<u>Answer:</u>
<em>Equivalence point and end point are terminologies in pH titrations and they are not the same.
</em>
<u>Explanation:</u>
In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.
At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the <em>pH at equivalence point. </em>
<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution.
</em>
Elements<span> in the same </span>group<span> in the periodic table </span>have similar chemical properties<span>. This is because their atoms </span>have<span> the same number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.
</span>
The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.
Answer:
B. Marginal cost equals long-run average total cost.
Explanation:
The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.
Answer:
a) 0.0007326
b) 0.03223
c) 0.2418
d) 0.2418
Explanation:
To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.
a) 
b) 
c) 
d) 
