Answer:
The answer is in the explanation.
Explanation:
The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.
The reaction that explain this use is:
KHP + NaOH → KNaP + H2O
<em>where 1 mole of KHP reacts per mole of NaOH</em>
That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added
With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.
The moles of KHP are obtained from the volume and the concentration as follows:
Volume(L)*Concentration (Molarity,M) = moles of KHP
If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.
Answer:
The amount of NaOH required to prepare a solution of 2.5N NaOH.
The molecular mass of NaOH is 40.0g/mol.
Explanation:
Since,
NaOH has only one replaceable -OH group.
So, its acidity is one.
Hence,
The molecular mass of NaOH =its equivalent mass
Normality formula can be written as:
Substitute the given values in this formula to get the mass of NaOH required.
Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g
Answer:
should be A)Melts.Because magma is liquid i think
H :) (I tried sorry if im wrong)
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
