Explanation Section1: To determine: To determine the total attractive force of three molecule combinations. Answer: Approximately, the total attractive force of three combinations is 2.06 × 10 − 8 N Explanation: Given Info: The distance of ( H-N ) and ( O-H ) is 0.110 nm . (i) Consider O-H-O combination of the force: Refer the figure.1, Distance between the O − -H + is, r O − -H + = ( r O-H-O − r H-O ) = ( 0.290 − 0.110 ) nm =0 .180 nm Formula to calculate the Force of O − -H + , F O − − H + = k ( q 1 q 2 ( r O − − H + ) 2 ) In the given system, the charges q 1 and q 2 are same therefore, for each protons charge is e Substitute e for q 1 and e for q 2 , F O − -H + = k ( e 2 ( r O − -H + ) 2 ) F O − -H + is force of O − -H + , e is a charge, k is proportionality constant, r O − -H + is distance between O − and H + . Substitute 1.60 × 10 − 19 C for e , 0.180 × 10 − 9 m for r O − -H + and 8.99 × 10 9 N ⋅ m 2 / C 2 for k to find F O − -H + . F O − -H + = ( 8.99 × 10 9 N ⋅ m 2 / C 2 ) ( ( 1.60 × 10 − 19 C ) 2 ( 0.180 × 10 − 9 m ) 2 ) = ( 8.99 × 10 9 N ) ( 7.9012 × 10 − 19 ) = 7.10 × 10 − 9 N (I) Since, the unlike charges get attractive force. Approximately, the force of O − -H + is 7.10 × 10 − 9 N (ii) Consider N-H-N combination of the force: Refer the figure.1, Distance between the N − -H + is, r N − -H + = ( r N-H-N − r H-N ) = ( 0.300 − 0.110 ) nm =0 .190 nm Formula to calculate the Force of N − -H + , F N − -H + = k ( q 1 q 2 ( r N − -H + ) 2 ) In the given system, the charges q 1 and q 2 are same therefore, for each protons charge is e Substitute e for q 1 and e for q 2 , F N − -H + = k ( e 2 ( r N − -H + ) 2 ) F N − -H + is force of N − -H + , e is a charge, k is proportionality constant, r N − -H + is distance between N − and H + . Substitute 1.60 × 10 − 19 C for e , 0.190 × 10 − 9 m for r O − -H + and 8.99 × 10 9 N ⋅ m 2 / C 2 for k to find F N − − H + . F N − -H + = ( 8.99 × 10 9 N ⋅ m 2 / C 2 ) ( ( 1.60 × 10 − 19 C ) 2 ( 0.190 × 10 − 9 m ) 2 ) = ( 8.99 × 10 9 N ) ( 7.0914 × 10 − 19 ) = 6.37 × 10 − 9 N (II) Since, the unlike charges get attractive force. Approximately, the force of N − -H + is 6.37 × 10 − 9 N (iii) Consider O-H-N combination of the force: Refer the figure.1, Distance between the O − − H + is, r O − -H + = ( r O-H-N − r H-N ) = ( 0.290 − 0.110 ) nm =0 .180 nm Formula to calculate the Force of O − -H + , F O − -H + = k ( q 1 q 2 ( r O − -H + ) 2 ) In the given system, the charges q 1 and q 2 are same therefore, for each protons charge is e Substitute e for q 1 and e for q 2 , F O − -H + = k ( e 2 ( r O − -H + ) 2 ) F O − -H + is force of O − -H + , e is charge, k is proportionality constant, r O − -H + is distance between O − and H + . Substitute 1.60 × 10 − 19 C for e , 0.180 × 10 − 9 m for r O − -H + and 8.99 × 10 9 N ⋅ m 2 / C 2 for k to find F O − -H + . F O − -H + = ( 8.99 × 10 9 N ⋅ m 2 / C 2 ) ( ( 1.60 × 10 − 19 C ) 2 ( 0.180 × 10 − 9 m ) 2 ) = ( 8.99 × 10 9 N ) ( 7.9012 × 10 − 19 ) = 7.10 × 10 − 9 N (III) Since, the unlike charges get attractive force. Approximately, the force of O − -H + is 7.10 × 10 − 9 N Consider the total attractive force: From equation (III), the force of O − -H + is 7.10 × 10 − 9 N , From Equation (II), the force of N − -H + is 6.37 × 10 − 9 N and From Equation (I), the force of O − -H + is 7.10 × 10 − 9 N . Formula to calculate the total attractive force is, F = F O − -H + + F N − -H + + F O − -H + = ( 7.10 × 10 − 9 N ) + ( 6.37 × 10 − 9 N ) + ( 7.10 × 10 − 9 N ) = 2.057 × 10 − 8 N ≈ 2.06 × 10 − 8 N Approximately, the total attractive force of three molecule combinations is 2.06 × 10 − 8 N Section2: To determine: To determine the total repulsive force of three molecule combination. Answer: Approximately, the total repulsive force of three combinations is 8.04 × 10 − 9 N Explanation: Given Info: The distance of ( H-N ) and ( O-H ) is 0

University Physics with Modern Physics Technology Update, Volume 2 (CHS. 21-37)

13th Edition

ISBN: 9780321898098

Author: Hugh D. Young

Publisher: PEARSON

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Chapter 21, Problem 24E

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University Physics with Modern Physics Technology Update, Volume 2 (CHS. 21-37)

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