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2 years ago

4

.An alkane reacts with a halogen to form a halocarbon. This is an example of _____. an elimination reaction a substitution reac

tion a condensation reaction a dehydration reaction

2 answers:

svet-max [94.6K]2 years ago

6 0

I think it is an example of a substitution reaction. Alkanes are saturated hydrocarbon which means each carbon atom has a maximum number of bonds, they lack double or triple bond. A substitution reaction involves substituting hydrogen atom by another element such as halogens. For example Ethane is the second member of alkane when reacted with a halogen such as chlorine, one chlorine atom replaces one hydrogen atom to form chloroethane and the other bromine combines with hydrogen to form hydrogen bromide.
CH3CH3 + Br2 = CH3CH2Br +HBr

allochka39001 [22]2 years ago

5 0

Answer:

the answer your looking for is substitution reaction

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In this, the physical state of ice is changing from solid to liquid state.

$H_2O(s)\rightleftharpoons H_2O(l)$

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2 years ago

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Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres

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2 years ago

SEP Use Models Why does the number of electrons in each principal energy shell increase as the number of the shell increases?

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Answer:

See explanation

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1 year ago

Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A & B B &am

Rina8888 [55]

Answer:

A) homotopic and B) enantiotopic

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6 0

2 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

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

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

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

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

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago