.ll touch ?7:52 PM© 25%X 5CC07162-8D85-4F61-B3BC-3FB5B9...Extra exercise:pain medication,withmolecularMorphine is aformula Cx Hy Oz NtI Knowing that I molecule of morphine Contains 17Carbon atoms. And the %.aHydrogen, and71.57%oxygen.6,75 %, and 16.8%mass of Carbenbyrespectivelyareラof nitrogen.tL Calculate the %1.2 Determine the molar mass of morphinc, then deduceits molecular formula.Giren: M():12 g.mat, MCH)-ly.melbymassM(0). 16 g.mul", M(N)- 14 g.mor"2. Determine the number of Hydrogen atims presentin 31 mg 6f merphineGiven NA-6.02 xlo molecules /mol.

Since 1 molecule contains 17 carbon atoms. So mass of carbon = 17 x 12 = 204 Percentage of carbon =…

Extra exercise: pain medication , with a molecular Morphine is a formula Cx Hy Oz Nt I Knowing that I molecule af morphine Contains 17 Carbon atoms. And the % by s of Carben are respectively 6.75 %, and 16.8 % mass of nitrogen. Hydragen , and 71.57% oxygen h Calculate the % by 1.2 Determine the molar mass of morphine, then deduo its molecular formula. Given. MC):12 g.mt, MCH)=ly.mal M(0). 16 g.mel, M(N)= 14g.mor" 2. Determine the number of Hydrogen ahms present. in 31 mg 6 Given NA - 602 xlo molecules mol. f merphine, 23 Ing-to

Extra exercise: pain medication , with a molecular Morphine is a formula Cx Hy Oz Nt I Knowing that I molecule af morphine Contains 17 Carbon atoms. And the % by s of Carben are respectively 6.75 %, and 16.8 % mass of nitrogen. Hydragen , and 71.57% oxygen h Calculate the % by 1.2 Determine the molar mass of morphine, then deduo its molecular formula. Given. MC):12 g.mt, MCH)=ly.mal M(0). 16 g.mel, M(N)= 14g.mor" 2. Determine the number of Hydrogen ahms present. in 31 mg 6 Given NA - 602 xlo molecules mol. f merphine, 23 Ing-to

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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Thank you for the answer, but I'm confused for part b why do you multiply by 2? Isn't it 1 mol of Na2CO3 for every 2 moles of AgNO3? If that's the case wouldn't you divide by 2 like in part a? Cause I tried it mself, got 0.03 moles AGNO3 like answer... My Stoichiometry Calc...0.03 moles AGNO3 x (1 mol Na2CO3)/(2 mol AgNO3) x (105.99 Na2CO3)/(1 mol Na2CO3)=1.5898g I don't think my calc is correct answer but confused why multiply by the 2 instead...
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