Answer:
119.7 mL.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas.
V is the volume of the container.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas (K).
- For the same no. of moles of the gas at two different (P, V, and T):
<em>P₁V₁/T₁ = P₂V₂/T₂.</em>
- P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.
- P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).
<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂) </em>= (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) = 121.4 <em>mL.</em>
Answer:
3.80x1024 grams containing if not that then above 80 grams 6 grams
Explanation:
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
It is referred as a covalent bond
Answer is (3) protons and neutrons.
(P.S. The outer part contains the electrons)
