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℃.
Answer:
because the acid properties of aspirin may be problematic.
Part B question 1 Answer: C
that’s the on that makes most sense
Answer:
Molar mass X = 18.2 g/mol
Explanation:
Step 1: Data given
Mass of compound X = 231 mg = 0.231 grams
Mass of benzene = 65.0 grams
The freezing point of the solution is measured to be 4.5 °C.
Freezing point of pure benzene = 5.5 °C
The freezing point constant for benzene is 5.12 °C/m
Step 2: Calculate molality
ΔT = i*Kf*m
⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C
⇒i = the Van't hoff factor = 1
⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m
⇒m = the molality = moles X / mass benzene
m = 1.0 / 5.12 °C/m
m = 0.1953 molal
Step 3: Calculate moles X
Moles X = molality * mass benzene
Moles X = 0.1953 molal * 0.065 kg
Moles X = 0.0127 moles
Step 4: Calculate molar mass X
Molar mass X = mass / moles
Molar mass X = 0.231 grams / 0.0127 moles
Molar mass X = 18.2 g/mol
I'd say no because the only pure substances are elements
