Missing question:
Chemical reaction: H₂ <span>+ 2ICl → 2HCl + I</span>₂.
t₁ = 5 s.
t₂ = 15 s.
c₁ = 1,11 M = 1,11 mol/L.
c₂ = 1,83 mol/L.
rate of formation = Δc ÷ Δt.
rate of formation = (c₂ - c₁) ÷ (t₂ - t₁).
rate of formation = (1,83 mol/L - 1,11 mol/L) ÷ (15 s - 5 s).
rate of formation = 0,72 mol/L ÷ 10 s.
rate of formation = 0,072 mol/L·s.
Explanation:
c. h. o
66.7%. 11.1%. 22.2%
____. ____. ____
12. 1. 16
1.558. 11.1. 1.39. (divide by the smallest)
1. 8. 1
empirical formula=ch8o
Answer:
[H+] = 1.74 x 10⁻⁵
Explanation:
By definition pH = -log [H+]
Therefore, given the pH, all we have to do is solve algebraically for [H+] :
[H+] = antilog ( -pH ) = 10^-4.76 = 1.74 x 10⁻⁵
Na is a cation (positive) and F is a anion (negative) this is an ionic bond. In a ionic bond the cation loses its valence electrons to give them to the anion. In this case Na will give its valence electron to F so F can complete its octet:)
