1 AM Wed Nov 3 令@ 68%| T Chapter 7 Section 5 in your lecture textbook covers the calculations needed for this week's lab. They are very similar to calculations we covered previously. Complete the following before coming to lab. It will be GRADED! Given the balanced equation: HBr(ag) + NaOH(aq) → NaBr(aq) + H2OM and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR (0.250 MBr)(l00 ml) :25.0 mmoles of HBr (b) Calculate the number of mmoles of NaOH in 100.0 ml of 0.250 MNAOH (100 mL) (1.250 MNA PH) =e5.0 mmoles of NaOH) © When these two solutions are mixed the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid or base, calculate the number of mmoles of salt produced by the reaction. I mol Inol NaOH 25 x LoS mol Na OH = 25 mmol Na Br O Using the mmoles of salt produced and the total volume in ml of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. : 25 x 10 mol - D. 025 nol Na Br 6.025 nol (NABr] - = (6.125 M 0.2 L (6) Concidor this examnle problem:
1 AM Wed Nov 3 令@ 68%| T Chapter 7 Section 5 in your lecture textbook covers the calculations needed for this week's lab. They are very similar to calculations we covered previously. Complete the following before coming to lab. It will be GRADED! Given the balanced equation: HBr(ag) + NaOH(aq) → NaBr(aq) + H2OM and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR (0.250 MBr)(l00 ml) :25.0 mmoles of HBr (b) Calculate the number of mmoles of NaOH in 100.0 ml of 0.250 MNAOH (100 mL) (1.250 MNA PH) =e5.0 mmoles of NaOH) © When these two solutions are mixed the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid or base, calculate the number of mmoles of salt produced by the reaction. I mol Inol NaOH 25 x LoS mol Na OH = 25 mmol Na Br O Using the mmoles of salt produced and the total volume in ml of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. : 25 x 10 mol - D. 025 nol Na Br 6.025 nol (NABr] - = (6.125 M 0.2 L (6) Concidor this examnle problem:
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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I completed this page of the titration worksheet. Can you please check my answers
![11:51 AM Wed Nov 3
*O 68%
T
Chapter 7 Section 5 in your lecture textbook covers the calculations needed for this
week's lab. They are very similar to calculations we covered previously. Complete the
following before coming to lab. It will be GRADED!
Given the balanced equation:
HBr(aq) + NaOH(aq) →
NaBr(ag) + H2O)
and remembering that
Molarity = moles/liter
OR
mmoles/mL
(1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR
(0.256 MBr)(l00 mb) :25.0 mmoles of HBr
(b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH
(100 mL) (1.25O MN. PH) =25.0 mmoles of NAOH)
(c) When these two solutions are mixed the acid and base should neutralize one
another exactly. This means that all of the acid and base are completely used up;
either one could be considered a “limiting reactant". Starting with the mmoles of
either the acid or base, calculate the number of mmoles of salt produced by the
reaction.
I mol
25 x Lo
mol NaOH
Inol NaOH
- (25 mmol Na Br
(d) Using the mmoles of salt produced and the total volume in ml of solution (from the
mixing of the acid and base solutions), calculate the molarity of the salt solution
produced by this reaction.
: 25 x 1D mol - 0. 025 nol
Na Br
6.025 nol
(NA Br] -
: (0.125 M
0.2 L
4
(o) Concidor this example prohlem:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdbf8765a-c0ab-46fe-8663-61e338d72b7c%2F146fc99b-0ddc-4e11-a127-92a48e469c44%2Fj352qvr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:11:51 AM Wed Nov 3
*O 68%
T
Chapter 7 Section 5 in your lecture textbook covers the calculations needed for this
week's lab. They are very similar to calculations we covered previously. Complete the
following before coming to lab. It will be GRADED!
Given the balanced equation:
HBr(aq) + NaOH(aq) →
NaBr(ag) + H2O)
and remembering that
Molarity = moles/liter
OR
mmoles/mL
(1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR
(0.256 MBr)(l00 mb) :25.0 mmoles of HBr
(b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH
(100 mL) (1.25O MN. PH) =25.0 mmoles of NAOH)
(c) When these two solutions are mixed the acid and base should neutralize one
another exactly. This means that all of the acid and base are completely used up;
either one could be considered a “limiting reactant". Starting with the mmoles of
either the acid or base, calculate the number of mmoles of salt produced by the
reaction.
I mol
25 x Lo
mol NaOH
Inol NaOH
- (25 mmol Na Br
(d) Using the mmoles of salt produced and the total volume in ml of solution (from the
mixing of the acid and base solutions), calculate the molarity of the salt solution
produced by this reaction.
: 25 x 1D mol - 0. 025 nol
Na Br
6.025 nol
(NA Br] -
: (0.125 M
0.2 L
4
(o) Concidor this example prohlem:
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