This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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Answer:
See below
Explanation:
<u> Name </u> <u>Formula </u> <u> Major species </u> <u> </u>
Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),
Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)
Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)
Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)
Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)
- Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
- The compounds containing metals are ionic. They produce ions in solution.
- ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.
a) To find the mass after t years:we will use this formula:
A = Ao / 2^n when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)b) Mass after 90 y :by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mgC) Time for 1 mg remaining:when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
It will only be valuable if proved correct so it's D