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3 years ago

Propose the structure of the following HNMR data.

Chemistry

1 answer:

Varvara68 [4.7K]3 years ago

7 0

Answer:

1-Chloropropane is likely the answer (attached a picture)

Explanation:

First off there are 3 peaks and 3 carbons which indicates to me that this will be a chain without any symmetry and that each carbon has hydrogens on it.

Second the triplet at 1.0 that integrates to 3 likely correlates to a CH3 (methyl) group. Peaks are very upfield triplets that integrate to three are almost always methyl peaks.

Third the triplet at 3.7 is indicative of being next to the halogen. Hydrocarbons by themselves do not have peaks that far downfield meaning that its shift could only be explained by the chlorine being involved. Also we know that this can't be next to the methyl group since its multiplicity is to low to be next to it.

That leaves the multiplet at 1.75 being the hydrogens on the middle carbon which also makes sense since it is more downfield then the methyl group (due to being closer to the chlorine) but is not far enough downfield to say the chlorine is there. It also makes sense that it is a multiple since it would be a hextet due to the adjacent 5 hydrogens which can't always be resolved.

I hope this helps and let me know if anything is unclear or needs further explanation.

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A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The e

kumpel [21]

Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

$q_{a} = -q_{b}$

$m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b}$

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

$m_{a}$ : is the water's mass = 72.0 g

$C_{a}$ : is the specific heat of water = 4.184 J/(g°C)

$T_{i_{a}}$ : is the initial temperature of the water = 19.2 °C

$T_{f_{a}}$ : is the final temperature of the water = 25.5 °C

$m_{b}$ : is the metal's mass = 141 g

$C_{b}$ : is the specific heat of metal =?

$T_{i_{b}}$ : is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$ : is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

$72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)$

$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

7 0

3 years ago

Which scenario involves a

olga_2 [115]

Answer:

Amount of products and reactants is constant...Although,if any of the factors that interfere this scenerio is involved then there will be a shift or change...

6 0

3 years ago

Last one i think.......

MatroZZZ [7]

Answer is c i believe

4 0

3 years ago

Read 2 more answers

Help me pretty please

Natalka [10]

Answer:

The top answer