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frutty [35]
3 years ago
14

•6.5 L of a gas has a pressure of 840 mmHg and temperature of 84 0C. What will be its volume at STP

Chemistry
1 answer:
EastWind [94]3 years ago
5 0

Answer:

5.5 L

Explanation:

Step 1: Given data

  • Initial volume (V₁): 6.5 L
  • Initial pressure (P₁): 840 mmHg
  • Initial temperature (T₁): 84 °C
  • Final volume (V₂): ?
  • Final pressure (P₂): 760 mmHg (standard pressure)
  • Final temperature (T₂): 273.15 K (standard temperature)

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C + 273.15

K = 84 °C + 273.15 = 357 K

Step 3: Calculate the final volume of the gas

We will use the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 840 mmHg × 6.5 L × 273.15 K / 357 K × 760 mmHg = 5.5 L

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Which of the following is ranked from smallest to largest?
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Answer:

The order is: electron, carbon, water,  glucose, glycogen.    

Explanation:

The electron is a negatively charged subatomic particle and is therefore the smallest.

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6 0
3 years ago
A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat
Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (1.0 × 294.65) / 258.15

P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

3 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
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