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lutik1710 [3]
3 years ago
14

When 1.04 g of an unknown non-electrolyte is dissolved in 50.0 g of cyclohexane, the freezing point decreased by 3.91 degrees C.

If the Kfp of the solvent is 20.1 K/m, calculate the molar mass of the unknown solute.
Chemistry
1 answer:
Advocard [28]3 years ago
8 0

Answer:

The molar mass of the unknown solute is 106,9 g/m

Explanation:

Cryoscopic descent formula to solve this

ΔT = Kf . m

Be careful because units in Kfp are K/m, so let's get the ΔT degrees °C in K

3,91°C  = 3,91 K

It's a difference, in the end it does not matter

For example you can have 5° C as the final temperature and as initial, 1,09 °C -- ΔT is 5 - 1.09 = 3.91

What happens in Kelvin?

5°C + 273 = 278 K

1,09° C + 273 = 274,09 K

ΔT = 278 K - 274,09 K = 3,91 K

3,91 K = 20,1 K/m * m

3,91 K / 20,1 m/K = m

0,194 = m (molality)

Molality means moles from solute in 1 kg of solvent.

1kg = 1000 g

1000 g ________  0,194 moles

50 g _________ x

x = (50 g * 13,77 moles) / 1000 g = 9,72 *10-3 moles

Moles = mass / molar mass

Molar mass = mass / moles

Molar mass = 1,04 g / 9,72 *10-3 moles

Molar mass = 106,9 g/m

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Which 2 resonance forms destablize the carbocation intermediate if bezonitrile undergoes chlronation at the ortho or para positi
Anna11 [10]

The question is incomplete, the complete question is shown in the image attached

Answer:

A and B

Explanation:

The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.

The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.

Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.

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3 years ago
For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No
Pachacha [2.7K]

Answer:

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The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
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The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_2 = final pressure of O_2 gas at STP= 760 torr

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T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

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3 years ago
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1.38 moles of oxygen

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2.76: x

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