The melting point for ice is at 32 degrees <span>Fahrenheit or O degrees Celsius.</span>
(sample g/1) X (1 mole/40.078(MW of Ca)) = moles of sample (moles of sample)(6.022 x 10^23( no of atoms)/ 1 mole) = # of atoms in a 120 g sample of calcium Avogadro's number=6.022x 10^23 atoms in 1 mole
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al is required.
Therefore for 2 moles of Al,
Total F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
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Answer:
It would be compound.
Explanation:
It is this way because if it adds another proton it becomes more positive that nuetral, and if you add an electron it just makes the atom more dense. That is why the answer is compound. Hope this helped :)
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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