Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
Answer:
[OH-] = 3.0 x 10^-19 M
Explanation:
[H3O+][OH-] = Kw
Kw = 1.0 x 10^-14
[H3O+][OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / 3.3 x 10^4 = 3.0 x 10^-19
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
Answer: the basic difference is Exergonic reactions release energy and an endergonic reactions absorb energy .
HOPE THIS HELPS!!!
