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Afina-wow [57]
2 years ago
8

Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcoho

l with two other carbons attached to the carbon with the hydroxyl group_____.An alcohol with one other carbon attached to the carbon with the hydroxyl group____.An alcohol with three other carbons attached to the carbon with the hydroxyl group____.
Chemistry
1 answer:
pochemuha2 years ago
7 0

Answer:

1). 1-pentanol - <u>Primary</u>

2). 3-ethyl-3-pentanol - <u>Tertiary</u>

3). 2-hexanol - <u>Secondary</u>

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - <u>Secondary</u>

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - <u>Primary</u>

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - <u>Tertiary</u>

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example:  3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

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

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

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

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

8 0
3 years ago
Consider a reactant with an order of 1: What effect does that reactant have on the rate equation when the concentration is doubl
____ [38]

Answer:

1) The reaction rate is double with respect to that reactant

Explanation:

Hello,

By considering the rate law:

-r_A=kC_A

If we double the reactant A concentration, by definition, the rate will be doubled as well since the C_A power is one (order 1), this could be proved just by checking it out in the rate law.

Best regards.

6 0
3 years ago
What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution
sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

Molariy = 0.207 moles / 0.0917L → 2.26M

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the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)

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Suppose you carry out a titration involving 2.50 molar NaOH and an unknown concentration of HBr. To bring the reaction to its en
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Answer:

B

Explanation:

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