Atomic weight of a Platinum:
195.078 u = 195.078 * 1.66 * 10^(-27) kg = 3.2382948 * 10^(-25) kg
m ( Pt ) = 3.2382948 * 10^(-25) * 5.35 * 10^(21) =
= 1.732488 * 10^(-3) kg = 1.732448 g
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
= 7.55 - 1.95 = 5.6
Answer:
20 L of Cl₂
Solution:
The reaction is as follow,
H₂C₂ + 2 Cl₂ → H₂C₂Cl₄
According to equation,
167.84 g (1 mole) H₂C₂Cl₄ is produced by = 44.8 L (2 mole) of Cl₂
So,
75 g of H₂C₂Cl₄ will be produced by = X L of Cl₂
Solving for X,
X = (44.8 L × 75 g) ÷ 167.84 g
X = 20 L of Cl₂
Answer:
P³⁻ + Cl⁻ --> PCl₃
Explanation:
PCl₃: phosphorus trichloride. prefix in front of chloride is "tri"–meaning three.
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