The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCI up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C=12 Na H = 1X1 C = 12x1 = 23x1 =23 =1 =12 O 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters ] = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI
The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by the student chemist. The burette was filled with 0.22 M HCI up to 0.52 ml one- 2 drops of the indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until the endpoint has been observed. The final reading was 13.18 mL 1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C=12 Na H = 1X1 C = 12x1 = 23x1 =23 =1 =12 O 16x3 =4 2. Solve for the net volume of 0.22 M HCI in Liters ] = 84 g/mol NaHCO3 3. What is the mole ratio of sodium bicarbonate to HCI
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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![The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by
the student chemist. The burette was filled with 0.22 M HCl up to 0.52 ml one- 2 drops of the
indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until
the endpoint has been observed. The final reading was 13.18 mL
1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C-12
Na
H = 1X1
C = 12x1
=1
=12
O
10
16x3 =4
2. Solve for the net volume of 0.22 M HCI in Liters
= 23x1 = 23
?].
= 84 g/mol NaHCO3
3. What is the mole ratio of sodium bicarbonate to HCI
1:1
NaHCO3 + HCl NaCl + H₂ C03
4. Solve for the number of grams of sodium carbonate needed to get 83.68%
5. Solve for the number of moles sodium bicarbonate
6. Solve for the number of moles of HCI present in the mixture](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5d2da4ec-87c6-41ab-93ae-805a8cd35c01%2F25a11887-7c7d-4dda-bb16-7f49a78da3a7%2Fzdunjtdn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The sodium bicarbonate NaHCO3, sample is only 72.56% pure after it has been analyzed by
the student chemist. The burette was filled with 0.22 M HCl up to 0.52 ml one- 2 drops of the
indicator were added to the dissolved sample and were then titrated with 0.24 M HCI until
the endpoint has been observed. The final reading was 13.18 mL
1. Solve for the molecular weight of the sample. Mass H=1, Na=23, O=16, C-12
Na
H = 1X1
C = 12x1
=1
=12
O
10
16x3 =4
2. Solve for the net volume of 0.22 M HCI in Liters
= 23x1 = 23
?].
= 84 g/mol NaHCO3
3. What is the mole ratio of sodium bicarbonate to HCI
1:1
NaHCO3 + HCl NaCl + H₂ C03
4. Solve for the number of grams of sodium carbonate needed to get 83.68%
5. Solve for the number of moles sodium bicarbonate
6. Solve for the number of moles of HCI present in the mixture
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