Iodine 131 and iodine 126 are the same in the sense that, they both have the same number of electrons and protons in their atoms, it is only the number of their neutrons that is different. Iodine 131 has 78 neutrons while iodine 126 has 73 neutrons.
Answer:
43 mole
Explanation:
Given data:
Number of atoms of Li = 2.6× 10²⁵ atoms
Number of moles = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
2.6× 10²⁵ atoms × 1 mole / 6.022 × 10²³ atoms
0.43 × 10² mole
43 mole
Answer:
25.6g de HF son producidos
Explanation:
<em>...¿Cuánto HF es producido?</em>
Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:
<em>Moles CaF2:</em>
Masa molar:
1Ca = 40g/mol
2F = 19*2 = 38g/mol
40+38 = 78g/mol
50g CaF2 * (1mol/78g) = 0.641 moles CaF2
<em>Moles H2SO4:</em>
Masa molar:
2H = 2g/mol
1S = 32g/mol
4O = 64g/mol
98g/mol
100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4
Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.
<em>Moles HF usando la reacción:</em>
0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF
<em>Masa HF:</em>
Masa molar:
1g/mol + 19g/mol = 20g/mol
1.282 moles HF * (20g/mol) =
<h3>25.6g de HF son producidos</h3>
The mass of nickel, Ni deposited when 0.2 faraday of electrical charge is passed through a solution of NiSO₄ is 5.9 g
