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Pavel [41]
2 years ago
7

How much energy must be added to a bowl of 125popcorn kernels in order for them to reach a popping temperature of 175°C? Assume

that their initial temperature is 21°C, that the specific heat capacity of popcorn is 1650 J/kg•°C, and that each kernel has a mass of 0.105 g.
Chemistry
1 answer:
Kitty [74]2 years ago
5 0

Answer:

3,335.06 Joules of an energy must be added to a bowl of 125 popcorn kernels in order for them to reach a popping temperature of 175°C.

Explanation:

Number of popcorn kernels = 125

Mass of single popcorn kernel = 0.105 g

Mass of 125 popcorn kernel = m = 125 × 0.105 g  = 13.125 g = 13.125 × 0.001 kg

1 g = 0.001 kg

Specific heat of popcorn = c = 1650 J/kg°C

Initial temperature of the popcorn kernels = T_1=21^oC

Final temperature of the popcorn kernels = T_2=175^oC

Heat added to popcorn kernel to rise their temperature to popping temperature be Q.

Q=m\times c\times (T_2-T_1)

Q=13.125\times 0.001 kg\times 1650 J/kg^oC(175^oC-21^oC)

Q = 3,335.06 J

3,335.06 Joules of an energy must be added to a bowl of 125 popcorn kernels in order for them to reach a popping temperature of 175°C.

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And, in order to attain stability it readily loses an electron and thus it become Na^{+} ion.

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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
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Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and 1s^22s^22p^6

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1

This element will easily loose 1 electron and form Rb^+ ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

1s^22s^22p^3

This element will easily gain 3 electrons and form N^{3-} ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

1s^22s^22p^6

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^5

This element will easily gain 1 electron and form Br^- ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

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