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djyliett [7]
3 years ago
7

Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?

Chemistry
1 answer:
Eddi Din [679]3 years ago
3 0
Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates  only partially. 

So, most particles of the solute is going to be in the solution of NaCl, 
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
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How much 2.0 M phenylmagnesium chloride solution is needed if adding 2.9 mmol of Grignard reagent to a reaction? Your answer sho
Vladimir [108]

Answer:

1,45 mL

Explanation:

The Grignard reaction is a very important organometallic chemical reaction where the Grignard reagent ( alkyl, vinyl, or aryl-magnesium halides) acts as nucleophile in order to the formation of Carbon-Carbon bonds.

In the problem, the phenylmagnesium chloride is the grignard reagent. The volume of 2.0 M phenylmagnesium chloride solution you need to add 2.9 mmol is:

2,9 mmol × \frac{1mL}{2,0mM} = <em>1,45 mL</em>

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I hope it helps!

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3 years ago
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7 0
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NEED HELP with 7. 8. 9. 10. 11. 12. And 14 !!
aleksley [76]

Answer:

1.      0.00040 calories

2.   8.57 calories

3.   0.196 calories

4.  68 calories

5. 243 calories

6.  83680 joules

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9. 556 (it says calories to calories so it wouldn't change)

10. 28367.52 joules

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13.  0.00234 calories

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Explanation:

6 0
3 years ago
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

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6 0
3 years ago
A piston in a heat engine does 500 joules of work, and 1400 joules of heat are added to the system. Determine the change in inte
san4es73 [151]

Answer : The change in internal energy is, 900 Joules.

Solution : Given,

Heat given to the system = +1400 J

Work done by the system = -500 J

Change in internal energy is equal to the sum of heat energy and work done.

Formula used :

\Delta U=q+w

where,

\Delta U = change in internal energy

q = heat energy

w = work done

As per question, heat is added to the system that means, q is positive and work done by the system that means, w is negative.

Now put all the given values in the above formula, we get

\Delta U=(+1400J)+(-500J)=900J

Therefore, the change in internal energy is 900 J.

The change in internal energy depends on the heat energy and work done. As we will change in the heat energy and work done, then changes  will occur in the internal energy. Hence, the energy is conserved.

8 0
3 years ago
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