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Feliz [49]
3 years ago
9

Choose the solvent below that would show the greatest freezing point lowering when used to make a 0.20 m nonelectrolyte solution

.
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
8 0
Missing question:
a) benzene, Kf = 5.12°C/<span>m.
</span><span>b) ethanol, Kf = 1.99°C/<span>m.
c) </span></span><span>chloroform, Kf = 4.70°C/<span>m.
d) </span></span><span>carbon tetrachloride, Kf = 29.9°C/<span>m
e) </span></span>diethyl ether, Kf = 1.79°C/m.
Answer is: d) carbon tetrachloride, Kf = 29.9°C/m.
b(solution) = 0,2 m.
ΔT = Kf · b(solution).
Because molality (b) is constant, greatest freezing point lowering is solution with highest Kf (<span>the freezing point depression constant).</span>


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Draw the major organic product in the reaction scheme below. Be sure to clearly show stereochemistry (if applicable).
Juliette [100K]

Answer:

see explanation below

Explanation:

You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.

According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:

R: cyclopentane

R - C ≡ C (-)

Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:

R - C ≡ C - CH3

Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:

R - CH = CH - CH3

In picture 2, you have the reaction and mechanism.

6 0
3 years ago
What is the best statement about the data collected in Amir’s table?
Strike441 [17]

Answer: The answer is D :)

Explanation:

3 0
3 years ago
Read 2 more answers
A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

6 0
3 years ago
The surface of what he can act like a sort of skin due to a Liquid called
PIT_PIT [208]
 A Liquid called Surface Tension. 
8 0
4 years ago
30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?
USPshnik [31]
Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 &deg; C = 23+273.15 = 296.15 &deg; K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

</span>
5 0
3 years ago
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