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Nimfa-mama [501]
3 years ago
11

How many electrons can the n=4 shell hold? a. 8b.16c.32d.64?

Chemistry
2 answers:
ra1l [238]3 years ago
6 0
The maximum the fourth shell can have is 32.
belka [17]3 years ago
3 0

Answer:

c.32

Explanation:

Hello,

In this case, it is possible to determine it by building the subshells organization for the first four shells as shown below:

1s^2\\2s^2\ 2p^6\\3s^2\ 3p^6\ 3d^{10}\\4s^2\ 4p^6\ 4d^{10}\ 4f^{14}

In such a way, counting the electrons at the n=4 shell, the result is 32 (2+6+10+14), so the answer is c.

Best regards.

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A substance with which bond type would have a high melting point?
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Read 2 more answers
A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris
Blababa [14]

<u>Answer:</u> The heat capacity of calorimeter is 15.66J/^oC

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.473 moles

\Delta H_{rxn} = enthalpy change of the reaction  = -1235 kJ/mol = -1235\times 10^3J/mol     (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J

To calculate the heat capacity of calorimeter, we use the equation:

q=c\Delta T

where,

q = heat absorbed by the calorimeter = 584.16\times 10^3J

c = heat capacity = ?

\Delta T = change in temperature = T_2-T_1=62.3^oC-25^oC=37.3^oC

Putting values in above equation, we get:

584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC

Hence, the heat capacity of calorimeter is 15.66J/^oC

4 0
4 years ago
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