Which of the following represents six molecules of water? 6HO 2 6H 2O 1 6H 2O H 6O
2 answers:
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Answer:
- Nitrogen has four pairs of electrons: 3 bonds and 1 lone pair in the valence shell;
- Electrons repel one another based on the VSEPR theory;
- Nitrogen has a total of 7 protons (its atomic number is 7) in its nucleus.
Explanation:
The shape and the bond orientation of molecules and ions are both explained by the valences shell electron pair repulsion theory (VSEPR).
Ammonia, , is a molecule which contains three N-H bonds, as well as one lone pair on nitrogen. According to the VSEPR theory, molecules try to acquire a shape which would minimize the repulsion exhibited by the electron clouds present, that is, between the bonding (shared in a bond) and non-bonding (lone pair) electrons.
In VSEPR, our main step is to calculate the steric number, this is the sum of the number of bonds (ignoring the multiplicity of any bond) and the lone pairs on a central atom. In ammonia, we have 3 bonds and 1 lone pair, totaling to a steric number of 4. A steric number of 4 without any lone pairs on a central atom and just bonds would yield a tetrahedral shape with bond angles of .
Now, in this case, since we have a lone pair instead of a bond, it is repelling stronger decreasing the bond angles to about .
The greater the number of lone pairs, the lower the angle becomes.
To summarize:
- Nitrogen has four pairs of electrons: 3 bonds and 1 lone pair in the valence shell;
- Electrons repel one another based on the VSEPR theory;
- Nitrogen has a total of 7 protons (its atomic number is 7) in its nucleus.
<u>Answer:</u> The concentration of required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of , we use the equation:
Moles of = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
For the given chemical equations:
Net equation:
To calculate the equilibrium constant, K for above equation, we get:
The expression for equilibrium constant of above equation is:
As, is a solid, so its activity is taken as 1 and so for
We are given:
Putting values in above equations, we get:
Hence, the concentration of required will be 0.285 M.