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3 years ago

Predict the number of peaks that you would expect in the proton-decoupled 13C spectrum of each comound.

Chemistry

1 answer:

dangina [55]3 years ago

4 0

Answer:

To interpret a 13C-NMR spectrum we will use some standards very simple. A 13C-NMR spectrum gives us the following information:

1. Indicates the number of non-equivalent carbons in the molecule.

2. Measuring the chemical shift we can intuit the environment

electronic and determine the next functional groups.

3. In this case we cannot count on integration since the different

carbons have different relaxation times.

The number of peaks in the spectrum indicates the number of types of carbon present in the analyzed substance.

The factors that influence the chemical shift of the signals in the 13C NMR are:

electronegativity of carbon bound groups
carbon hybridization

Explanation:

The nuclear magnetic resonance of C13 is complementary to that of H1. This technique is used to determine the magnetic environment of carbon atoms.

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2 years ago

Anything in red is the question

laila [671]

Answer:

Question 19: the three are molecular compounds.

Question 20: CuSO₄.5H₂O

Explanation:

Question 19.

C₂H₄

H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

C₂H₄: two carbon atoms per four hydrogen atoms

HF: one hydrogen atom per one fluorine atom

H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

Question 20. Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

$\dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36$

From which we can solve for x:

$18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5$

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

CuSO₄.5H₂O

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3 years ago

so how many moles of Fe are in 3.10 mole of iron oxide FeO still waiting for answer please, my grandchild needs help that I can'

alisha [4.7K]

The answer is 3.10 because it's still the same amount moles of iron.

7 0

2 years ago

A/an bonds allows metals to conduct electricity

dalvyx [7]

If I understand you correctly, you should fill the gap and decide which kind og bonds allows metals to conduct electricity. If so, without any doubt I can say that the metalic bonds allows metals to conduct electricity. I'm pretty sure it will help you! Regards.

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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