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katovenus [111]

3 years ago

11

PLEASE HELP ONLY ONE QUESTION

2 answers:

rosijanka [135]3 years ago

7 0

When reactant particles collide with proper energy and proper orientation

ad-work [718]3 years ago

6 0

A chemical reaction occurs when reactant particles collide with proper energy and proper orientation.

Hope this helps! Have a great day! <3

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Consider the chemical equation in equilibrium.

Kruka [31]

By increasing the P reaction moves toward less number of moles in this case
<span>The equilibrium will shift to the left to favor the reverse reaction.</span>

4 0

3 years ago

Read 2 more answers

What is the ph of a solution that contains 1.0 l of 0.10 m ch3cooh and 0.080 m nach3coo after 0.03 moles of naoh added?

zimovet [89]

Answer: pH = 4.996

Explanation:

No of moles = molarity x volume

:• no of moles of CH3COOH = 0.1M x 0.1L

n(CH3COOH) = 0.1mol

Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.

Therefore, Moles of CH3COOH becomes,

0.1 - 0.03 = 0.07 mol

Subsequently, the moles of CH3COONa increases and becomes,

0.08 + 0.03 = 0.11 mol

Using the Hendersom-Hasselbach equation,

pH = pKa + log [Moles of conjugate÷ moles of Ch3COOH]

From literature, pKa of Ch3COOH is 4.8

Thus,

pH = 4.8 + log [0.11/0.07]

pH = 4.8 + 0.1963

pH = 4.996

4 0

3 years ago

Which of the following best defines the term oxidation state?

eimsori [14]

C seems to be the best answer

4 0

3 years ago

Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the

Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If you know pH and pka:

10^(pH-pka) = $\frac{[A^-]}{[HA]}$

The ratio will be: 10^(pH-pka)

At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

0 = log₁₀ $\frac{[A^-]}{[HA]}$

10^0 = $\frac{[A^-]}{[HA]}$

1 = $\frac{[A^-]}{[HA]}$

As ratio is 1, [conjugate base] = [acid] in solution.

At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0

3 years ago

When iodine solution is mixed with an unknown sample of food, the mixture turns a blue black color. this color change indicates

andrezito [222]

It indicates the presence of starch

8 0

3 years ago