Answer:
1.50 g
Explanation:
The heat absorbed by the aluminum in this case is:
q = m x C x ΔT m= q/ (C x ΔT)
q= 9.86 J
C = 0.90 J/g-K
ΔT = ( 30.5 ºC - 23.2 ºC ) = 7.3 ºC = 7.3 K (this is a range of temperature)
m = 9.86 J / ( 0.90 J/g-K ) x 7.3 K ) = 1.50 g
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Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.
