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3 years ago

A carbon-carbon triple bond is found in a molecule of

2 answers:

5 0

The answer is (4) butyne. The carbon-carbon triple bond is found in alkynes. And the alkynes has -yne at the end of the name. "-ene" end means that it is alkene and has carbon-carbon double bonds.

jolli1 [7]3 years ago

5 0

Answer: Option (4) is the correct answer.

Explanation:

A hydrocarbon in which all atoms are bonded by single bond and have a general formula $C_{n}H_{2n+2}$ is known as an alkane.

A suffix "ane" adds at the end of the name of this hydrocarbon.

A hydrocarbon in which an atom is bonded by a double bond and have a general formula $C_{n}H_{2n}$ is known as an alkene.

A suffix "ene" adds at the end of the name of this hydrocarbon.

A hydrocarbon in which an atom is bonded by triple bond and have a general formula $C_{n}H_{2n-2}$ is known as an alkene.

A suffix "yne" adds at the end of the name of this hydrocarbon.

Thus, we can conclude that a carbon-carbon triple bond is found in a molecule of butyne.

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Calculate the weight of 5 atoms of Mg.

kumpel [21]

Answer:

2.0179701e-22

Explanation:

8 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

What is the molecular polarity of the following lewis structure

puteri [66]

Pretty sure non polar bc it’s equal

3 0

3 years ago

Which type(s) of matter would need to be separated by chemical methods?

Marta_Voda [28]

Answer:

solids, liquids, gases, or plasma

Explanation:

In chemistry, a chemical substance is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components without breaking chemical bonds. Chemical substances can be solids, liquids, gases, or plasma

3 0

3 years ago

What is reduction potential?

Shtirlitz [24]

Answer:

A reduction potential measures the tendency of a molecule to be reduced by taking up new electrons. ... Standard reduction potentials can be useful in determining the directionality of a reaction. The reduction potential of a given species can be considered to be the negative of the oxidation potential.

Explanation:

8 0

3 years ago

Read 2 more answers