Answer:
• One mole of oxygen is equivalent to 16 grams.
→ But at STP, 22.4 dm³ are occupied by 1 mole.

Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is
= 283.725 kJ ⋅ mol − 1
Answer:
ΔT = 0.78 °C
Explanation:
Given data:
Mass of Al = 9.5 g
Specific heat capacity of Al = 0.9 J/g.°C
Temperature change = ?
Heat added = 67 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
67 J = 9.5 g × 0.9 j/g.°C × ΔT
67 J = 85.5 j/°C × ΔT
ΔT = 67 J / 85.5 j/°C
ΔT = 0.78 °C
It is 100 because a 5 and up means you round up. also the 9 makes it a 6 rounded up