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STARTING AMOUNT How many moles of hydrogen are in 3.06 x 10 of glycine, CHNO,? ADFACTOR CJANO, 2 75.00 8.15 x 10 1 3.06 x 10 6.022 ECHNO H MOR PEMY 29.07 1.01 4.08 x 10" 10 2.04 10 mel CHNO molecules CJ,NO,

STARTING AMOUNT How many moles of hydrogen are in 3.06 x 10 of glycine, CHNO,? ADFACTOR CJANO, 2 75.00 8.15 x 10 1 3.06 x 10 6.022 ECHNO H MOR PEMY 29.07 1.01 4.08 x 10" 10 2.04 10 mel CHNO molecules CJ,NO,

Chemistry for Engineering Students

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Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter1: Introduction To Chemistry

Section: Chapter Questions

Problem 1.76PAE: 1.76 How does Gorilla Glass differ from more commonly found alumina silicate glass?

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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

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Step 1: Determine the number of moles of H in given amount of glycine:

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How did you get this? k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs 0.693 a 102.0 nanogram(ng) sample of sodium-24 was stored in a lead-lined cabinet for 2.5 days How much sodium-24 remained? The half-life os sodium -24 is 15 hours _________ ng sodium -24 remain check_circle Expert Answer thumb_up thumb_down Step 1 Radioactive decay is a first order reaction. Conversion of days into hours: 1 day2.5 days===24 hrs2.5×2460 hrs1 day=24 hrs2.5 days=2.5×24=60 hrs Calculation of k: k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs-1 Step 2 Radioactive decay is a first order reaction, therefore: [A]=[Ao] e−kt[A]=[Ao] e-kt where [A] is the final amount remained at time t, [Ao] is the initial amount. Putting values: [A]====102.0 ng×e−0.0462 hrs−1×60 hrs102.0 ng×e−2.772102.0 ng×0.062546.38 ng[A]=102.0 ng×e-0.0462 hrs-1×60 hrs=102.0 ng×e-2.772=102.0 ng×0.06254=6.38 ng Thus 6.38 ng of sodium-24 remained.