H₂SO₄:
V=0,95L
Cm=0,420mol/L
n = CmV = 0,42mol/L * 0,95L = 0,399mol
KOH:
V=0,9L
Cm=0,26mol/L
n = CmV = 0,26mol/L * 0,9L = 0,234mol
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
1mol : 2mol
0,399mol : 0,234mol
limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol
n = 0,282mol
V = 0,950L + 0,900L = 1,85L
Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
Freezing, condensation, Deposition.
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d answer is correct that help
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
