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3 years ago

What is the difference between atomic elements and molecular elements?

2 answers:

6 0

The molecular element describes the amount of protons, neutrons, and electrons found in an atom

An atomic element is the subject in the formula such as ... sulfur, carbon, or oxygen

kramer3 years ago

4 0

Woooooooobbbbbbbbbbbb

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What is responsible for the larger size of an anion in comparison with the atom from which it is form?

GarryVolchara [31]

Answer:

electron-electron repulsion

Explanation:

When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.

In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.

To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.

Download pdf

4 0

3 years ago

Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no

nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True. Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1

for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)

D) True. Standard conditions refer to

T= 298 K

pH= 7

P= 1 atm

C= 1 M for all reactants

Water = 55.6 M

5 0

3 years ago

1 Flower color D. What color are the parents?

timofeeve [1]

Answer:

what is the baby flowers color?

Explanation:

if the baby flower is purple the parents were purple

6 0

2 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

Which law relates temperature, pressure, volume, and the number of moles of a gas under perfect conditions? Boyle's law Charles'

Shalnov [3]

Answer: Avogadro's Law

Explanation: It states that the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature of the gas.

7 0

3 years ago

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