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ludmilkaskok [199]
3 years ago
8

Draw the structure of the compound C9H12 from its proton (1H) NMR spectrum below. First-order spin-spin splitting rules and equa

l coupling constants can be assumed. (Detailed analysis of any non-first order portions of the spectrum will not be required.) Integral ratios to the nearest whole number are (left to right) 5:2:2:3.
Chemistry
2 answers:
Vikki [24]3 years ago
4 0

Answer:

Isopropylbenzene

Explanation:

If you draw the structure, you can see that there are two methyl groups and in between there.

Adjacent to CH3, there are four neighbouring hydrogens, therefore, n=4+1 = 5. The same is for methyl on other side. For carbon present in benzene ring,  there is 2, since there is one hydrogen on benzene per carbon.

Temka [501]3 years ago
4 0

Answer:Draw the structure of the compound C9H12 from its proton (1H) NMR spectrum below. First-order spin-spin splitting rules and equal coupling constants can be assumed. (Detailed analysis of any non-first order portions of the spectrum will not be required.) Integral ratios to the nearest whole number are (left to right) 5:2:2:3.

from A Solvent CDCls 9.0 7.5 6.0 1.5 4.5 Chemical shift, δ (ppm) 3.0 0.0

Explanation: As applicable above

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

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I believe the answer to that is c!!!
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fenix001 [56]
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% v/v = (25 mL ethanol / 25 mL + 150 mL ) x 100
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