Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar
1.062 mol/kg.
<em>Step 1</em>. Write the balanced equation for the neutralization.
MM = 204.22 40.00
KHC8H4O4 + NaOH → KNaC8H4O4 + H2O
<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)
Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)
= 4.035 mmol KHP
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)
= 4.035 mmol NaOH
<em>Step 4</em>. Calculate the mass of the NaOH
Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)
= 161 mg NaOH
<em>Step 5</em>. Calculate the mass of the water
Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g
= 37.973 g
<em>Step 6</em>. Calculate the molal concentration of the NaOH
<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg
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