Part A Potassium cyanide is a toxic substance, and the median lethal dose depend

Question

Part A Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it. The median lethal dose of KCN for a person weighing 145 lb (65.8 kg ) is 9.29×103 mol . What volume of a 0.0520 M KCN solution contains 9.29×103 mol of KCN? Express the volume to three significant figures and include the appropriate units.

Volume =
SubmitHintsMy AnswersGive UpReview Part Part B Gastric acid pH can range from 1 to 4, and most of the acid is HCl. For a sample of stomach acid that is 3.16×103 M in HCl, how many moles of HCl are in 19.2 mL of the stomach acid? Express the amount to three significant figures and include the appropriate units.

amount of HCl Part A Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it. The median lethal dose of KCN for a person weighing 145 lb (65.8 kg ) is 9.29×103 mol . What volume of a 0.0520 M KCN solution contains 9.29×103 mol of KCN? Express the volume to three significant figures and include the appropriate units.

Volume =
SubmitHintsMy AnswersGive UpReview Part Part B Gastric acid pH can range from 1 to 4, and most of the acid is HCl. For a sample of stomach acid that is 3.16×103 M in HCl, how many moles of HCl are in 19.2 mL of the stomach acid? Express the amount to three significant figures and include the appropriate units.

amount of HCl Part A Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it. The median lethal dose of KCN for a person weighing 145 lb (65.8 kg ) is 9.29×103 mol . What volume of a 0.0520 M KCN solution contains 9.29×103 mol of KCN? Express the volume to three significant figures and include the appropriate units.

Volume =
SubmitHintsMy AnswersGive UpReview Part Part B Gastric acid pH can range from 1 to 4, and most of the acid is HCl. For a sample of stomach acid that is 3.16×103 M in HCl, how many moles of HCl are in 19.2 mL of the stomach acid? Express the amount to three significant figures and include the appropriate units.

amount of HCl

Explanation / Answer

part A) Mole of KCN = 9.29×10-3 mole

Molarity of Kcn = 0.0520 M

Volume of KCN = mole of KCN/concentration of KCN

Volume of KCN = 9.29×10-3/0.0520 = 0.179 L = 179 mL

PartB ) concentration of HCl = 3.16 x 10-3 M

Volume of HCl = 19.2 mL= 0.0192 L

Mole of HCl = concentration of HCl x Volume of HCl

mole of HCl = 3.16 x10-3 x 0.0192 = 6.0672 x 10-5 moles

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