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2 years ago

Compare the following in the three states of matter

Chemistry

1 answer:

Vilka [71]2 years ago

5 0

Explanation:

Anything that occupies space and has mass is called matter.

(a) Inter particle space

Solids - Atoms of a solid substance are closely packed with each other due to which there is very less or no inter particle space.
Liquids - Atoms of a liquid are little loosely packed due to which there is some inter particle space.
Gases - Atoms of a gas are very loosely packed due to which there is large inter particle space.

(b) Inter particle force of attraction

Solids - Atoms of a solid are attached to each other by strong force of attraction.
Liquids - Atoms of a liquid have slightly less force of attraction as compared to a solid substance.
Gases - Atoms of a gas have Vander waal forces. Hence, there is no inter particle force of attraction between atoms of a gas.

(c) Particle motion

Solids - There is low or negligible particle motion between atoms of a solid.
Liquids - There is high particle motion as compared to that present in a solid.
Gases - There is very high particle motion as compared to solids and liquids.

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Both hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you open

wlad13 [49]

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass.

Explanation:

4 0

3 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?

KengaRu [80]

Answer:

d. Copper (II) sulfate

Explanation:

Given data:

Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 1.25 g/ 27 g/mol

Number of moles = 0.05 mol

Number of moles of CuSO₄:

Number of moles = mass/molar mass

Number of moles = 3.28 g/ 159.6 g/mol

Number of moles = 0.02 mol

now we will compare the moles of reactant with product.

Al : Al₂ (SO₄)₃

2 : 1

0.05 : 1/2×0.05=0.025 mol

Al : Cu

2 : 3

0.05 : 3/2×0.05 = 0.075 mol

CuSO₄ : Al₂ (SO₄)₃

3 : 1

0.02 : 1/3×0.02=0.007 mol

CuSO₄ : Cu

3 : 3

0.02 : 0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

4 0

3 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago

Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the

stiv31 [10]

Molarity of solution is mathematically expressed as,
M = $\frac{x\text{weight of solute(g)}}{\text{Molecular weight X Volume of solution(l)}}$

We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l

Also, we know that molecular weight of sucrose = 342.3 g/mol
∴M = $\frac{x\text{211.4}}{\text{342.3 X 0.08849}}$
= 6.979 M

Thus, molarity of solution is 6.979 M

8 0

2 years ago

How do you separate alcoh0l from water?

Xelga [282]

Answer:

Fractional distillation

Explanation:

7 0

2 years ago