B.Elements
Explanation: they cannot be separated
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
P = nRT/V
P = 3.5 x 10^-3 x 0.082 x 298 /0.5
P = 0.171 m Hg
P = 171 mm Hg
hope this helps
