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Masteriza [31]
3 years ago
15

(a) The 1H NMR spectrum of DMF exhibits three signals. Upon treatment with excess LAH followed by water, DMF is converted into a

new compound that exhibits only one signal in its 1H NMR spectrum. Explain. DMF, like most , exhibits restricted rotation about the bond between the carbonyl group and the nitrogen atom. This restricted rotation causes the groups to be in different electronic environments. They are not chemically equivalent, and will therefore produce different signals (in addition to the signal from the other proton in the compound). Upon treatment with excess LAH followed by water, DMF is to an amine that exhibit restricted rotation. As such, the methyl groups chemically equivalent and will together produce . (b) Based on your answer to part a, how many signals do you expect in the 13C NMR spectrum of DMF? Restricted rotation causes the methyl groups to be in electronic environments. As a result, the 13C NMR spectrum of DMF should have .

Chemistry
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

Explanation:

check below for the solution

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Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)
vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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3 years ago
We can preserve natural resources by
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The answer is c, relying on renewable energy sources
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3 years ago
What are the horizontal parts of the periodic table called
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The Answer is Periods .
5 0
3 years ago
Can you please help me and can you show your work please
natali 33 [55]

Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

4. 340.6 grams of ammonia.

5. 4.5x10²³ molecules of sulfur dioxide.

Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

Best regards!

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3 years ago
At what point on the hill will the car have zero gravitational potential energy? A) Half-way down the hill. B) At the top of the
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The answer is C...........
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