Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total
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Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total mass of 11.1 ± 0.1 g. What is the average mass of calcium in each sample? (Assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.)
. University of Massachuset × Determine The Molecular -)C fi Dwww.saplinglearning.com/ibiscms/mod/ibis/view.php?id=2524754 Sapling Learning Jump to... d) Logout Help Sapling Learning University of Massachusetts, Dartmouth-CHM 204-Spring16-BLANCHETTE Activities and Due Dates Ch3 My Assignment Resources o 3/4/2016 1 1 :55 PM 0 78/1003/3/2016 06:35 PM Assignment Information Available From 2/27/2016 10:00 PM Due Date: Points Possible: Grade Category: Graded Description: Policies: Gradebook #AttemptsScore Print CalculatorPeriodic Table Question 18 of 20 3/4/2016 11:55 PM 98 100 sapling learning 93 Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085), had a total mass of 11.1 ± 0.1 g What is the average mass of calcium in each sample? (Assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.) Homework You can check your answers. You can view solutions when you complete or give up on any question. Number Number (Input the answer in terms of absolute uncertainty.) 95 You can keep trying to answer each question until you get it right or give up. 98 You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer eTextbook Help With This Topic Web Help &Videos; Technical Support and Bug Reports 12 13 4 Previous & Give Up & View Solution O Check Answer Next Exit Hint 20 Copyright © 2011-2016 Sapling Learning, Inc.-151 about us careers partners privacy policy terms of use contact ushelp Ask me anything 5:10 PM 3/4/2016Explanation / Answer
5 same samples mass is 11.1 +/- 0.1 g
Hence 5 X = 11.1 + /-0.1
X = average sample mass or mass of one sample = ( 11.1+/0.1) / 5 = 2..22 +/- 0.02 g
= 2220 mg +/- 20 mg
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