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scZoUnD [109]
3 years ago
11

Number of protons for oxygen 16

Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0
8 protons in oxygen bc it has an atomic number of 8
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What is the net charge of a nitrogen atom with 8 electrons
DaniilM [7]

Answer:

– 1

Explanation:

From the question given above, we obtained the following:

Electron = 8

Net charge of nitrogen =.?

Nitrogen has atomic number of 7. This also means that nitrogen has 7 proton because atomic number of an element is the equal to number of protons in the atom of the element.

Thus, we can obtain the net charge of nitrogen with 8 electrons by calculating the difference between the protons and electrons of the nitrogen atom. This can be obtained as follow:

Proton = 7

Electron = 8

Net charge = Proton – Electron

Net charge = 7 – 8

Net charge = – 1

Therefore, the net charge of the nitrogen atom with 8 electrons is – 1

7 0
2 years ago
In the bohr model of the hydrogen atom, how does the radius of the electron’s orbit depend on the principal quantum number?
wolverine [178]
The radius of the electron's or basically the energy level for which the electron is found orbiting the nucleus of he hydrogen atom, as the principal quantum number tells us primarily the energy level that the electron will be found, is it the 1st, 2nd, and 3rd. The other quantum numbers tells us more specifically as per the subshell of the main shell the electron is in, the spin of the electron etc.
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3 years ago
Which three groups of the periodic table contain the most elements classified as metalloids (semimetals)?
attashe74 [19]
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).

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8 0
3 years ago
Read 2 more answers
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(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_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

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

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

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_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

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

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

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 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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3 years ago
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erik [133]

i dont know because im stupid

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3 years ago
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