Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
What are you trying to ask??????
Answer:
b. E = 2,28V
Explanation:
The maximum work is the same than ΔG. As ΔG could be written as:
ΔG = nFE <em>(1)</em>
Where n is moles of electrons transferred, F is faraday constant (96485 J/Vmol) and E is the voltage of the cell.
For the reaction:
CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l)
The oxidation state of C in CH₃OH is -2 but in CO₂ is +4, that means transferred electrons are +4 - -2 = <em>6e⁻</em>
Replacing in (1):
1320x10³ J = 6mol e⁻×96485J/Vmol×E
<em>E = 2,28V</em>
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I hope it helps!
Answer:
C.
increasing level of carbon dioxide