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2 years ago

Which of the following is not a type of structural isomerism? A. Functional group B. Optical C. Chain D. Positional

Chemistry

1 answer:

guapka [62]2 years ago

8 0

Answer:

B. Optical

Explanation:

First, let's go through what isomerism is. Isomerism is the relation of two or more compounds, radicals, or ions that are composed of the same kinds and numbers of atoms but differ from each other in structural arrangement (structural isomerism), as CH3OCH3 and CH3CH2OH, or in the arrangement of their atoms in space and therefore in one or more properties. There are 5 types of structural isomerism. They are chain isomerism, position isomerism, functional group isomerism, metamerism, and tautomerism. This would mean that these types of structural isomerism that are in the options would be eliminated as the question ask which of the "following" is not a type of structural isomerism. This would leave the only option available which is Option B. Optical Isomerism. So instead of being a structural isomerism, it is a type of stereoisomerism. Stereoisomerism is the arrangement of atoms in molecules whose connectivity remains the same but their arrangement in space is different in each isomer.

Therefore, the answer is Option B. Optical.

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The unit cell edge length for the alloy is 0.405 nm

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Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

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Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

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$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

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