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3 years ago

Are these gases lighter than or denser than air? How can you tell?

1 answer:

3 0

Some Gases are lighter than air
Remember,Air is a mixture of gases
So you cant tell!
It depends on the gas
Hope i helped

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What pressure will be exerted by 0.57 moles of CO2 at a temperature of 25°C and a volume of 500 ML? _____atm

svetlana [45]

The pressure exerted by 0.57 moles of CO2 at a temperature of 25°C and a volume of 500 ml is 28 atm.

<u>Explanation:</u>

According to ideal gas law,

PV = nRT

where P represents the pressure of a gas,

V represents the volume of a gas,

n represents the number of moles,

R represents the gas constant = 0.0821 L atm / mol K.

T represents the temperature of a gas.

Given V = 500 ml = 0.5 l, T = 25°C = 298 K, n = 0.57 mol

PV = nRT

P = nRT / V

= (0.57 $\times$ 0.0821 $\times$ 298) / 0.5

P = 28 atm.

The pressure of a gas is 28 atm.

4 0

4 years ago

Emperical formula of carbon dioxide

Stells [14]

CO2 is the emperical formula of carbon dioxide

6 0

3 years ago

Read 2 more answers

The most ideal way to perform a two-point turn is by

Stells [14]

Is by turning a few seconds before

5 0

4 years ago

Where must the moon be in its orbit for the full moon phase to be visible?

ElenaW [278]

Answer:

<h2>A.EXACTLY CROSSING EARTH'S ORBIT OF THE SUN</h2>

5 0

3 years ago

A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the ch

kolezko [41]

Answer:

The change in entropy of gas is $\Delta S= nC_{P}ln3$

Explanation:

n= Number of moles of gas

Change in entropy of gas = $ds= \int \frac{dQ}{T}$

$dQ= nC_{p}dT$

From the given,

$V_{i}=V$

$V_{f}=3V$

Let "T" be the initial temperature.

$\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}$

$\frac {V}{T}=\frac {3V}{T_{f}}$

${T_{f}} = 3T$

$\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}$

$\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})$

$\Delta S = nC_{p}ln3$

Therefore, The change in entropy of gas is $\Delta S= nC_{P}ln3$

3 0

3 years ago