How does the total volume of gas particles compare to the volume of the

How does the total volume of gas particles compare to the volume of the space between gas particles?

2 months ago

Solution 1

Guest Guest #4170
2 months ago
             There is more space between gas particles than the size of the particles.

                   This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules. 

📚 Related Questions

Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?
Solution 1
Answer is: volume of oxygen is 14.7 liters.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.

A gas occupies 2240.0 l at 373 k. what are the volumes at standard temperature answers
Solution 1
Answer is: the volumes at standard temperature is 1639.46 L.
V₁(gas) = 2240.0 L.
₁(gas) = 373 K.
₂(gas) = 273 K, standard temperature.
₂(gas) = ?
Charles' Law:  The Temperature-Volume Law - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:
₁/T₁ = V₂/T₂.
2240 L/373 K = V
₂/273 K.
₂ = 1639.46 L.
Question 13 unsaved the nonmetals in groups 5a, 6a, and 7a: question 13 options: lose electrons when they form ions. form ions with charges of 3-, 2-, and 1-, respectively. form positively charged ions. form ions with a numerical charge equal to their group number.
Solution 1

How does the law of conservation of mass apply  to this reaction: C2H4 + O2 → H2O + CO2?

Question 3 what geometric arrangement of charge clouds is expected for an atom that has five charge clouds? trigonal bipyramidal square planar octahedral tetrahedral
Solution 1

AX5E0 - trigonal bipyramidal - zero lone pairs;
AX4E1 - seesaw - 1 lone pair;
AX3E2 - T-shaped - 2 lone pairs;
AX2E3 - linear - 3 lone pairs;

The only option that matches is trigonal bipyramidal.

Question 10 unsaved in a chemical reaction, the mass of the products question 10 options: has no relationship to the mass of the reactants. is less than the mass of the reactants. is greater than the mass of the reactants. is equal to the mass of the reactants.
Solution 1
What are youre answer choices????
What volume of O2( g. at 810. mmHg pressure is required to react completely with a 4.50g sample of C(s) at 48°C? 2 C(s) + O2( g. → 2 CO( g.
Solution 1
Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.

V(O₂) = n·R·T / p.
V(O₂) = 0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.

V(O₂) = 4.63 L.

What is the purpose of the catalyst? A. to start the chemical reaction B. to stop the chemical reaction C. to speed up the chemical reaction D. to make the chemical reaction safe to observe
Solution 1
A catalyst either increases (positive catalyst) or decreases (negative catalyst) the rate of the reaction.
Ammonia (NH3(g), es001-1.jpgHf = –46.19 kJ/mol) reacts with hydrogen chloride (HCl(g), es001-2.jpgHf = –92.30 kJ/mol) to form ammonium chloride (NH4Cl(s), es001-3.jpgHf = –314.4 kJ/mol) according to this equation: NH3(g) + HCl(g) es001-4.jpg NH4Cl(s) What is es001-5.jpgHrxn for this reaction? kJ
Solution 1
Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.
Solution 2

The value of heat of the reaction for the given chemical reaction is equal to -175.91 kJ.

How do we calculate the change in enthalpy of the reaction?

Change in enthalpy of the reaction is calculated by substracting the total sum of enthalpies of reacatnts from the the total sum of the enthalpies of products.

Given chemical reaction is:

NH₃(g) + HCl(g) → NH₄Cl(s)

According to the equation total enthalpy of the reaction calculated as:

ΔHrxn = ΔfH(NH₄Cl) - [(ΔfH(NH₃) + ΔfH(HCl)]

On putting values from the question to the equation, we get

ΔHrxn = -314.4 kJ/mol - [-46.19 kJ/mol + (-92.30 kJ/mol)]

ΔHrxn = -314.4 kJ + 138.49 kJ.

ΔHrxn = -175.91 kJ

Hence the heat of the reaction is -175.91 kJ.

To know more about total heat of reaction, visit the below link:

How many molecules are in 13.5g of sulfur dioxide, so2?
Solution 1

Taking into account the definition of avogadro's number, 0.21 moles of sulfur dioxide contain 1.26482×10²³ molecules.

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

First you must determine the number of moles that 13.5 g of sulfur dioxide contains. For that, I use the molar mass of the compound, which is defined as the amount of mass that a substance contains in one mole.

In this case, the molar mass of sulfur dioxide is 64 g/mole. So the number of moles that 13.5 grams of the compound contain can be calculated as:

13.5 gramsx\frac{1 mole}{64 grams} = 0.21 moles

Then you can apply the following rule of three: if 1 mole of sulfur dioxide contains 6.023×10²³ molecules, then 0.21 moles contain how many molecules of sulfur dioxide?

amount of molecules of sulfur dioxide= (6.023×10²³ molecules× 0.21 mole)÷ 1 mole

amount of molecules of sulfur dioxide=1.26482×10²³ molecules

Finally, 0.21 moles of sulfur dioxide contain 1.26482×10²³ molecules.

Learn more about Avogadro's Number:

Solution 2
Answer is: there are 1.27·10²³ molecules of sulfur dioxide.
m(SO₂) = 13.5 g.
n(SO₂) = m(SO₂) ÷ M(SO₂).
n(SO₂) = 13.5 g ÷ 64 g/mol.
n(SO₂) = 0.21 mol.
N(SO₂) = n(SO₂) ·Na.
N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
N(SO₂) = 1.27·10²³.
n - amount of substance.
M - molar mass of substance.
Na - Avogadro number.
How many valence electrons does a neutral atom of aluminum have? A. 3 B. 10 C. 13 D. More information is needed to figure this out.
Solution 1
A. Valence electrons are the electrons on the outermost energy level. If you write out the electron configuration you add all of the electrons on the 3rd energy level.