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alina1380 [7]
4 years ago
9

14. Which group of diamagnetic transition metals exhibits trends in density and melting points that don't match the same trends

seen in
other groups?
A. Group 3
B. Group 12
C. Group 7
D. Group 11​
Chemistry
1 answer:
inn [45]4 years ago
6 0

Answer:

Group 12

Explanation:

Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.

Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.

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soldier1979 [14.2K]

<u>Answer:</u> The heat released for the given process is -1892 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

q=m\times L_{f,v}       ......(1)

where,

q = amount of heat released = ?

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L_{f,v} = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

q=m\times C_{p,m}\times (T_{2}-T_{1})        .......(1)

where,

q = amount of heat released = ?

C_{p,m} = specific heat capacity of medium

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T_2 = final temperature

T_1 = initial temperature

Calculating the heat absorbed for each process:

  • <u>For process 1:</u>

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, (\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg

We are given:

m=2.6kg\\L_f=-334\times 10^3J/kg

Putting values in equation 1, we get:

q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J

  • <u>For process 2:</u>

We are given:

m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K

Putting values in equation 2, we get:

q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J

Total heat absorbed = q_1+q_2

Total heat absorbed = [-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ

Hence, the heat released for the given process is -1892 kJ

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victus00 [196]

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