Answer:
Hydrogen spectrum
Explanation:
Balmer series - Observed in the visible region
Brackett series - Observed in the infrared region
Paschen series - Observed in the infrared region
Lyman series - Observe in the Ultraviolet region.
Based on the information I would assume B, 73 degrees...
It shouldn't be A, 4 minutes on the burner should increase the temperature.
If it were D, it would be beyond boiling, and water takes a decent amount of energy to heat, D should be all vapor.
Same logic for C, it's basically almost boiling.
I would say 73 degrees seems most reasonable for 4 minutes.
The following is the introduction to a special e-publication called Determining the Age of the Earth (click the link to see a table of contents). Published earlier this year, the collection draws articles from the archives of Scientific American. In the collection, this introduction appears with the title, “Stumbling Toward an Understanding of Geologic Timescales.”
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!
Answer:
It is heterogeneous as it is made up of visibly different substances or phases.