Chemistry help?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka for acetic acid (CH₃COOH) = 1.8 × 10⁻⁵

Let y% be the percent dissociation of acetic acid.

           CH₃COOH(aq) + H₂O(ℓ)  ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)   Ka = 1.8 × 10⁻⁵
Initial:         0.35 M                               0 M                 0 M
Change:  -0.35y% M                      +0.35y% M       +0.35y% M
Eqm:   0.35(1 - y%)  M                    0.35y% M         0.35y% M
               ≈ 0.35 M                          = 0.0035y M     = 0.0035y M

At equilibrium:
Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = (0.0035y)² / 0.35
1.8 × 10⁻⁵ = 0.0035²y² / 0.35
y = √[0.35 × (1.8 × 10⁻⁵) / 0.0035²]
y = 0.72
Percent dissociation of acetic acid = 0.72%

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I spend a lot of time on each problem, I just want to make sure it's for a worthwhile cause.

Use the Ka equation to solve for [H+] . The Ka is small enough to avoid the use of the quadratic route
 Ka = [H+]² / 0.35
 [H+]² = (1.8*10^-5) * 0.35
[H+]² = 6.30*10^-6
[H+] = 2.51*10^-3M
% dissociation = (2.51*10^-3) / 0.35 * 100% = 0.72%