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bija089 [108]
3 years ago
8

Classify these structures as hemiacetal, acetal, or other.

Chemistry
2 answers:
saveliy_v [14]3 years ago
8 0
Hemiacetals are compounds that are derived from aldehydes and ketones while acetals are compounds with a central carbon atom with 4 bonds attached to it. It must be saturated and has a tetrahedral geometry.
Pepsi [2]3 years ago
6 0

Explanation:

Hello!

Let's solve this!

A hemiacetal is the product of a reaction between an aldehyde and an alcohol.

The structure is

      R

       |

RO-C-H

       |

      OH

An acetal has two alkoxy groups. The structure is :

        H

         |

 OH-C-OR '

         |

        OR ''

Of the compounds given, the first two are hemiacetals.

The next three are acetals and the last two are others.

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A liquid-phase isomerization is carried out in a 1000-gal CSTR that has a single impeller located halfway down the reactor. The
igomit [66]

Answer:

The conversion achieved for the first CSTR impeller is 0.382

Discrepancy = 0.188

Explanation:

The impeller divides the CSTR into 2 equal reactors of volume 500gal

Using V = FaoX/ (-ra)

500gal = Fao×Xa/[(KCao^2( 1 -X1)^2]

500gal = CaoVoX1/ KCao^2(1-X1)

500gal= 500gal × X1'/(1 - X1)^2

(1 -X1)^2 = X1

X1^2 - 3X1 + 1 = 0

X1= 0.382

Conversion achieved in the first CSTR is 0.382

Actual measured CSTR = 57% =57/100=0.57

Discrepancy in the conversions= 0.57 -0.383 =0.188

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3 years ago
Select the single best answer. Which mechanism properly shows the movement of electrons in the reaction? 2xsafari
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Answer:

D

Explanation:

We must study the reaction pictured in the question closely before we begin to attempt to answer the question.

Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.

Hence we must show electron movements in both species using a half arrow.

5 0
3 years ago
An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH . The pKa
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<u>Answer:</u> The pH of the solution is 12.61

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ......(1)

  • <u>For hydrazoic acid:</u>

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol

  • <u>For NaOH:</u>

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol

The chemical reaction for hydrazoic acid and NaOH follows the equation:

                   HN_3+NaOH\rightarrow NaN_3+H_2O

<u>Initial:</u>           0.291        0.342

<u>Final:</u>                  0          0.051                 0.291      0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L    (Conversion factor:  1 L = 1000 mL)

  • <u>For NaOH left:</u>

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.0408M

Putting values in above equation, we get:

pOH=-\log(0.0408)\\\\pOH=1.39

To calculate pH of the solution, we use the equation:

pH+pOH=14\\\\pH=14-1.39=12.61

Hence, the pH of the solution is 12.61

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