Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Answer:
0.0000098 should be the answer
Explanation:
Answer:
765.0 grams CuS
Explanation:
The limiting reagent is the reactant which completely reacts before the other reactant(s) is used up. When 9.0 moles Na₂S and 8.0 moles CuSO₄ react, it appears that CuSO₄ is the limiting reagent. You can tell because it results in the production of less product.
You can determine the mass of CuS by multiplying the moles by the molar mass. It is important to arrange the ratio in a way that allows for the cancellation of units.
Molar Mass (CuS): 95.62 g/mol
8.0 moles CuS 95.62 g
------------------------- x ----------------------- = 765.0 grams CuS
1 mole
