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CaHeK987 [17]
3 years ago
5

Nickel has chemical symbol Ni and the atomic 28. How many protons,neutrons,and electronic would be found in an atom of nickel-78

Chemistry
2 answers:
ira [324]3 years ago
8 0
In nickel-78, there are 28 protons, 28 electrons, and 50 neutrons.

The protons is always the atomic number.
The electron would always be the atomic number when it's a neutral atom. (Different for isotopes)
The neutrons is always the number that makes up the mass with protons. You can figure it out by subtracting atomic mass by protons. (78-28 = 50)
Andreas93 [3]3 years ago
8 0

Explanation:

Atomic number is the sum of only total number of protons present in an element. Whereas mass number or atomic mass is the sum of total number of both protons and neutrons present in an element.

For example, given atom of Ni has atomic number 28.  And, it is known that atomic mass of Ni is 58 g/mol.

Therefore, number of neutrons present in it will be calculated as follows.

                    Mass number = no. of protons + no. of neutrons

                         58 = 28 + no. of neutrons

                    no. of neutrons = 58 - 28  

                                               = 30

Also, when an atom is neutral in nature then its number of protons equal to the number of electrons present in it.

Hence, we can conclude that in a neutral atom of Ni there are 28 protons, 30 neutrons and 28 electrons.

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A gas container has a volume of 446.9 with a temp of 14c. When the volume is decreased to 238.7l the new temp is what
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Answer:

\frac{V _{1}}{T _{1}}  =  \frac{V _{2}}{T _{2} }  \\  \frac{446.9}{(14 + 273)}  =  \frac{238.7}{T _{2} } \\ {T _{2}} =  \frac{238.7 \times 287}{446.9}  \\ {T _{2}} = 153.3 \: kelvin \\  = 119.7  \degree \: c

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15. List the substances A-E in order from most dense to least dense based on the facts provided.
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Answer:

So 1st it is B then D then E then a then C

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3 years ago
According to bohr's model of the atom in which orbitals do electrons have the least energy
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7 0
3 years ago
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo
dimaraw [331]

<u>Answer:</u> The value of K_c for given reaction is 0.465

<u>Explanation:</u>

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

                           A+2B\rightarrow C

<u>Initial:</u>                0.65     1.35     0.30

<u>At eqllm:</u>        0.65-x   1.35-2x   0.30+x

Evaluating the value of 'x'

0.650-x=0.550\\\\x=0.100

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of K_c for above equation follows:

K_c=\frac{[C]}{[A][B]^2}

Putting values in above equation, we get:

K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465

Hence, the value of K_c for given reaction is 0.465

7 0
3 years ago
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