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romanna [79]
4 years ago
11

What is Cu,As,Ca,Br listed from most metallic to least metallic?

Chemistry
1 answer:
Ilia_Sergeevich [38]4 years ago
5 0

Answer:

LEAST METALLIC.....F, S, P, Zn, Co, Cr, Ca, Rb....HIGHEST METALLIC

Rb is the highest and F is the lowest.

Explanation:

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3.10 mol of an ideal gas with CV,m=3R/2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial
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Answer : The value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

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First we have to calculate the final temperature of the gas.

Using poission equation:

(\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}

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P_1 = initial pressure = 14.7 bar

P_2 = final pressure = 1.00 bar

T_1 = initial temperature = 305 K

T_2 = final temperature = ?

\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67

Now put all the given values in the above expression, we get:

(\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}

T_2=103.7K

Now we have to calculate the q, w, ΔU and ΔH for the process.

As we know that, at adiabatic process the value of q=0. So,

dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)

w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K

w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K

w=dU=-7782.278J=-7.78kJ

Now we have to calculate the value of C_p

C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}

Now we have to calculate the value of ΔH.

dH=n\times C_p\times (T_2-T_1)

dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K

dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K

dH=-12970.5J=-12.97kJ

Thus, the value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

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