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nikitadnepr [17]
3 years ago
11

. A container with a volume of 0.50 L contains air at a pressure of 1.0 atm. If the volume is reduced to 0.10 L at constant temp

erature, what will be the resulting pressure of the air inside? 5.0
Chemistry
1 answer:
marin [14]3 years ago
3 0

Answer:

The answer to your question is Pressure = 5 atm

Explanation:

Data

Volume 1 = V1 = 0.5 l

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = 0.1 l

Pressure 2 = P2 = x

Formula

To solve this problem use the Boyle's equation

                      V1P1 = V2P2

Solve for P2

                     P2 = V1P1/V2

Substitution

                     P2 = (0.5 x 1) / 0.1

Simplification

                     P2 = 0.5/0.1

Result

                     P2 = 5 atm          

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The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

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Answer:

2.275 M

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Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

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8 0
3 years ago
PLEASE HELP!!
grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

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