Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.
The term isotope is formed from the Greek roots isos ("equal") and topos ("place"), meaning "the same place".
Answer:
n=6 to n=3 (B)
Explanation:
Energy of an electron present in the 
orbit is directly proportional to 
.Hence a transistion from one orbit to another orbit emits an energy proportional to the difference of their squares of the orbits. that is if an electron travels from orbit n1 to orbit n2 then it emits an energy corresponding to 
.So in the above question the highest energy emission occurs when an electron moves from n=6 to n=3.(Highest difference of energy levels).
Answer: 7
Explanation:
Before a number but after a decimal. The zeros at the end would usually mean that it doesn't count but since the numbers are before the zeros and after a decimal it's 7 sig figs
Answer:
B
Explanation:
Step 1 occurs at a slower rate than Step 2.
