Answer: RbI (aq) and Zn(CN)2 precipitate
Explanation: zinc cyanide is insoluble in water.
<span>Based on solubilities you can separate naphthalene and NaCl by adding the mixture to a volume of water. NaCl is highly soluble in water, so it will dissolve completely, while the insoluble naphthalen will remain solid. Next your filter the solution, washing the solid with pure water to eliminate all the NaCl from it. The NaCl will remain is solution, and you can obtain it by evaporating the water. The other method to separate naphthalene and NaCl is based on sublimation property of naphthalene: you can permit that naphthalene sublimes freely or use vacuum to accelerate the process. You use a closed vessel to catch all the gas, while the NaCl will remain solid.</span>
Answer:
28.0mL of the 0.0500M NaOH solution
Explanation:
<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>
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The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:
H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
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You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.
the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:
0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>
To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:
1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =
<h3>28.0mL of the 0.0500M NaOH solution</h3>
The answer to the question is a
