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1 year ago

Give explanations for the large drops in melting point from C to Si and from Ge to Sn.

1 answer:

3 0

As we move down the group, the metallic bond becomes more stable and the formation of forming covalent bond decreases down the group due to the large size of elements.

Covalent and metallic bonding leads to higher melting points. Due to a decrease in attractive forces from carbon to lead there is a drop in melting point.

Carbon forms large covalent molecules than silicon and hence has a higher melting point than silicon.

Similarly, Ge also forms a large number of covalent bonds and has a smaller size as compared to that of Sn. Hence melting point decreases from Ge to Sn.

The order will be C>Si>Ge>Pb>Sn.

To learn more about the covalent bond, visit: brainly.com/question/10777799

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3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

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2 years ago

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Ilia_Sergeevich [38]

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3 years ago

Read 2 more answers