The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :
Mass of metal iodide formed : 97.12 g, so mass of Iodine :
Then mol iodine (MW=126.9045 g/mol) :
mol ratio of Cobalt and Iodine in the compound :
The answer is true. According to the second law of thermodynamics, energy tends to become more spread out
Answer:
a. 7.8*10¹⁴ He⁺⁺ nuclei/s
b. 4000s
c. 7.7*10⁸s
Explanation:
I = 0.250mA = 2.5 * 10⁻³A
Q = 1.0C
1 e- contains 1.60 * 10⁻¹⁹C
But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C
(A).
No. Of charge per second = current passing through / charge
1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C
1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei
(B).
I = Q / t
From this equation, we can determine the time it takes to transfer 1.0C
I = 1.0 / 2.5*10⁻⁴ = 4000s
(C).
Time it takes for 1 mol of He⁺⁺ to strike the target =?
Using Avogadro's ratio,
1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)
Note : ions cancel out leaving the value of the answer in mols.
1.0 mol of He = 7.7 * 10⁸s
