The answer to this item depends entirely to the chemical reaction. If the compound, NH4Cl, is in the left hand side of the reaction, when it is added, the reaction will shift to the left. In the same manner, when the compound is in the right-hand side of the reaction, the reaction will shift to the right.
This happens because initially the reaction is in equilibrium and adding another compound to it will most likely lead to the shifting of the reaction.
There are four types of chemical bonds essential for life to exist: Ionic Bonds, Covalent Bonds, Hydrogen Bonds, and van der Waals interactions. We need all of these different kinds of bonds to play various roles in biochemical interactions. These bonds vary in their strengths.
To play a variety of roles in biochemical interactions, we require all of these diverse sorts of linkages. The tensile strength of these linkages varies. In chemistry, we consider the range of strengths between ionic and covalent bonds to be overlapping. This indicates that in water, ionic bonds usually dissociate. As a result, we shall consider these bonds from strongest to weakest in the following order:
Covalent is followed by ionic, hydrogen, and van der Waals.
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Answer:
0.45 M
Explanation:
Now we need to make use of the dilution formula
C1V1= C2V2
C1= initial molarity of the solution which is the unknown
V1 = initial volume of the solution which is 175 ml
C2 =final molarity of the solution= 0.315 M
V2= final volume of solution= 250 ml
From C1V1= C2V2 we have;
C1= C2V2/V1
C1= 0.315 × 250/ 175
C1= 0.45 M
<span>To determine the pH of the solution given, we make
use of the acid equilibrium constant (Ka) given. It is the ratio of the
equilibrium concentrations of the dissociated ions and the acid. The
dissociation reaction of the CH3COOH acid would be as follows:
</span>CH3COOH = CH3COO- + H+<span>
The acid equilibrum constant would be expressed as follows:
Ka = [H+][</span>CH3COO-] / [CH3COOH] = 1.8× 10^–5
<span>
To determine the equilibrium concentrations we use the ICE table,
CH3COOH H+ </span>CH3COO<span>-
I 1.60 0 0
C -x +x +x
----------------------------------------------------------------
E 1.60-x x x
</span>1.8× 10^–5 = [H+][CH3COO-] / [CH3COOH] <span>
1.8 x 10^-5 = [x][x] / [0.160-x] </span>
Solving for x,
x = 1.69x10^-3 = [H+] = [F-]
pH = -log [H+] = -log [1.69x10^-3] = 2.8
1: Decomposition reaction
2: Combination reaction
3: product
4: Reactant
