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1 year ago

Which electron is, on average, close to the nucleus: an electron in a 2s orbital or an electron in a 3s orbital?

1 answer:

6 0

electron is, on average, close to the nucleus: an electron in a 2s orbital or an electron in a 3s orbital is electrons . The electron in 2s orbital are closer to the nucleus than those of 3s orbital.

The 3s orbital electron posses more energy than those electrons in 2s orbital. The size of orbital is determine by the principal quantum number n. The higher the value of n , the farther the orbital from the nucleus . And same as the electron situated to orbital.

Thus, electron is, on average, close to the nucleus: an electron in a 2s orbital or an electron in a 3s orbital is electrons . The electron in 2s orbital are closer to the nucleus than those of 3s orbital.

To learn more about orbital here

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What depends on public opinion and support? A. regulations at a local level B. enforcement of laws C. success of a policy D. rev

Nastasia [14]

i believe the answer is success of a policy. if this was correct please mark brainliest and lmk if you have any more questions x

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4 years ago

1 If 5.80 L of gas is collected at a pressure of 92.0 kPa, what volume will the same gas occupy at 101.3 kPa if the temperature

Anni [7]

Answer:

The volume that the same gas will occupy at 101.3 kPa if the temperature is kept constant is 5.27 L.

Explanation:

As the volume increases, the particles (atoms or molecules) of the gas take longer to reach the walls of the container and therefore collide with them less times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls. In this way pressure and volume are related, determining Boyle's law that says:

"The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

P * V = k

Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:

P1 * V1 = P2 * V2

In this case, you have:

P1= 92 kPa
V1= 5.80 L
P2= 101.3 kPa
V2= ?

Replacing:

92 kPa* 5.80 L= 101.3 kPa* V2

and solving, you get:

$V2=\frac{92 kPa* 5.80 L}{101.3 kPa}$

V2= 5.27 L

The volume that the same gas will occupy at 101.3 kPa if the temperature is kept constant is 5.27 L.

4 0

3 years ago

Which of these is an oxidation half-reaction?

Zepler [3.9K]

Answer:

E= mc sqared

Explanation:

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4 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

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

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

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

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

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

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

4 years ago

1. Except for special cases such as peroxides, oxygen has an oxidation number of _______ in compounds. A. +2 B. –1 C. +1 D. –2

egoroff_w [7]

Answer:

The Typical Oxidation number of Oxygen is –2.

And as the question said... its Oxidation changes to –1 when it is in Peroxides and –½ when its in SuperOxides.

Correct Answer : Option D.

7 0

3 years ago