₁ = 7.74 L                           V₂ = 0.23 L

P₁V₁ = P₂V₂                                                      <----- Boyle's Law

(3.88 atm)(7.74 L) = P₂(0.23 L)                       <----- Insert values

30.0312 = P₂(0.23 L)                                      <----- Simplify left side

131 = P₂                                                          <----- Divide both sides by 0.23

6 0
1 year ago
Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative
masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

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How many nucleons are in an Iodine element? Okay
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Answer:

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Explanation:

3 0
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The volume is 100mL.

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