Answer:
The nswer to the question is
The maximum fraction of the air in the room that could be displaced by the gaseous nitrogen is 0.548 or 54.8 %
Explanation:
To solve the question we note that
The density of the liquid nitrogen = 0.808g/mL and the volume is 195 L tank (vaporised)
Therefore since density = mass/volume we have
mass = Density × volume = 0.808 g/mL × 195 L × 1000 ml/L =157560 g
In gaseous form the liquid nitrogen density =1.15 g/L
That is density = mass/volume and volume = mass/density = 157560 g/(1.15g/L) or
volume = 137008.69565 L
The dimension of the room = 10 m × 10 m × 2.5 m = 250 m³ and
1 m³ is equivalent to 1000 L, therefore 250 m³ = 250 m³ × 1000 L/m³ = 250000L
Therefore fraction of the volume occupied by the gaseous nitrogen =
137008.69565 L/250000 L = 0.548
Therefore the gaseous nitrogen occpies 54.8% of the room
Answer:
Part 1. When the balloon is filled half of the way, and placed into the freezer, it will shrink. This happens because kinetic molecular theory tells us that a decrease in temperature decreases the kinetic energy of the gas molecules in the balloon. Viscous gases like hydrogen are less likely to shrink.
Part 2. When the balloon is placed out in the hot sun, most likely the balloon will swell and grow. This happens because the kinetic energy of the gas molecules increases due to solar radiation transforming into heat energy and then transforming into kinetic energy. Sticky gases like neon are more likely to grow.
Explanation:
For this problem we use the wave equation. It is expressed as the speed (c) is equal to the product of frequency (f) and wavelength (v).
c = v x f
We know the wavelength of the an red light which is 6.5 x 10^-7 m. Now, we solve for the wavelength of the unknown wave to see the relation between the two waves.
2.998 X 10^8 = 5.3 X 10^15 X v
v = 2.998 X 10^8 / (5.3 X 10^15) = 5.657 X 10^-8 m
Therefore, the wavelength of the unknown wave is less than the wavelength of the red light.
Answer:
There will be one Al3+ ion.
There will be 3 NO3- ions
Explanation:
Dissociation equation:
Al(NO₃)₃ → Al³⁺ + 3NO₃¹⁻
When aluminium nitrate dissociate it produces one silver ion (Al³⁺) and three (NO₃¹⁻) ions.
Properties of Al(NO₃)₃:
It is inorganic compound having molecular mass 169.87 g/mol.
It is white odor less compound.
Its density is 4.35 g/mL.
Its melting and boiling points are 120°C and 440°C.
It is soluble in water.
It is sued to treat infections.
It is used in the photographic films.
It s toxic and must be handled with great care.
