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2 years ago

A 7.5 L cylinder contains 5 moles of gas at a temperature of 274°C. What is its pressure in kiloPascals (kPa)?

Chemistry

1 answer:

AleksAgata [21]2 years ago

3 0

<h3><u>Answer;</u></h3>

= 3032.15 kPa

<h3><u>Explanation;</u></h3>

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Volume = 7.5 L, T = 274 +273 = 547 K, N = 5 moles

Therefore;

Pressure = nRT/V

= (5 × 0.08206 × 547)/7.5 L

= 29.925 atm

But; 1 atm = 101325 pascals

Hence; Pressure = 3032150.63 pascals

<u>= 3032.15 kPa</u>

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Answer:

O-H bond

Explanation:

Let us work out the electronegativity difference between the elements in each bond in order to decide which of them is most polar.

For the C-O bond

2.55 - 2.2 =0.35

For the F-F bond

3.98 - 3.98 = 0

For the O-H bond

3.44 - 2.2 = 1.24

For the N-H bond

3.04 - 2.2 = 0.84

The O-H bond has the highest electronegativity difference, hence it is he most polar bond.

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2 years ago

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Novosadov [1.4K]

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3 years ago

A 400 ml sample of gas is heated from -20 c to 60

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Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
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2 years ago

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ludmilkaskok [199]

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2 years ago

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How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

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2 years ago