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amm1812
3 years ago
7

A sample of gas has a volume of 100.0 L at 135°C. Assuming the pressure remains constant, what is the volume of the gas if its t

emperature decreases to 72.0 °C?
Chemistry
1 answer:
lina2011 [118]3 years ago
8 0

Answer: 84.56L

Explanation:

Initial volume of gas V1 = 100L

Initial temperature T1 = 135°C

Convert temperature in Celsius to Kelvin

( 135°C + 273 = 408K)

Final temperature T2 = 72°C

( 72°C + 273= 345K)

Final volume V2 = ?

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

100L/408K = V2/345K

To get the value of V2, cross multiply

100L x 345K = V2 x 408K

34500 = V2 x 408K

V2.= 34500/408

V2 = 84.56L

Thus, the volume of the gas becomes 84.56 liters

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Answer:

los metales alcalinotérreos: berilio (Be), magnesio (Mg), calcio (Ca), estroncio (Sr), bario (Ba) y radio (Ra).

o simplemente llamado grupo 2A

Explanation:

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3 years ago
What is the average atomic mass listed for nitrogen in the periodic table?
wel
Average atomic mass listed for nitrogen in the periodic table is 14

Hope this helps!
3 0
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Why does burned toast taste so bad? Is it a chemical or physical change?
german

Because things that are burnt have been changed into something else like moldy bread thats cooked.

It is chemical change because you cannot unburn something.

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The combustion of carbohydrates and the combustion of fats are both exothermic processes, yet the combustion of carbohydrates is
saveliy_v [14]

Explanation:

Combustion of a compound is the reaction with oxygen , hence , the process of combustion is an oxidation reaction.

The carbohydrates contain more amount of oxygen as compared to the fats ,

Hence ,

carbohydrates , have a lot of oxygen contents , are are already partially oxidized , but fats have lower oxygen content .

Therefore ,

The partially oxidized carbohydrates are very difficult to oxidized in comparison to fats .

4 0
3 years ago
A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

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

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0
2 years ago
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