<h3>
Answer:</h3>
Lead-205 (Pb-205)
<h3>
Explanation:</h3>
<u>We are given;</u>
We are supposed to identify its product after an alpha decay;
- Polonium-209 has a mass number of 209 and an atomic number of 84.
- When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
- Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
- The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
- The equation for the decay is;
²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He
- Note; An alpha particle is represented by a helium nucleus, ⁴₂He.
When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
8.68 moles of BaI₂
Explanation:
Given data:
Number of moles of BaI₂ = ?
Number of formula units = 5.23× 10²⁴
Solution:
By using Avogadro number,
1 mole of any substance contain 6.022× 10²³ formula units.
5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units
0.868 × 10¹ moles of BaI₂
8.68 moles of BaI₂
Thus, 5.23× 10²⁴ formula units of BaI₂ contain 8.68 moles of BaI₂
