Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: <span>[He] 2s</span>²<span> 2p</span>⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. <em>Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.</em>
Answer: Option (e) is the correct answer.
Explanation:
A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.
For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.
Whereas a bond that is formed by sharing of electrons is known as a covalent bond.
For example, 
will be a covalent compound as there is sharing of electron between carbon and bromine atom.
Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.
For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.
Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.
Answer:
The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.
Explanation:
The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:
H2O (s) ⇒ H2O (l)
The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:
Entropy change (ΔSsys) = ΔSproduct - ΔSreactant
= (69.9 - 47.89) J mol/K
= 22.0 J mol/K
Therefore, the value of entropy change is positive.
Now the value of entropy for surrounding ΔSsurr will be,
ΔSsurr = -ΔHfusion/T
= -6012 j/mol/273
= -22.0 J/molK
Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.
Guess and check, test, trial and error, completion.
