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:))
Answer:
B
B
Explanation:
the both have the same thing or something
<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O
It’s 819. UWu Jan thanks s whhakswn shoaakb
<span>ANSWER:
Electrical energy accelerates the electrons in the neon gas. The gas ionizes and becomes plasma, containing both positive and negative ions.</span>
