Answer:
A lot of them.
Explanation:
It would take hundreds of thousands of trees to clear all of the emmisions.
I think the answer is ‘repulsion’
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

In terms of mass, specific heat and temperature change is:

Now, solve for the final temperature, as follows:

Then, plug in the masses, specific heat and temperatures to obtain:

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
Regards!
I think there is a lack of information in the given problem above such as the grams of copper sulfate and sodium hydroxide that was used in the experiment. Kindly resubmit the question with the complete details so that we can help you. Thank you.