How much faster does nitrogen monoxide (NO) effuse in comparison to dinitrogen tetraoxide (N2O4)?
1 answer:
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Answer: 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.
Explanation:
According to the dilution law,
where,
= concentration of ist progesterone gel = 8%w/w
= weight of ist progesterone gel = x g
= concentration of another progesterone gel = 4% w/w
= weight of another progesterone gel = 1.45 g
= concentration of resulting progesterone gel = 5.5%w/w
= weight of resulting progesterone gel = (x+1.45) g
Thus 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation
We noe have:
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96