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?
📚 Related Questions
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.
In this reaction Particle that loose H⁺ is A. NH4⁺ ion.
n(O₂) = 0.00825 mol, amount of substance.
V = 174 mL ÷ 1000 mL/L = 0.174 L, volume of gas.
T = 15°C = 288.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.00825 mol · 0.08206 L·atm/mol·K · 288.15 K / 0.174 L.
p = 1.12 atm.