5F2 + 2NH3 --> N2F4 + 6HF
<span>60.1g NH3 / 17g/mole = 3.54moles NH3 </span>
<span>3.54moles NH3 x (5 F2 / 2NH3) x 38g/mole = 335.85g required </span>
<span>5.25g HF / 20g/mole = 0.262moles HF </span>
<span>0.262moles HF x (2NH3 / 6HF) x 17g/mole = 1.49g required </span>
<span>209g / 38g/mole = 5.5moles F2 </span>
<span>5.5moles F2 (1 N2F4 / 5F2) x 66g/mole = 72.6g produced </span>
<span>Li3N + 3H2O --> NH3 + 3LiOH </span>
<span>(37.7g / 34.7g/mole) x (3H2O / 1 Li3N) x 18g/mole = 58.67g required </span>
<span>1.08moles Li3N (1NH3 / 1Li3N) x 6.022x10^23molecules/mole = 6.54x10^23 molecules </span>
<span>10.3L at STP: 10.3L / 22.4L/mole = 0.46moles NH3 produced </span>
<span>0.46moles NH3 x (1Li3N / 1NH3) x 34.7g/mole = 15.96g</span>
Answer:
2-nitrophenol is more acidic than 3-nitrophenol
benyl amine is more basic than aniline
dimethylamine is more basic than triethylamine
Explanation
Mesomeric effect stabilizes the anion formed when hydrogen of the the -OH group in the 2-nitrophenol is lost. This mesomerism is most pronounced at the 2 and 4 position.
Stronger bases have more available lone pair of electrons. The lone pair of electrons in aniline is delocalized over the ring. Hence it is unavailable for protonation while the lone pair on benzyl amine is available for protonation
Diethylamine is a stronger base because it has a balance of both inductive electron donation by alkyl groups, and solvation of the conjugate acid unlike triethylamine. Hence Diethylamine is a stronger base than triethylamine
Answer:
Chiral:
3-Methyl-2-butanol
Achiral:
1-Pentanol
3-Pentanol
3-Methyl-1-butanol
Explanation:
Only the molecule that bears a carbon atom surrounded by 4 different groups can be said to be Chiral, the other molecules don't satisfy this property.
The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M
<h3>Balanced equation </h3>
H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 1
- The mole ratio of the base, KOH (nB) = 3
<h3>How to determine the molarity of H₃PO₄ </h3>
- Volume of acid, H₃PO₄ (Va) = 10.2 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
- Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
- Molarity of acid, H₃PO₄ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 10.2) / (0.2 × 53.5) = 1 / 3
(Ma × 10.2) / 10.7 = 1 / 3
Cross multiply
Ma × 10.2 × 3 = 10.7
Ma × 30.6 = 10.7
Divide both side by 30.6
Ma = 10.7 / 30.6
Ma = 0.35 M
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1.2*104 is 12000 it is easy just try ur best
