Answer:
One: <u>Selenium</u> is Paramagnetic
Explanation:
Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.
Electronic configuration of Scandium;
Sc = 21 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹
Sc³⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶
Hence in Sc³⁺ there is no unpaired electron.
Electronic configuration of Bromine;
Br = 35 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵
Br⁻ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶
Hence in Br⁻ there is no unpaired electron.
Electronic configuration of Magnesium;
Mg = 12 = 1s², 2s², 2p⁶, 3s²
Mg²⁺ = 1s², 2s², 2p⁶
Hence in Mg²⁺ there is no unpaired electron.
Electronic configuration of selenium;
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
Or,
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹
Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
Answer: I would go with B
Explanation: The motor in a circuit isn't moving. That's very vague, but it doesn't show any evidence that an electrical current is going through it, likewise it doesn't show that an electrical current ISN'T going through it. However in regards to this question I would go with B.
Answer:
22 mol
Explanation:
Given data:
Number of atoms of Cl = 2.65×10²⁵ atom
Number of moles of Cl = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms
0.44×10² mol
22 mol
Answer:
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Answer:
29Si has a natural abundance of 4.68%.
30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.
Explanation:
The atomic mass of silicon is given by:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
Where:
Si: atomic mass of silicon (28.086)
Si²⁸: relative atomic mass of 28Si (27.97693)
A₁: natural abundance of 28Si (92.23%)
Si²⁹: relative atomic mass of 29Si (28.97649)
A₂: natural abundance of 29Si
Si³⁰: relative atomic mass of 30Si
A₃: natural abundance of 30Si
We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.
We have to set up a system of three equations in three unknowns:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
A₃=0.6592×A₂
A₁+A₂+A₃=1
First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:
A₁+A₂+0.6592×A₂=1
A₁+1.6592×A₂=1
1.6592×A₂=1-A₁
A₂=
=0.0468
Then, we find the value of A₃:
A₃=0.6592×A₂
A₃=0.6592×0.0468=0.0309
Finally, we find the value of Si³⁰ in the first equation:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309
28.086=27.15922+Si³⁰×0.0309
28.086-27.15922=Si³⁰×0.0309
=Si³⁰
Si³⁰=29.99288
