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EleoNora [17]
3 years ago
14

PJ’s unknown solid 1) dissolves in hot ethanol, 2) is essentially insoluble in hexane, and 3) is insoluble in cold water, but sp

aringly soluble in warm water. Outline the recrystallization procedure you would suggest she use here.
Chemistry
1 answer:
Gnoma [55]3 years ago
5 0

Answer:

1. The solid is dissolved in the hot solvent (ethanol or water);

2. If the impurities are not dissolved, they are separated by filtration;

3. The solution is then cold, and the crystals of the solid are formed;

4. The solution is filtrated and the pure solid is obtained.

Explanation:

The recrystallization is a process to separate impurities from a solid. The solid with the impurities is dissolved in a hot solvent, and, is insoluble in the cold one. But the impurities must be soluble in the cold solvent, or insoluble in the hot one.

If the impurities are soluble in the cold solvent, then, the crystals of the analyte will be removed, and if they are insoluble in the hot solvent, then its crystal is removed first.

So let's assume that the solid is insoluble in hot ethanol, and soluble in hot water. Because its insoluble in hexane, the recrystallization is not possible with it. So the procedure would be:

1. The solid is dissolved in the hot solvent (ethanol or water);

2. If the impurities are not dissolved, they are separated by filtration;

3. The solution is then cold, and the crystals of the solid are formed;

4. The solution is filtrated and the pure solid is obtained.

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A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

6 0
3 years ago
A train travels at a speed of 30 miles per hour. If 1 mile = 1.6 kilometers, how fast is the train traveling in kilometers per m
Alina [70]
The answer 
 the speed of <span>the train traveling in kilometers per minute can be found by using the following method:
v = </span><span>30 miles per hour
</span><span>1 mile = 1.6 kilometers

just do the calculus by changing miles to kilometers
 </span>1 mile = 1.6 kilometers, so 30 miles= 1.6 km x 30 = 48 km
 so the speed of the train is
v=30 miles / hour = 48km / h = 48km / 60 mn  = 0.8 km/ mn
the answer is
<span>C- 0.8 km/min</span> 
7 0
3 years ago
Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi
lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

Moles of propane =\frac{1.860 g}{44 g/mol}=0.0422 mol

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol

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Answer:

He used Velocity and Radius.

Explanation:

The uncertainty truths contradicts Bohr's thoughts of electrons.

4 0
3 years ago
What main chemicals are in space?
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Hi!

I believe there are 4 main chemicals in space uracil, cytosine, thymine, and pyrimidine :)
5 0
3 years ago
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