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Sonja [21]
3 years ago
6

The force between two atoms is the result of _____________________ repulsion, nucleus-nucleus ________________, and nucleus-elec

tron _______________. At the point of _____________________, the ______________ forces balance the ____________ forces. The most stable arrangement of atoms exists at the point of __________________________, when the atoms bond covalently and a __________ forms.
Chemistry
1 answer:
Elanso [62]3 years ago
4 0

Answer:

<h2>(1). electron electron repulsion</h2><h2>(2). repulsion </h2><h2>(3). attraction </h2><h2>(4). maximum attraction  </h2><h2>(5). attractive </h2><h2>(6). repulsive </h2><h2>(7). maximum attraction </h2><h2>(8). molecule </h2>

Explanation:

The same charges repel each other while opposite charges attract each other. During electron-electron interaction repulsion take palace because the electron has negative charges. Nucleus has positive charges so the interaction between two nucleus results in the form of repulsion. When interaction takes place between nucleus and electron then attraction takes place between nucleus and electrons due to opposite charges.

          The formation of a bond that takes place due to the sharing of the electrons is known as a covalent bond and thus the covalent molecule is formed.

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Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G
dangina [55]

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change or heat required = ?

m = mass of water = 45 g

c_{p,s} = specific heat of solid water = 2.09J/g^oC

c_{p,l} = specific heat of liquid water = 4.18J/g^oC

c_{p,g} = specific heat of liquid water = 1.84J/g^oC

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

\Delta H_{vap} = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]

\Delta H=139287.75J=139.28775KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

3 0
3 years ago
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