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ludmilkaskok [199]
4 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]4 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]4 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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How many atoms of nitrogen (N) are in this compound? <br> 3 Pb(NO3)4
Studentka2010 [4]

Answer:

12 nitrogen atoms

Explanation:

This is essentially 3 molecules of Lead (IV) Nitrate, And since there were originally 4 nitrogen atoms in this single ionic compound, there are now 12 total

4 0
3 years ago
Why are solutions considered mixtures?
Alex_Xolod [135]
Solute and solvent form a solution, which means two different kinds of molecules/atoms/compounds are mixed together, therefore, it is a mixture.
6 0
3 years ago
La columna de la izquierda corresponde a los tipos de sales y la columna derecha a los tipos de fórmula que presentan. Relaciona
Galina-37 [17]

Answer:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas  b. MXO

4. Basicas d. M(OH)X

Explanation:

¡Hola!

En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.

Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas  b. MXO

4. Basicas d. M(OH)XO

En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.

¡Saludos!

3 0
3 years ago
What are the major problems associated with the production of nuclear energy?
Sindrei [870]

Answer:

• long time lag between planning and operation.

• cost.

• weapons proliferation risk.

• meltdown risk.

• mining lung cancer risk.

• carbon-equivalent emissions and air pollution.

• waste risk.

Explanation:

hope this help <33

7 0
3 years ago
Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+
DIA [1.3K]

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
  • X - 4 = +1
  • X = +1 + 4
  • X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

6 0
3 years ago
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