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3 years ago

What splits into carbon and oxygen atoms

Chemistry

1 answer:

OLga [1]3 years ago

8 0

When light breaks apart CO2, the molecule normally splits into carbon monoxide (CO) and an oxygen atom (O).

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Which ion has the lower ratio of charge to volume? Explain. (a) Br⁻ or I⁻

Leto [7]

Due to its bigger size compared to bromide ions, iodide ions have a poorer charge-to-volume ratio.

Iodide ion has a lower ratio of charge to volume because iodine is bigger and has the same charge as both other ions. Ca ions have a lower ratio of charge to volume than Sc ions because they are larger in size and have less charge than Sc ions. Due to its bigger size compared to potassium ions, bromide ions have a poorer charge-to-volume ratio. Sulfate ions have a larger charge density than ClO4 ions do because sulfate ions have more charge than ClO4 ions do.

Learn more about Iodide ion here-brainly.com/question/20533826

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4 0

2 years ago

Which of the following cycles does not involve living organisms?

Aneli [31]

The correct answer is letter C. Rock Cycle. Living organisms involved in carbon cycle, oxygen cycle, and in nitrogen cycle. These are involved in the air that living organisms are taking in and out.

7 0

3 years ago

Read 2 more answers

A chemist fills a reaction vessel with 3.82 atm methanol (CH,OH) gas, 7.56 am oxygen (O2) gas, 5.29 atm carbon dioxide (CO2) gas

alisha [4.7K]

Answer: The Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

Explanation:

The equation used to calculate Gibbs free energy change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})]$

We are given:

$\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{p}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$Q_{p}$ = Ratio of concentration of products and reactants = $\frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}$

$p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm$

Putting values in above expression, we get:

$\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ$

Hence, the Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

5 0

3 years ago

Help please help!!!!

stiks02 [169]

Answer:

Explanation:

6 0

3 years ago

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Cyclohexane (δh° = 936 kcal/mol ) and methylcyclopentane (δh° = 941 kcal/mol) have the same molecular formula but different stra

nadya68 [22]

The difference of the structures of the two isomers are shown in the picture. Generally, cyclic alkanes are much easier to break than straight-chained alkanes. When the molecules are cyclic, they are close to each other, thus lesser bond angles. Because they are closer, repulsion forces could be greater than attractive forces. That is why cyclohexane needs lesser energy of 936 kcal/mol compared to 941 kcal/mol because there is an additional straight-chained methyl substituent to break in methylcyclopentane.

4 0

3 years ago