You use the equation M1V1=M2V2 for dilution problems
(250)3=.5x
x= 1500 mL
1. Molarity : 0.25 M
2. mol CH₄ = 7.4 moles
mol O₂ = 14.8 moles
<h3>Further explanation</h3>
1.
Given
83.2 g CuCl2 in 2.5 liters of water
Required
the molarity
Solution
Molarity : mol solute per liter of solution(not per liter of solvent)
mol solute = mol CuCl₂
mol CuCl₂ = mass : MW CuCl₂
mol CuCl₂ = 83.2 : 134.45
mol CuCl₂ = 0.619
Molarity(M) = mol : V
Assume density CuCl₂ = 3.39 g/cm³
volume CuCl₂ = 8.32 g : 3.39 g/cm³ = 2.45 cm³=2.45 x 10⁻³ L
With this small volume value of CuCl₂, the volume of the solute is sometimes neglected in calculating molarity
volume of solution = 2.5 L + 2.45 x 10⁻³ L = 2.50245 L
Molarity(M) = mol : V
M = 0.619 : 2.50245 L = 0.247≈0.25
2.
Given
Reaction
The correct balanced reaction:
CH4 + 2O2 → CO2 + 2H2O
7.4 moles CO2
Required
moles of methane (CH4) and oxygen gas (O2)
Solution
From the equation, mol ratio of CO₂ : CH₄ : O₂ = 1 : 1 : 2
mol CH₄ = mol CO₂ = 7.4 moles
mol O₂ = 2 x mol CO₂ = 2 x 7.4 moles = 14.8 moles
Answer:
group 1, 2 and 3 tend to get rid of electrons and start to form compounds with groups 7, 6, and 5.
Explanation:
Answer:
Sodium chloride
Sodium chloride is by a huge margin the most common chlorine compound, and it is the main source of chlorine and hydrochloric acid for the enormous chlorine-chemicals industry today.
ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
