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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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What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
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Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

Thus, the de broglie wavelength is 2.037 \times 10^{-34} m.

                               

6 0
3 years ago
Read 2 more answers
Write the chemical symbol for an element in Period 6 and Group 2A
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5 0
3 years ago
What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor
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Answer:

1.75272\ \text{m}^3

Explanation:

The breakdown reaction of ozone is as follows

CF_3Cl + UV \rightarrow CF_3 + Cl

Cl + O_3 \rightarrow ClO + O_2

O_3 + UV \rightarrow O_2 + O

ClO + O \rightarrow Cl + O_2

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

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M = Molar mass of CF_3Cl = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = 62.363\ \text{L mmHg/K mol}

Number of moles is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}

20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3

For 20 cycles of the reaction the volume of the ozone is 1.75272\ \text{m}^3.

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3 years ago
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