19. When 0.5g of calcium trioxocarbonate (IV) was added to excess dilute
1 answer:
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Answer:
empirical formula:
2 g H
32.7 g S
65.3 g O
Explanation:
Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:
- 2.00% * 100g = 2 g H
- 32.7% * 100g = 32.7 g S
- 65.3% * 100g = 65.3 g O
Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:
- the molar mass of H is 1.01 g/mol
- the molar mass of S is 32.06 g/mol
- the molar mass of O is 16 g/mol
2 g H ÷ 1.01 g/mol = 1.98 mol H
32.7 g S ÷ 32.06 g/mol = 1.02 mol S
65.3 g O ÷ 16 g/mol = 4.08 mol O
Our ratio of H : S : O is now:
1.98 mol : 1.02 mol : 4.08 mol
Divide them all by the smallest number, which is 1.02:
1.98/1.02 : 1.02/1.02 : 4.08/1.02
1.94 : 1 : 4
1.94 ≈ 2
So:
2 : 1 : 4
Thus, the empirical formula is: .
The ground-state electron configurations of transition metal ions are diamagnetic [Kr]
. The ion is diamagnetic because there all electrons are paired.
- A magnetic field repels diamagnetic materials because it induces an opposing magnetic field in them when it is applied, which produces a repelling force.
- In contrast, a magnetic field draws paramagnetic and ferromagnetic materials together.
- All materials experience the quantum mechanical phenomenon known as diamagnetism, which is the only source of magnetism in a material.
- The magnetic dipoles within paramagnetic and ferromagnetic materials exert an attracting force that outweighs the modest diamagnetic force.
- Diamagnetic materials have a magnetic permeability that is less than vacuum, or 0.
- Although superconductors behave as strong diamagnets, diamagnetism is often a modest effect that can only be observed by sophisticated laboratory equipment.
To learn more about Diamagnetic with the given link
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