The hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.
<h3>How to calculate [H3O+]?</h3>
The hydrogen ion concentration of a solution can be calculated as follows:
pOH = - log [OH-]
pOH = - log [6.3 × 10-⁷M]
pOH = - [-6.2]
pOH = 6.2
Since pOH + pH = 14
pH = 14 - 6.2
pH = 7.8
pH = - log [H3O+]
7.8 = - log [H3O+]
[H3O+] = 10-⁷:⁸
[H3O+] = 1.5 × 10-⁸M
Therefore, the hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.
Learn more about hydrogen ion concentration at: brainly.com/question/15082545
Answer:
KJ8RT898TGHO7-6734354546746R476
Explanation:
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>
