Determine the molar mass of NaHCO3. Use 22.99, 1.01, 12.01, and 16.00 g mol-¹. (1X22,99 9/mol) + (1x1.01 g/mo) + 1x12,019/mol) + 3x10.5 Molar mass of Na+1CD3 = 84,01 g/mol b. Determine the number of moles of NaHCO3 used in balloon #1 on the far right. (report 3 sig. fig.) a. C. 1gx 1 2gx- mole 9 Determine the number of moles of NaHCO3 used in balloon #2. (report 3 sig. fig.) 1 mole = mole 9 mole =

Determine the molar mass of NaHCO3. Use 22.99, 1.01, 12.01, and 16.00 g mol-¹. (1X22,99 9/mol) + (1x1.01 g/mo) + 1x12,019/mol) + 3x10.5 Molar mass of Na+1CD3 = 84,01 g/mol b. Determine the number of moles of NaHCO3 used in balloon #1 on the far right. (report 3 sig. fig.) a. C. 1gx 1 2gx- mole 9 Determine the number of moles of NaHCO3 used in balloon #2. (report 3 sig. fig.) 1 mole = mole 9 mole =

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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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".

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B, 2g, 3g, 4g, and 5g of sodium bicarbonate, NaHCO3.
a. Determine the molar mass of NaHCO3. Use 22.99, 1.01, 12.01, and 16.00 g mol¹¹.
(1X22,99 9/mol) + (1x1.01 g/mo) + 1x12.01g/mol) + 3x16.9/mol
Molar mass of Na+1CD3 = 84,01 g/mol
b. Determine the number of moles of NaHCO3 used in balloon #1 on the far right. (report 3 sig. fig.)
C.
1gx
2 gx-
Determine the number of moles of NaHCO3 used in balloon #2. (report 3 sig. fig.)
1
mole
9
Limiting Reactant Lab - Take Home
mole =
mole
9
mole =
d. Determine the number of moles of NaHCO3 used in balloon #3. (report 3 sig. fig.)
3/25/2023 9:27 PM

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