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QveST [7]
3 years ago
9

Enter the number of unique signals in the 1H NMR spectrum of ethanol. Assume that the hydroxy proton does NOT exchange with the

NMR solvent.
Chemistry
1 answer:
ss7ja [257]3 years ago
5 0

Answer:

In the 1H NMR spectrum of ethanol three different signals are observed, this is due to the existence of 3 types of hydrogens with different chemical environment. Hydrogens A (3.57 ppm) are more screened than C (1.10 ppm) due to the presence of oxygen (electonegative atom that removes electron density). The chemical environment of hydrogen B (4.78 ppm), attached directly to oxygen, is also different by resonating at a frequency different from the previous ones.

C1HC_{3} -C2HA_{2} -OHB

The hydroxylic hydrogen produces a singlet, the pair of carbon hydrogens one give rise to a quadruplet and the three hydrogens of carbon two produce a triplet.

Explanation:

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1. If you have 5g of pennies how many dozen pennies do you have?
likoan [24]

Answer:

15.69 dozen

Explanation:

Mass of penny = 5 g

Dozens of penny =..?

Next, we shall convert 5 g to gross. This can be obtained as follow:

3824 g = 1000 gross

Therefore,

5 g = 5 g × 1000 gross / 3824 g

5 g = 1.3075 gross

Thus, 5 g is equivalent to 1.3075 gross.

Finally, we convert 1.3075 gross to dozen. This can be obtained as follow:

1 gross = 12 dozen

Therefore,

1.3075 gross = 1.3075 gross × 12 dozen / 1 gross

1.3075 gross = 15.69 dozen

Thus, 5 g of penny is equivalent to 15.69 dozen

4 0
3 years ago
How does the process of waste removal in a unicellular organism compare to the waste removal in a multicellular organism
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The digestive system of multicellular is made up of several organs that work together to break down food so that it may be used in the body. Unicellular organisms have organelles, similar to organs, which help to digest food that will be used by the cell
6 0
3 years ago
Which pair of dispersed phases and dispersing media can never form a colloid?
lana66690 [7]

Answer:

Option B Liquid and Gas

7 0
3 years ago
What can you do differently for
drek231 [11]

Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

3 0
2 years ago
An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the
Brrunno [24]

Answer:

Approximately 81.84\%.

Explanation:

Balanced equation for this reaction:

{\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm  NaCl}\, (aq) + {\rm CaCO_{3}}\, (s).

Look up the relative atomic mass of elements in the limiting reactant, \rm CaCl_{2}, as well as those in the product of interest, \rm CaCO_{3}:

  • \rm Ca: 40.078.
  • \rm Cl: 35.45.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass for both the limiting reactant and the product of interest:

\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}.

\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the quantity of the limiting reactant (\rm CaCl_{2}) available to this reaction:

\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}.

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant (\rm CaCl_{2}) and the product ({\rm CaCO_{3}}) are both 1. Thus:

\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1.

In other words, for every 1\; \rm mol of \rm CaCl_{2} formula units that are consumed, 1\; \rm mol\! of \rm CaCO_{3} formula units would (in theory) be produced. Thus, calculate the theoretical yield of \rm CaCO_{3}\! in this experiment:

\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}.

Calculate the theoretical yield of this experiment in terms of the mass of \rm CaCO_{3} expected to be produced:

\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}.

Given that the actual yield in this question (in terms of the mass of \rm CaCO_{3}) is 13.19\; \rm g, calculate the percentage yield of this experiment:

\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}.

6 0
2 years ago
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