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3 years ago

13

All of the covalent carbon-carbon bonds in unsaturated hydrocarbons share 2 pairs of electrons.

2 answers:

Evgen [1.6K]3 years ago

5 0

False. carbon-carbon bonds that share 2 pairs of electrons are double bonds. An unsaturated hydrocarbon isnt necessary to only have double bonds. they can also have single bonds or triple bonds.

IgorLugansk [536]3 years ago

5 0

Answer: The given statement is false.

Explanation:

An unsaturated hydrocarbon is a chain of carbon and hydrogen atoms in which adjacent carbon atoms are attached together through double or triple bond.

Since it is known that carbon atom forms covalent compounds so, an unsaturated hydrocarbon not only contains double or triple bonds it also has single bonds in between.

Thus, we can conclude that the statement all of the covalent carbon-carbon bonds in unsaturated hydrocarbons share 2 pairs of electrons, is false.

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In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

<u>Answer:</u> The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

<u>Reaction 1:</u> $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 2:</u> $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 3:</u> $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 4:</u> $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

<u>Reaction 5:</u> $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

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A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o

Andre45 [30]

Answer:

* $x_{CH_3OH}=0.0425$

* $\%m/m_{CH_3OH}=7.31\%$

* $m=2.46m$

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

$x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}$

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

$n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O$

Hence, mole fraction is:

$x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425$

Next, mass percent is:

$\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%$

And the molality, considering the mass of water in kg (0.185 kg):

$m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m$

Regards.

7 0

3 years ago