How does the total volume of gas particles compare to the volume of the space between gas particles?
Solution 1
There is more space between gas particles than the size of the particles.
Explanation:
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
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
m(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.
V₁(gas) = 2240.0 L.
T₁(gas) = 373 K.
T₂(gas) = 273 K, standard temperature.
V₂(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:
V₁/T₁ = V₂/T₂.
2240 L/373 K = V₂/273 K.
V₂ = 1639.46 L.
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
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.
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.
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.
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:
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:
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:
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.