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℃.
Not very much because half the time the water is purified but just normal water in my area probably has serious levels of scarcity.
Answer : The value of reaction quotient, Q is 0.0625.
Solution : Given,
Concentration of 
= 2.00 M
Concentration of 
= 2.00 M
Concentration of 
= 1.00 M
Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.
The balanced equilibrium reaction is,
The expression of reaction quotient for this reaction is,
![Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BProduct%5D%5Ep%7D%7B%5BReactant%5D%5Er%7D%5C%5CQ%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5E1%5BH_2%5D%5E3%7D)
Now put all the given values in this expression, we get
Therefore, the value of reaction quotient, Q is 0.0625.
Don’t click that link EVER, they try to use your camera fsr
Answer:
82.0343 g/mol
Explanation:
Count each element and the number of atoms for each element. Add them all together. Use the periodic table.
