Trial 1 Trial 2 Trial 3 Moles HCI (mols) (Use the balanced equation to 7. create a mole ratio to convert moles of NaOH to moles of HCI) 0.00267168 0.00271584 0.00272688 Part 6 of 10 Trial 1 Trial 2 Trial 3 0.02420 0.0284 Volume of HCL (L) 8. |(1L = 1000 mL) 0.02500 * 0.02500 e 0.02500 a 0.02500 Part 7 of 10 Trial 1 Trial 2 Trial 3 Molarity HCl (M) moles solute 9. (molarity liters solution

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Find Molarity for HCl for Trials one through three

0.02420
0.02460
0.02470
Volume NaOH Used (L)
5.
(1L = 1000 mL)
Part 4 of 1o
Trial 1
Trial 2
Trial 3
Moles NaOH (mols)
2.671
0.00271584
0.00272688
6.
(The concentration of the
NaOH solution is 0.1104 M)
Part 5 of 10
Trial 1
Trial 2
Trial 3
Moles HCI (mols)
(Use the balanced equation to
7. create a mole ratio to convert
moles of NaOH to moles of
HCI)
0.00267168
0.00271584
0.00272688
Part 6 of 10
Trial 1
Trial 2
Trial 3
0.02420
0.0284'
x
0.02500
8.
Volume of HCL (L)
(1L = 1000 mL)
d 0.02500
o 0.02500
O 0.02500
Part 7 of 10
Trial 1
Trial 2
Trial 3
Molarity HCl (M)
moles solute
9.
molarity
liters solution
Transcribed Image Text:0.02420 0.02460 0.02470 Volume NaOH Used (L) 5. (1L = 1000 mL) Part 4 of 1o Trial 1 Trial 2 Trial 3 Moles NaOH (mols) 2.671 0.00271584 0.00272688 6. (The concentration of the NaOH solution is 0.1104 M) Part 5 of 10 Trial 1 Trial 2 Trial 3 Moles HCI (mols) (Use the balanced equation to 7. create a mole ratio to convert moles of NaOH to moles of HCI) 0.00267168 0.00271584 0.00272688 Part 6 of 10 Trial 1 Trial 2 Trial 3 0.02420 0.0284' x 0.02500 8. Volume of HCL (L) (1L = 1000 mL) d 0.02500 o 0.02500 O 0.02500 Part 7 of 10 Trial 1 Trial 2 Trial 3 Molarity HCl (M) moles solute 9. molarity liters solution
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