Elements<span>simplest form of matter that can exist under normal laaboratory conditions</span>
Answer: Option (b) is the correct answer.
Explanation:
In material bonding, there occurs Vander waal foces between the molecules in which their is either an induced or permanent dipole moment that attract molecules towards each other.
And, due to these forces the molecules are held together.
On the other hand, in a ionic bond there will always be transfer of electrons from one atom to another. This is because on atom which loses its valence electrons acquires a positive charge and another atom which gains the electrons acquires a negative charge.
Hence, these opposite charges strongly gets attracted towards each other forming a strong bond.
Whereas in a covalent bond, there will be sharing of electrons between the combining atoms.
In a metallic bond, there occurs a sea of electrons which is uniformly distributed throughout the solid substance or material.
Thus, we can conclude that the statement, Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other, about material bonding is correct.
As depth increases, the density of the layers decreases.
There are three types of artificial cloning:
Gene cloning: copies a few segments of DNA or genes.
Reproductive cloning: it copies the whole animal.
Therapeutic cloning: produces embryonic stem cells for experiments aimed at creating tissues to replace injured or diseased tissues.
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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