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Phantasy [73]
3 years ago
5

Consider solutions A and B

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
4 0

Answer: Option c. Solution A has a lower boiling point than solution B

Explanation:

For solution A, i = 1 means the solution is a non-electrolyte. This clearly indicates that solution A contains covalently bonded compound. Covalently bonded compounds have lower boiling when compared to those of the ionic.

For solution B, i = 2 means the compound is an electrolyte and it forms two ions. From this result, it is very clear that the compound in solution A is an ionic compound. Ionic compounds have higher boiling points than covalent compounds

Therefore, solution A will have a lower boiling point than solution B

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Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene
kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}, 2-bromo-<em>2</em>-methylbutane;
  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}, 2-bromo-<em>1</em>-methylbutane.

It is expected that \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3} would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions \text{H}^{+} and negatively-charged bromide ions \text{Br}^{-}.

\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbo<em>cation</em> with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3};
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}.

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has <em>three</em> carbon-carbon single bonds after the addition of the proton \text{H}^{+} as opposed to <em>two</em> as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.  

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.  

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-<em>2</em>-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.

3 0
3 years ago
Attempt 3 During an experiment, a student adds 2.90 g CaO 2.90 g CaO to 400.0 mL 400.0 mL of 1.500 M HCl 1.500 M HCl . The stude
kondor19780726 [428]

Answer:

Explanation:

Equation of the reaction:

CaO(s) + 2H+(aq) -----> Ca2+(aq) + H2O(g)

The ∆Hrxn would be for one mole of CaO reacted or 2 moles of H+, whichever is the limiting reactant.

Number of moles = mass ÷ molar mass

Molar mass of CaO = 40 + 16

= 56 g/mol

moles of CaO = 2.90/56

= 0.0518 mol

Number of moles = concentration × volume

moles of HCl = 400 × 10^-3 × 1.500 = 0.6 moles

Moles of HCl = moles of H+

From the equation, 1 mole of CaO reacted with 2 moles of H+ to give 1 mole of water.

To find the limiting reagent,

0.6 mole of H+/2 moles of H+ × 1 mole of CaO

= 0.3 moles of CaO(> 0.0518 moles)

So, CaO is limiting reactant.

∆H = m × Cp × ∆T

m = density × volume

= 400 × 1

= 400 g

Cp = 4.184 J/g-ºC

∆T = +6 ºC

∆H = 400 × 4.184 × 6

= 10041.6 J

Since the reaction is exothermic,

∆Hrxn = -∆H/mol(CaO)

= -10041.6/0.0518

= -193853 J

= -193.9 kJ/mol.

3 0
3 years ago
A substance is highly malleable and has a shiny luster. Which of the following best explains the probable position of the substa
Aleks04 [339]
The answer is D only metals are shiny and highly malleable
5 0
3 years ago
Read 2 more answers
Consute a tabela periódica e identifique o elemento químico ao qual corresponde cada item:
VMariaS [17]

Answer:

a

Explanation:

7 0
3 years ago
How sulphuric acid react with Glucose ? Give reaction​
Ierofanga [76]

Answer:

Concentrated sulfuric acid can perform a dehydration reaction with table sugar. After mixing, the color changes from white to brownish and eventually to black. The expansion of the mixture is the result of vaporization of water and CO2 inside the container.

8 0
3 years ago
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