Answer:
to the left
Explanation:
<u>If the concentration of products is increased for a reaction that is in equilibrium, the equilibrium would shift to the left side of the reaction (the reactant's side). </u>
For a reaction that is in equilibrium, the reaction is balanced between the reactants and the products. According to Le Cha telier's principle, if one of the constraints capable of influencing the rate of reactions is applied to such a reaction that is in equilibrium, the equilibrium would shift so as to neutralize the effects created by the constraint.
<em>Hence, in this case, if the concentration of the products of a reaction in equilibrium is increased, the equilibrium would shift in such a way that more reactants are formed so as to annul the effects created by the increase in the concentration of the products. Since reactants are always on the left side of chemical equations, it thus means that the equilibrium would shift to the left.</em>
Answer:
Option D. 3, 1, 3, 1
Explanation:
From the question given above,
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
The equation can be balance as follow:
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:
3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:
3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃
Now, the equation is balanced.
Thus, the coefficients are 3, 1, 3, 1
The answer would be C there would be 6 nonbonding electrons and 2 bonding electrons.
We are told we have an oxyacid of the formula HOFO. We will assume the atoms are in this order and will draw a proper lewis structure for this compound by first drawing bonds between each of the 4 atoms and then place the remaining electron pairs on each atom:
.. .. ..
H - O - F - O:
·· ·· ··
We can calculate the formal charge of an atom using the following formula:
Formal charge = [# of valence electrons] - [# of non-bonded electrons + # of bonds]
H: Formal charge = [1]-[0+1] = 0
O: Formal charge = [6]-[4+2] = 0
F: Formal charge = [7]-[4+2] = +1
O: Formal charge = [6]-[6+1] = -1
As we can see the overall charge of the molecule is neutral since the fluorine as a +1 charge and the oxygen a -1 charge.
