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LiRa [457]
4 years ago
6

13C NMR is a technique in which the total number of signals represents the number of unique carbon atoms in a molecule. Propose

a structure that is consistent with the following data.
a. The IR includes peaks at 1603 and 1495 cm^-1
b. The 13c NMR has a total of 7 signals
c. The compound has one acidic proton.
Chemistry
1 answer:
asambeis [7]4 years ago
4 0

Answer:

D. Poop Butt.

Explanation:

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Object A at 40ºC and object B at 80ºC are placed in contact with each other. Which statement describes the heat flow between the
Arada [10]
By the second law of thermodynamics:
Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system.
Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ). 
Answer:  B ) Heat flows from object B to object A.
6 0
3 years ago
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True or False: The atomic number is the number of neutrons in an element?
klio [65]

Answer:

true

Explanation:

5 0
3 years ago
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
Is particle size one of the factors that affect solubility? How?
Over [174]
Hello!

The reduction in particle size results in an increased rate of solution. It occur because is harder for a a solvent to surround bigger molecules.

Hugs!
4 0
3 years ago
Pls help, needed for chemistry
vovangra [49]

Answer:

go to a calculator and see the answer then make the hypothisis which is the answer

Explanation:

magic

7 0
2 years ago
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