CHEMISTRY HIGH SCHOOL

If 20.0 ml of glacial acetic acid (pure hc2h3o2 is diluted to 1.70 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

Answers

Answer 1

Answer:

The pH of the resulting solution is 2.72

Data;

volume of acid = 20mL
density of acid = 1.05 g/ml
mass = 1.05 * 20 = 21g

pH of a solution

The number of moles of acetic acid is

$n = 21/60 = 0.333 moles$

The concentration of acetic acid in 1.7L is

$c = (0.333)/(1.7) = 0.196M$

The equation for dissociation is

$CH_3COOH + H_2O \to CH_3COO^- +H_3O^+$

The Ka of this reaction is

$K_a = ([CH_3COO^-][H_3O^+)/([CH_3COOH)$

The Ka of acetic acid = 1.8*10^-5

$1.8*10^-^5 = (x*x)/(0.196) \nx = 0.187*10^-^2M\n$

The concentration of hydrogen ion is

$[H^+] = 1.87*10^-^3M$

The pH of this solution is

$pH = -log[H^+]\npH = -log(1.87*10^-^3)\npH = 2.72$

The pH of the resulting solution is 2.72

Learn more on pH of a solution here;

brainly.com/question/25639733

Answer 2

Answer: Given:
20.0 ml of glacial acetic acid
1.70 L with water
density of the glacial acetic acid is 1.05 g/ml ; Ka = 1.8 x 10⁻⁵

1) find out how many grams of HC2H3O2 is in this glacial matter. It says here it has a density of 1.05g/mL.
So we find the grams from that => 1.05 g/mL * (20.0 mL) = 21 grams of HC2H3O2.
Convert this to moles => 21 grams (1 mol / 60 g) = 0.35 mol HC2H3O2

We then get the Ka formula from that equation so it is
Ka = [H3O+] [C2H3O2] / [HC2H3O2] (Remember H2O has an activity of 1 so we exclude it)

We know the concentration of HC2H3O2 because we have moles of HC2H3O2
Finding concentration of HC2H3O2 => 0.35 mol HC2H3O2/1.70 L = 0.20588 M - initial concentration

1.8 x 10⁻⁵ = (x) (x) / (0.20588 M - x)
1.8 X 10⁻⁵ (0.20588 M) = x²
3.70584⁻ x 10⁻⁶ = x²
x = 1.9251 x 10⁻³

pH = -log (H₃O⁺) = - log (1.9251 x 10⁻³) = -(-2.7156) = 2.7156

pH of the resulting solution is 2.7156

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Which of the following summarizes photosynthesis?A. glucose + carbon dioxide → oxygen + water
B. carbon dioxide + water → glucose + oxygen + water
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Answers

Answer : The correct option is, (B) $\text{carbon dioxide}+\text{water}\rightarrow \text{glucose}+\text{oxygen}+\text{water}$

Explanation :

Photosynthesis : It is a type of process which takes place in the plant cell. In this process the light energy converted into chemical energy.

As we know that the plant needs carbon dioxide, water and light. During the photosynthesis the carbon dioxide, water and sunlight react to gives glucose, oxygen and water as a product.

The balanced reaction will be,

$6CO_2+12H_2O\overset{sunlight}\rightarrow C_6H_(12)O_6+6O_2+6H_2O$

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What is the mass of an aluminum cylinder that has a volume of 1.50m3

Answers

Answer: The mass of aluminium cylinder is 4065 kg.

Explanation:

Density is defined as the ratio of mass of the object and volume of the object.

Mathematically,

$\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}$

We are given:

Density of aluminium = $2710kg/m^3$

Volume of aluminium cylinder = $1.50m^3$

Putting values in above equation, we get:

$2710kg/m^3=\frac{\text{Mass of aluminium cylinder}}{1.50m^3}\n\n\text{Mass of aluminium cylinder}=4065kg$

Hence, the mass of aluminium cylinder is 4065 kg.

Aluminium density: 2700g/m^3

therefore; 2700g/m^3 X 1.50m^3 = 4050g = 4.05Kg

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Answers

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Answers

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Answers

When calculating total valence electrons, extra electron pairs are added to the Outside of an atom. The answer is letter C. Suppose you have a compound of CCl4. You know that chlorine can only share 1 electron because 7 of its electrons are filled. Also, in carbon, it can only share 4 electrons because 4 of it are already filled. That is why carbon needs four chlorine to form CCl4.

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Answers

The best answer to this question would be (d.) iodine. Iodine is the elements who has the ability to undergo sublimation. Iodine is an important compound in terms of nutrition, production of acetic acid and polymers. Iodine is capable of transitioning from solid to gas without any specialized process.

The answer to your question is

d. idoine

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