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Zarrin [17]
4 years ago
12

Similarities and differences between solids liquids and gases

Chemistry
1 answer:
Masteriza [31]4 years ago
8 0

Answer:

Detail is given below.

Explanation:

Similarities:

All are the forms of matter.

Solid, liquid and gases are made up of ions, atoms and molecules.

They have shape and volume.

Molecules in Liquid and solid are closer to each others.

Liquid and gases can flow very easily.

Liquid and solid both can not compressed very easily.

Differences:

Molecule of gases randomly move everywhere and occupy all available space but liquid and solid's molecules are not move as much easier as molecules of gases can.

The density of gases are very low as compared to the liquid and solid.

Gas molecules are at long distance from each other while in liquid and solid they are closer to each other.

The very weak inter molecular forces are present between gas molecules as compared to the liquid and solids.

Solids have definite volume and shape but liquid and gases acquire the shape of container.

In solids molecules are tightly pack and very close to each other while in liquid molecules are not tightly pack like in case of solid.

The densities of solids are also very high as compared to the liquid and gas.

There are very strong inter molecular forces are present between solid molecules.

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Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a
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Answer : The mass of NaN_3 at temperature 20^oC 28.47 g.

The mass of NaN_3 at temperature 10^oC 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = 20^oC=273+20=293K     (0^oC=273K)

Volume of gas = 25\times 25\times 20cm=12500cm^3=12.5L   (1L=1000cm^3)

Molar mass of NaN_3 = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature 20^oC. The gas produced in the given reaction is N_2.

Using ideal gas equation,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)

By rearranging the terms, we get the value of 'n'

n=0.7015moles

The moles of N_2 = 0.7015 moles

The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.7015 moles of N_2 produced from \frac{20}{32}\times 0.7015=0.438 moles of NaN_3

Now we have to calculate the mass of NaN_3.

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g

Therefore, the mass of NaN_3 needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature 10^oC and same volume & pressure.

Using ideal gas equation,

PV=nRT

Now put all the given values in this formula, we get

(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)

By rearranging the terms, we get the value of 'n'

n=0.726moles

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The given balanced reaction is,

20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)

As, 32 moles of N_2 produced from 20 moles of NaN_3

So, 0.726 moles of N_2 produced from \frac{20}{32}\times 0.726=0.454 moles of NaN_3

Now we have to calculate the mass of NaN_3.

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g

Therefore, the mass of NaN_3 needed are 29.51 g.

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