Ask question

Ask question

All categories

3 years ago

5

Explain why the acid-catalyzed dehydration of an alcohol is a reversible reaction, whereas the base-promoted dehydrohalogenation

of an alkyl halide is an irreversible reaction.

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

electrophile(H⁺) is needed to react with alkene in the first step and nucleophile (OH⁻) is not available in the first step

Explanation:

You might be interested in

Choose the element that IS NOT in the same period as Potassium.

Lesechka [4]

E. sodium is the answer

8 0

3 years ago

How to balance reaction equations

spayn [35]

They will become unequivalent

7 0

2 years ago

In a labratory activity the density of a sample of vanadium is determined to be 6.9g/cm^3 at room temperature. what is the perce

Snowcat [4.5K]

Answer:

c : 13%

Explanation:

Data Give:

Experimental density of vanadium = 6.9 g/cm³

percent error = ?

Solution:

Formula used to calculate % error

% error = [experimental value -accepted value/accepted value] x 100

The reported accepted density value for vanadium = 6.11 g/cm³

Put value in the above equation

% error = [ 6.9 - 6.11 / 6.11 ] x 100

% error = [ 0.79 / 6.11 ] x 100

% error = [ 0.129] x 100

% error = 12.9

Round to the 2 significant figure

% error = 13 %

So, option c is correct

7 0

3 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

4 0

2 years ago

Which statements are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain o

Dvinal [7]

Answer:A, C, D, E, F

Explanation:

A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces

B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.

C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.

D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)

E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.

F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)

4 0

3 years ago