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:
12.77 M
Explanation:
8.3 moles of NaCl in .65 L of water ? Looking for M ?
8.3 M / .65 L = 12.77 M
Answer:
The correct answer to this problem is B. 7.0 X 10^-8 meters
Explanation:
To solve this problem, we have to use the following equation:
c = λν, or speed of light = wavelength*frequency
If we substitute in the values we are given by the problem, we get:
3.00 * 10^8 m/s = (4.3 * 10^15 Hz)*(wavelength)
wavelength = 6.98 * 10^-8 m
Since the given value has 2 significant figures, our answer should similarly include two significant figures since the operation in the problem was multiplication.
Therefore, the answer is B. 7.0 X 10^-8 meters.
Hope this helps!
