Answer:
1) 0.0025 mol/L.s.
2) 0.0025 mol/L.s.
Explanation:
<em>H₂ + Cl₂ → 2HCl.</em>
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<em>The average reaction rate = - Δ[H₂]/Δt = - Δ[Cl₂]/Δt = 1/2 Δ[HCl]/Δt</em>
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<em>1. Calculate the average reaction rate expressed in moles H₂ consumed per liter per second.</em>
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The average reaction rate expressed in moles H₂ consumed per liter per second = - Δ[H₂]/Δt = - (0.02 M - 0.03 M)/(4.0 s) = 0.0025 mol/L.s.
<em>2. Calculate the average reaction rate expressed in moles CI₂ consumed per liter per second.</em>
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The average reaction rate expressed in moles Cl₂ consumed per liter per second = - Δ[Cl₂]/Δt = - (0.04 M - 0.05 M)/(4.0 s) = 0.0025 mol/L.s.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Answer:
Rain shadow
Explanation:
Climate is the average weather condition of a place over a long period of time.
The image shows a rain shadow effect around a mountainous region. The type of rainfall controlled this way is known as an orographic rainfall.
- On the leeward side which is the opposite side of where the rain cloud forms, there is dryness and lack of rainfall.
- As moist air rises along the edge of the windward side, cloud forms.
- This windward side receives a significant amount of rainfall
- The windward side is hit with cold and dry wind and it is barren.
You would weigh the zinc with a weight, Zinc is a mineral so wether it's a solid or a liquid you would measure it with a zinc.
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
