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grigory [225]
3 years ago
5

Infrared (IR) and Nuclear Magnetic Resonance (NMR) are two spectroscopic techniques you've encountered in organic chemistry I, C

HM2210. In organic chemistry laboratory, IR and NMR are common tools used to characterize a given product. Consider what you learned in CHM22110, and select the concepts that you feel confident about: Group of answer choices Assigning 1H NMR signals to a given molecule The theory behind NMR Identify splitting patterns I do not recall/understand much about NMR correlate the intensity of a 1H NMR signal to the number of protons The theory behind IR I do not recall/understand much about IR Assigning 13C NMR signals to a given molecule Identify functional groups based on IR absorptions Understanding (de)sheilding Deduce the structure of an unknown given its molecular formula, IR and NMR spectra
Chemistry
1 answer:
zalisa [80]3 years ago
8 0

Answer:

The solution to this question can be defined as follows:

Explanation:

Please find the attached file for the solution:

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Which function allows a user to add another item to a list in Python? order() print() sort() append()
vovikov84 [41]

Answer:

Probably (print)

Explanation:

I think

6 0
3 years ago
Calculate the rms speed of co molecules at 320 k .
mezya [45]
<span>rms = (3RT/M)^1/2 Therefore, the rms of CO is: rms(CO) = [(3)(8.3145 J/mol*K)(320 K)/(0.028 kg/mol)]^1/2 rms = [2.65x10^5 m^2/s^2]^1/2 rms = 533.9181 m/s</span>
8 0
3 years ago
The standard cell potential (E°cell) for the reaction below is +1.10V. The cell potential for this reaction is ________ V when t
alexandr402 [8]

Answer: 0.94 V

Explanation:

For the given chemical reaction :

Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +1.10 V

E_{cell} = emf of the cell = ?

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

E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}

E_{cell}=+1.10-0.16V=0.94V

The cell potential for this reaction is 0.94 V

5 0
3 years ago
Which element is located in group 2 period 2 on the periodic Table?
AnnZ [28]

Answer:

Berilium is located in period 2 group2

Explanation:

you are welcome

4 0
2 years ago
Read 2 more answers
The ph of a 0.24 m solution of dimethylamine is 12.01. calculate the kb value for dimethylamine.
vlada-n [284]
1- First we will get the value of POH from:

PH + POH = 14 

and we have PH = 12.01

∴ POH = 14 - 12.01 = 1.99 

2- Then we need to get the concentration of OH:

when POH = -㏒[OH]

           1.99 = -㏒[OH]

∴[OH] = 0.01 M

now we have the concentration at Equ by subtracting from the initial concentration of OH  = 0.24 M

∴ [OH] = 0.24 - 0.01 = 0.23 M

∴ Kb = (0.01)^2 /  0.22977 M

∴ Kb value = 4.4 x 10^-4

8 0
3 years ago
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