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katovenus [111]
3 years ago
13

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

lution with an optimum pH of 9.2−9.3? CH3COONa/CH3COOH (Ka = 1.8 x 10^−5) NH3/NH4Cl (Ka = 5.6 x 10^−10) NaOCl/HOCl (Ka = 3.2 x 10^−8) NaNO2/HNO2 (Ka = 4.5 x 10^−4) NaCl/HCl
Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

pH=pKa+log\frac{[base]}{[acid]}

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

  • CH₃COONa/CH3COOH (pKa = 4.74)
  • NH₃/NH₄Cl (pKa = 9.25)
  • NaOCl/HOCl (pKa = 7.49)
  • NaNO₂/HNO₂ (pKa = 3.35)
  • NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

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How many moles are in 16 grams of nitrogen molecule
vekshin1
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≈ 1.142 moles

Explanation:
formula is “grams given/molar mass = moles”
nitrogen’s molar mass is 14.0067g
solve:
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How many grams of sodium hydroxide are needed to completely react with 50.0 grams of H2SO4?
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48.0 grams of NaOH I believe

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One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compressi
oksian1 [2.3K]

Answer:

  • final temperature (T2) = 748.66 K
  • ΔU = w = 5620.26 J
  • ΔH = 9367.047 J
  • q = 0

Explanation:

ideal gas:

  • PV = RTn

reversible adiabatic compression:

  • δU = δq + δw = CvδT

∴ q = 0

∴ w = - PδV

⇒ δU = δw

⇒ CvδT = - PδV

ideal gas:

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP = - CvδT

⇒ RδT - RTn/PδP = - CvδT

⇒ (R + Cv,m)∫δT/T = R∫δP/P

⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303

∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5

⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212

⇒ T2/T1 = 2.512

∴ T1 = 298 K

⇒ T2 = (298 K)×(2.512)

⇒ T2 = 748.66 K

⇒ ΔU = Cv,mΔT

⇒ ΔU = (3/2)R(748.66 - 298)

∴ R = 8.314 J/K.mol

⇒ ΔU = 5620.26 J

⇒ w = 5620.26 J

  • H = U + nRT

⇒ ΔH = ΔU + nRΔT

⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)

⇒ ΔH = 5620.26 J + 3746.787 J

⇒ ΔH = 9367.047 J

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beryllium has a higher ionization energy because its radius is smaller. boron has a higher ionization energy because its radius is smaller.

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