Answer:
A) Polar molecules have a non-zero net dipole moment. Both CO2 and H2O have two polar bonds. However the dipoles in the linear CO2 molecule cancel each other out, meaning that the CO2 molecule is non-polar. The polar bonds in the bent H2O molecule result in a net dipole moment, so H2O is polar.
B) In PCl5, in addition to equitorial bonds which lie in the same plane, there are axial it suffers more repulsion it is longer than the other bonds; hence PCl5 decomposes to form PCl3. So PCl5 is highly reactive. Source(s): 5 Heavy ElectroNegative atoms in One Molecule Lone Pair.
C) In h2o there is hydrogen bonding because oxygen has a high electronegativity (only second to flourine). so, the hydrogen atoms from other molecules of water forms a hydrogen bonds with oxygen resulting in intermolecular hydrogen bonding. therefore at room temperature h2o is a liquid and h2s is a gas.
Hope this helps also can you answer my question after since I helped you prepare for your exam.
Explanation:
The answer is D because there is no forces of attraction or repulsion between gas particles .
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
<span>Cellulose strengthens plant cells.</span>
Converting the specific heat into kJ/kg°C, we get that the specific heat capacity is 3.52 kJ/kg°C
Now, we use the equation:
Q = mcΔT, where Q is the heat energy, m is the mass of substance, c is the specific heat capacity and ΔT is the change in temperature of the mass of substance. Substituting these values, we get:
Q = 1.4 x 3.52 x (27.45 - 28.5)
Q = -5.17 kJ
The reaction absorbs 5.17 kilojoules of energy.
