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CaHeK987 [17]
3 years ago
9

Quinine is a natural product extracted from the bark of the cinchona tree, which is native to South America. Quinine is used as

an antimalarial agent. When 3.28 g of quinine is dissolved in 25.0 g of cyclohexane, the freezing point of the solution is lowered by 8.43 °
c. The freezing point and Kf constant for cyclohexane can be found here.
Calculate the molar mass of quinine.
Chemistry
1 answer:
evablogger [386]3 years ago
5 0
For the answer to the question above, <span>The formula for freezing point depression is </span>
<span>ΔTf = mkfi </span>

<span>kf is the freezing point constant </span>
<span>i is the Van't Hoff factor which in this case is 1 </span>
<span>m is molality (moles of solute/kg of solvent) </span>
<span>ΔTf is temperature change </span>

<span>ΔTf is 2.17 °C, the molality is the amount of solute Quinine </span>
<span>in the solvent cyclohexane. We cannot calculate moles therefore we need to substitute moles with g/mm. </span>

<span>moles = g/mm so molality=(g/mm)/kg </span>
<span>molality = (0.845/mm)/0.025 = 33.8/mm </span>

<span>2.17 = 33.8/mm(20.8) rearrange </span>

<span>mm = (33.8/2.17)(20.8) = 324g/mol</span>
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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
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<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

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Half reactions for the given cell follows:

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<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t
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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.

<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

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Read more about moles:

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