The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
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Answer:</h3>
Lead-205 (Pb-205)
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Explanation:</h3>
<u>We are given;</u>
We are supposed to identify its product after an alpha decay;
- Polonium-209 has a mass number of 209 and an atomic number of 84.
- When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
- Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
- The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
- The equation for the decay is;
²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He
- Note; An alpha particle is represented by a helium nucleus, ⁴₂He.
Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
Anything less that 7, would be a base.
