LAB CHEM QUESTION NEED HELP?

2017-02-07 1:39 pm

Data and Calculations:
1. Volume of water in calorimeter: 100.0 mL
2. Mass of water in calorimeter (water's density 1.00g/mL): 100.0g
3. Initial Temperature of water in calorimeter:19.5 degrees Celsius
4. Final Temperature of water in calorimeter: 3.0 degrees Celsius
5. Final volume of water in calorimeter: 122.5mL
6. Volume of water from melted ice: 22.5mL
7. Mass of water from melted ice (water's density 1.00g/mL)
8. Temperature change of water in calorimeter: -16.5

9. Find the energy change (deltaH) of water in calorimeter (heat capacity for water 4.18J/gC:_____ (i got -6900J but im not so sure)
10. Energy change (delta H) of water from melted ice: ____ ( i got 140J)
11. Energy change of melted ice: _____ **(main one that i need help on)
12. Energy change per mole of melting ice (deltaH fusion): ____

回答 (1)

胡雪8°

2017-02-07 2:06 pm

✔ 最佳答案

9.
Mass = Density × Volume
Mass of water in calorimeter = (1.00 g/mL) × (100 mL) = 100 g

Energy change = m c ΔT
Energy change of water in calorimeter, ΔH₁ = (100 g) × (4.18 J/g°C) × [(3.0 - 19.5)°C] = -6900 J

10.
Mass = Density × Volume
Mass of water from melting ice = (1.00 g/mL) × (22.5 mL) = 22.5 g

Energy change = m c ΔT
Energy change of water from melted ice, ΔH₂ = (22.5 g) × (4.18 J/g°C) × [(3.0 - 0)°C] = 282 J

11.
In the calorimeter, total energy change = 0
(Energy change of melting ice) + ΔH₁ + ΔH₂ = 0
Energy change of melting ice = -(ΔH₁ + ΔH₂) = -(-6900 + 282) J = 6620 J

12.
Molar mass of water (H₂O) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of water = (22.5 g) / (18.0 g/mol) = 1.25 mol

Energy change of melting ice = n Lf
6620 J = (1.25 mol) × Lf
Energy change per mole of melting ice = (6620 J) / (1.25 mol) = 5300 J/mol

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