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3 years ago

How are real gases different from ideal gases?

1 answer:

3 0

Real gases have small attractive and repulsive forces between particles and ideal gases do not. Real gas particles have a volume and ideal gas particles do not. Real gas particles collide inelastically loses energy with collisions and ideal gas particles collide elastically.

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The list below includes some of the properties of ammonia

suter [353]

Ammonia is colorless gas with a characteristic smell. Its density is 0.589 times than air which makes it lighter than air. Ammonia can be easily liquefied due to the hydrogen bonding between the molecules. The boiling point is at -33.3 degrees Celsius and the freezing point is at -77.7 degrees Celsius.

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3 years ago

How do geologists use seismic waves to locate an earthquakes epicenter

GREYUIT [131]

The waves detect magnetic waves and the disturbances in the to locate where a earth quake will or may hit

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3 years ago

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Lilit [14]

It is A. Barium

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3 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

How many kj of heat are needed to completely melt 32.3 g of h2o, given that the water is at its melting point? the heat of fusio

timurjin [86]

Answer: fourth option, 10.8 kJ

Explanation:

The heat of fusion, also named latent heat of fusion, is the amount of heat energy required to change the state of a substance from solid to liquid (at constant pressure).

The data of the heat of fusions of the substances are reported in tables and they can be shown either per mole or per gram of substance.

In this case we have that the heat of fusion for water is reported per mole: 6.02 kJ/mole.

The formula to calculate how many kJ of heat (total heat) are needed to completely melt 32.3 g of water, given that the water is at its melting point is:

Heat = number of moles × heat of fusion

The calculations are:

number of moles = mass / molar mass

number of moles = 32.3 g / 18.015 g/mol = 1.79 mol

Heat = 1.79 mol × 6.02 kJ / mol = 10.8 kJ ← answer

5 0

3 years ago

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