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Molecular weight refers to the length of a polymer molecule. t/f?

          Molecular weight refers to the length of a polymer molecule. t/f?
        

Added by Harold G.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Molecular weight refers to the length of a polymer molecule. t/f?
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Molecular weight data for some polymer are tabulated here. Compute the following: (a) the number-average molecular weight (b) the weight-average molecular weight. (c) If it is known that this material's degree of polymerization is 477 , which one of the polymers listed in Table $14.3$ is this polymer? Why?

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molecular-weight-data-for-some-polymer-are-tabulated-here-molecular-weight-range-gmol-10000-25000-004-001-25000-40000-007-004-40000-55000-016-011-55000-70000-026-024-70000-85000-024-027-8500-36387

Molecular weight data for some polymer are tabulated here. Molecular Weight Range (g/mol) xi wi 10000 - 25000 0.04 0.01 25000 - 40000 0.07 0.04 40000 - 55000 0.16 0.11 55000 - 70000 0.26 0.24 70000 - 85000 0.24 0.27 85000 - 100000 0.12 0.16 100000 - 115000 0.08 0.12 115000 - 130000 0.03 0.05 Compute the following: (a) the number-average molecular weight. Mn = g/mol (b) the weight-average molecular weight. Mw = g/mol

Adi S.

molecular-weight-data-for-some-polymer-are-tabulated-here-compute-a-the-number-average-molecular-wei

Molecular weight data for some polymer are tabulated here. Compute (a) the number average molecular weight, and (b) the weight-average molecular weight. (c) If it is known that this material's degree of polymerization is $477,$ which one of the polymers listed in Table 14.3 is this polymer? Why? $$\begin{array}{ccc} \hline \text {Molecular Weight} & & \\ \text {Range }(\mathrm{g} / \mathrm{mol}) & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{w}_{\boldsymbol{i}} \\ \hline 8,000-20,000 & 0.05 & 0.02 \\ 20,000-32,000 & 0.15 & 0.08 \\ 32,000-44,000 & 0.21 & 0.17 \\ 44,000-56,000 & 0.28 & 0.29 \\ 56,000-68,000 & 0.18 & 0.23 \\ 68,000-80,000 & 0.10 & 0.16 \\ 80,000-92,000 & 0.03 & 0.05 \\ \hline \end{array}$$

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Transcript

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00:01 To solve this question we have the maximum allowable surface crack using the propagation equation as sigma c.
00:08 It is equal to klc divided by y under root of pi into a which is equal to which belongs to a equals to 1 .5 k lc divided by y sigma c whole square we got a value equals to 1 .5 into 82 .4 divided by 1 into 345 whole square we get this value goes as 0 .01815 meter which is equals to 18 .15 millimeter therefore minimum surface crack length at given condition is is 18 .15 millimeter...
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