Answer:
first; why is it a question if teacher never teached you it LOL anyways..
Explanation:
i would say its the second one
"The smae quality of each element is present on both sides of the equation."
Answer:
pKb = 10.96
Explanation:
Tartaric acid is a dyprotic acid. It reacts to water like this:
H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1
HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2
When we anaylse the base, we have
Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1
HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2
Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14
Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹
so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96
Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.
I would say #3 I’m sorry if it’s wrong tho
So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P
