Hello!
The concentration of the final solution when a<span> chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water is
0,3 MTo calculate that, you'll need to use the dilution law, where initial and final concentrations are M1 and M2 respectively, and initial and final volumes are V1 and V2, as shown below.
Keep in mind that the final volume is the sum of the 200 mL of water and the 50 mL of H</span>
₂SO₄ that were added by the teacher. 
Have a nice day!
#1...Evaporation
#2....Gas
First off, what is nonpoint-source pollution?
It comes from many sources BUT this type of pollution doesn't come from factories discharging their waster, nor sewage pipes, nor smokestack emissions; but rather it comes from rainfall or snow-melt moving through and over the ground. Taking away natural and human-made pollutants.
This being said, the answer, is A. Polluted Rainwater.
Answer:Acids. :H2CO3, H3PO4,H3OBR
CH3COOH,
Bases MgO,NH4OH,CaOH
Neutral species; CaC,CaCO3,ClO3
Explanation:
Answer:- 47.62 mL
Solution:- It is a dilution problem where we are asked to calculate the volume of 15.75 M perchloric acid solution required to make 500.0 mL of 1.500 M solution.
For solving this type of problems we use the dilution equation:

Where,
is the concentration of the concentrated solution and
is it's volume.
is the concentration of the diluted solution and
is it's volume. Let's plug in the values in the equation and solve it for
.

On rearranging this for
:


So, 47.62 mL of 15.75 M perchloric acid are required to make 500.0mL of 1.500 M solution.