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yuradex [85]
1 year ago
5

For each pair, choose the compound with the smaller lattice energy, and explain your choice: (b) K₂O or K₂S.

Chemistry
1 answer:
dlinn [17]1 year ago
5 0

the compound with the smaller lattice energy is potassium sulphide here the size of the molecule play a important role

The quantity of energy released by the electrostatic attraction of oppositely charged ions is known as lattice energy (L.E). The ion's size and charge affect the lattice energy.

lattice energy is inversely proportional to   size of ion and directly proportional to charge of the ion. They are each charged equally that is plus two and minus two However, because the Sulphur is larger in size  and the oxygen is lesser in this case, The lattice energy of potassium oxide is larger  the lattice energy of potassium sulphide  is smaller.

To learn more about lattice energy :

brainly.com/question/18222315

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A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod
sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

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