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:
The coefficient of Ca(OH)2 is 1
Explanation:
Step 1: unbalanced equation
Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O
Step 2: Balancing the equation
On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O
On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.
Now the equationis balanced.
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
The coefficient of Ca(OH)2 is 1
Answer:
No. of atom =
no.of moles x avagardro's number xatomicity
= weight /molar mass x No x atomicity
=8.2/142 x6.02x10^23 x 4
=0.346 x 10^23(approximately)
