Your answer would be, Sodium Chloride in Aqueous Solution is represented as, Letter choice(D), NaCI(aq), or as HCI + NaOH + H2O(I) + NaCI(aq)

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5 0

4 years ago

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Please help me i will give brain list

Strike441 [17]

Answer:

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3 0

3 years ago

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Which segment of this graph shows a decreasing velocity? F 80 D. El 60 Position (m) с B 20 10 20 30 40 50 60 70 80 90 100 Time (

Ivan

Answer:

E to F

Explanation:

We know that v = dx/dt

and a = d²x/dt²

Decreasing velocity means deceleration or negative acceleration,

therefore d²x/dt²<0

i.e the portion where the graph is having its curvature downwards like an inverted u, from graph E to F portion looks like that.

Therefore in the portion E to F velocity decreases.

3 0

3 years ago

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

3 years ago