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oee [108]
2 years ago
11

The melting points of alkaline earth metals are many times higher than those of the alkali metals. Explain this difference on th

e basis of atomic properties. Name three other physical properties for which Group 2A(2) metals have higher values than the corresponding 1A(1) metals.
Chemistry
1 answer:
7nadin3 [17]2 years ago
3 0

The correct answer is higher melting point, bound by metal metal bonds.

While alkali metals only have one valence electron, alkaline earth metals have two. Metal to metal connections hold the metals together. Alkaline earth metals have a stronger metallic connection and a higher melting point because they have two valence electrons.

the characteristics that Group 2 metals excel in over Group 1 metals.

  • Initial Ionization Potential
  • Group 2 items are more difficult than group 1 elements.
  • Strong propensity to produce bivalent compounds

As a result, group 2 metals have stronger metallic bonding, which leads to increased cohesive energy and compact atom packing. This explains why group 2 metals are harder and have higher melting and boiling temperatures than group 1 metals.

To learn more about  Group 2A(2) refer the link:

brainly.com/question/9431096

#SPJ4

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Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

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Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 2 mole of NaN_3 react to give 3 mole of N_2

So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

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