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3 years ago

What types of organic molecules have the same molecular formula, but atoms that are joined together in different ways?

Chemistry

2 answers:

hjlf3 years ago

8 0

Answer:

A. structural isomers :)

Explanation:

i just took the test & A was correct !

xenn [34]3 years ago

6 0

The main premise of an isomer is that it is a molecule with the same molecular formula but some sort of difference between the molecule and the isomer. Geometric isomers have the same structure and vary differently, the other non isomer answers are red herrings really and so the correct answer would be structural isomer as these are molecules with the same formula but are structurally different or the atoms are arranged in different ways. Hope that helps!!

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In an experimental investigation, the variable that the researcher changes or manipulates in order to see its effects is called

Ghella [55]

Independent variables don't change

Dependent variables change as a result of the alteration made during the experiment

4 0

3 years ago

Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi

sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Entropy change of the reaction ; ΔS

$\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]$

$=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0

3 years ago

Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(b) Hg₂²⁺(aq)

Olegator [25]

The given reaction is not spontaneous.

We must recognize changes in oxidation states that take place across elements in order to balance these equations. To accomplish this, keep in mind following guidelines:

A neutral element on its own has an oxidation number of zero.For a neutral molecule, the total number of oxidations must be zero.The net charge of an ion is equal to the sum of its oxidation numbers.In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.In a compound with no oxygen present the other halogens will also prefer -1.

One of the mercury atoms is oxidized from +1 to +2 in the simple aqueous ion, for a loss of 1 electron.

Oxidation half-reaction:

$0.5 Hg^{2+} _{2} (aq)$ → $Hg^{2+} (aq) + 1e^-$