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Inessa05 [86]
3 years ago
7

Describe the interactions between molecules in a mixture (analyte), the stationary phase and various mobile phases used in colum

n chromatography and the effect of these interactions on the separation of the mixture.
Chemistry
1 answer:
ratelena [41]3 years ago
4 0
<h2>Hydrophobic interactions. </h2>

Explanation:

  • Column chromatography is a technique used for separation in which a mixture having a mobile phase is caused to move in contact with the stationary phase.
  • The Stationary phase used is a solid and the mobile phase used is liquid.
  • There are interactions between the molecules of the analyte, stationary phase, and the various mobile phases like hydrogen bonding and dipole-dipole interactions.
  • The effect of these interactions is that the compound is retained for a longer time and as there are different interactions between stationary and mobile phases they are carried along with the mobile phases with varying degrees and the separation of the mixture is done.
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How many grams of water are produced from the decomposition of 63 grams of lithium hydroxide
Ymorist [56]

Answer:

18.65004 grams H2O

Explanation:

First, we need to write down the balanced chemical equation for the decomposition reaction:

2LiOH -> H2O + Li2O

Since we have grams of LiOH and we need to know the grams of water, we need to convert to moles since we can only compare moles to moles.

The amu of LiOH is 23.947.

The given grams of LiOH is 63.. To convert to moles, we will divide 63 by 23.947..

This gives us 2.6310 moles LiOH..

To convert to moles of H2O (and later grams of H2O), we will use the mole fractions from the balanced equation...

When we look at the balanced equation we can see that 2 moles of LIOH can produce 1 mol of Water, so:

2.6310 moles * \frac{1 molH_{2}O}{2 mol LiOH} = 1.3155 moles H2O

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3 years ago
A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

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