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℃.
<u>Answer:</u> The structure of the geometrical isomers are attached below.
<u>Explanation:</u>
Cis- and Trans- isomers are the geometrical isomers which have same chemical formula but different structural formula
According to CIP rule, the groups on the doubly bonded carbon atoms are given priorities based on the the atomic masses of first connected atom.
If the highest priority groups are on the same side, it is known as cis-form and if the highest priority groups are on opposite side, it is known as trans-form.
We are given a chemical compound, which is 2-pentene.
In this the highest priority groups are methyl and ethyl groups.
When the groups are on the same side, it forms cis-form and when the groups are on the opposite side, it forms trans-form
The structure of the geometrical isomers are attached below.
B) A chemical change because the nail reacts with water/oxygen to create rust (a type of oxide)
Answer:
How much heat energy required to convert following?
How much heat energy, in kilojoules, is required to convert 47.0 g of ice at -18.0 C to water at 25.0 C ?
Specific Heat of Ice - 2.09 j/g * c
This is how I did it and the answer is wrong...Please check and correct me
Q = m * Cice * Change in Temp
Q = (47.0 g)(2.09 J/g*c)(43) = 4222.6 J * 0.001 kj / j = 4.22 kj
