Answer: The final temperature of both the weight and the water at thermal equilibrium is 
.
Explanation:
The given data is as follows.
mass = 7.62 g, 
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = 
= 
Now, heat gained by lead will be calculated as follows.
q = 
= 
According to the given situation,
Heat lost = Heat gained
=
T =
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is .
Answer:
The temperature of the mixture is 60°C
Explanation:
We can write the energy of water as follows:
E = m×C×T
<em>Where E is energy in Joules, m is mass of water, C is specific heat of water = 4.184J/g°C and T is temperature</em>
<em />
Replacing fot both samples:
E = 20g×4.184J/g°C×30°C
E = 2510.4J
E = 40g×4.184J/g°C×75°C
E = 12552J
The total energy of the mixture is 12552J + 2510.4J = 15062.4J
Mass = 60g:
15062.4J = 60g×4.184J/g°C×T
60°C = T
<h3>The temperature of the mixture is 60°C</h3>
Answer:
Explanation:
There are two ways of looking at this problem. The first way, slightly more advanced, is to understand that the carbocation formed is an intermediate in this reaction: it is formed in one step and consumed in the subsequent step.
Secondly, we have hydroxide involved as our reactant, so it should be our second reactant in the second bimolecular step.
Thirdly, the product formed would be a combination of the anion and cation, one of our products, this means we have the following second step:
Another way is to verify this knowing that by adding all of the steps should yield a net equation, notice if we add the two steps together (reactants on one side and products on the other), we obtain:
Notice that the intermediate carbocation cancels out on both sides to yield the final net equation:
This means we have the correct second step.
Where is the reaction? The whole question didn't make any sense! Next time please put some time on what you want to ask.
