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2 years ago

7

Draw the major product for the reaction of 1-butyne with water in the presence of catalytic TfOH (i.e., CF3SO3H). Then answer th

e additional question regarding this transformation.

1 answer:

Mademuasel [1]2 years ago

4 0

Answer:

2-Butanone

Explanation:

From the given information:

The presence of mercury as an acid catalyst brings about the addition of water to the triple bond which yields enol. Then, according to Markownikov's rule and after tautomerism has occurred, we have a methyl ketone ( 2- Butanone) as the product.

The answer regarding the transformation is addition and hydration.

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Titanium dioxide, TiO₂, reacts with carbon and chlorine to give gaseous TiCl₄: TiO₂+2C+2CI₂−TiCI₄+2CO The reaction of 7.39 kg ti

Dima020 [189]

Answer:

17.57kg of $TiCl_{4}$ and its percentage yield is 81.0%

Explanation:

Through the reaction you can get the theoretical amount of $TiCl_{4}$ that must be produced.

$7.39kgTiO_{2}x\frac{1kmolTiO_{2} }{79.867kgTiO_{2}}x \frac{1kmolTiCl_{4}}{1kmolTiO_{2}}x\frac{189.867kgTiCl_{4} }{1kmolTiCl_{4}}=17.57kgTiCl_{4}$

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage

$yield=\frac{actual amount}{theoretical amount}x100= \frac{14.24kg}{17.57kg}x100$ =81.0%

3 0

2 years ago

A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p

Zepler [3.9K]

1) Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:

<span><span>Determine the Rf values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent.</span>Which compound, A, B, or C, is the most polar?<span>What would you expect to happen to the Rf values if you used acetone instead of hexanes as the eluting solvent?</span><span>How would the Rf values change if eluted with hexanes using an alumina TLC plate?</span></span>

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2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?

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3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

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4) A student spots an unknown sample on a TLC plate. After developing in hexanes/ethyl acetate 50:50, he/she saw a single spot with an Rf of 0.55. Does this indicate that the unknown material is a pure compound? What can be done to verify the purity of the sample?

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5) Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC plate and developed in a hexanes/ethyl acetate solvent mixture. Predict the relative Rf values for the three compounds in the sample.

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6) Plate A, below, represents the TLC chromatogram of a compound run in hexanes. The same compound was then spotted on a large TLC plate and again run in hexanes. Which TLC plate, B, C, or D, correctly represents how far the compound would run on the longer plate?

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Back to TLC

Original content © University of Colorado at Boulder, Department of Chemistry and Biochemistry.
The information on these pages is available for academic use without restriction.

3 0

1 year ago

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

Viktor [21]

Answer: Your friend is incorrect.

Explanation: If we have an object or something that isn’t moving, (let’s say a notebook on a desk). If there is change, and the notebook moves, there is acceleration. Force = Mass times acceleration, f = m*a. There has to be a force, first of all. If you touched the notebook and moved it, some of your energy is transferred and now the notebook has kinetic energy. If our system is you and the notebook, the total energy doesn’t change. the energy is transferred, but doesn’t change. Your friend is not correct. Please give brainliest hope this helped!

6 0

1 year ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t

-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒ ABCC = $\frac{4RBCC}{\sqrt{3} }$

⇒ = $\frac{4\times 0.12584}{\sqrt{3}}$

⇒ = $0.29062 \ nm$

Likewise AFCC as well as RFCC are interconnected by

√2AFCC = 4RFCC

⇒ AFCC = $\frac{4RFCC}{\sqrt{2}}$

⇒ = $\frac{4\times 0.12894}{\sqrt{2} }$

⇒ = $0.36470 \ nm$

Now,

The Change in Percent Volume,

= $\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent$

= $\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent$

= $\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent$

= $\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent$

= $97.62 \ percent (approximately)$

Note: percent = %

7 0

2 years ago