Nat(t)D) If you have 100.0 mL of solution, how many moles of NaCl are present? What i:the mass? (MM of NaCl is 58.44 g/mol)molM=Vzmol=(0050M) (O.100OL)mol= MVL= 0.0050 aol 5%.44,)-0,292%3DI molE) You have diluted 5.00 mL of the original solution to 25.00 mL. What is theconcentration now?Vi5.00mLWz=25.00ML0.050MVi Mi = Vz Mz 2(5.0010.050)- 125.00)(人)2525ME 0,010 M)

Given data, Volume of solution=100mL=0.100LMolarity of NaCl=0.050M (taken from solution)Molar mass…

D) If you have 100.0 mL of solution, how many moles of NaCl are present? Wha the mass? (MM of NaCl is 58.44 g/mol) mol M= mol=(0@5OM) CO.100OL) = 0.0050 ao( 5%.4)-0.2a mol= MVc 100.mL=01000L 2 mol E) You have diluted 5.00 mL of the original solution to 25.00 mL. What is the concentration now?

D) If you have 100.0 mL of solution, how many moles of NaCl are present? Wha the mass? (MM of NaCl is 58.44 g/mol) mol M= mol=(0@5OM) CO.100OL) = 0.0050 ao( 5%.4)-0.2a mol= MVc 100.mL=01000L 2 mol E) You have diluted 5.00 mL of the original solution to 25.00 mL. What is the concentration now?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

Nat
(t)
D) If you have 100.0 mL of solution, how many moles of NaCl are present? What i:
the mass? (MM of NaCl is 58.44 g/mol)
mol
M=
Vz
mol=(0050M) (O.100OL)
mol= MVL
= 0.0050 aol 5%.44,)-
0,292
%3D
I mol
E) You have diluted 5.00 mL of the original solution to 25.00 mL. What is the
concentration now?
Vi5.00mL
Wz=25.00ML
0.050M
Vi Mi = Vz Mz 2
(5.0010.050)- 125.00)(人)
25
25
ME 0,010 M)

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