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Tomtit [17]
2 years ago
13

student has two balloons, one filled with helium and one filled with argon. Each balloon has a volume of 22.4-L at STP. Which of

the following correctly describes these samples
Chemistry
1 answer:
slamgirl [31]2 years ago
4 0

The statement that can be made about the samples is that each balloon contains 1 mole of each gas.

<h3>What is molar volume?</h3>

Molar volume refers to the volume that is occupied by one mole of a gas. According to Avogadro, the volume occupied by one ole of  a gas is 22.4-L at STP.

Hence, if 22.4-L at STP is the volume of both the helium and argon filled balloons, then each balloon contains one mole of the gas.

Learn more about molar volume: brainly.com/question/4172228

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A 150 mL sample of hydrochloric acid (HCl) completely reacted with 60.0 mL of a 0.100 M NaOH solution. The equation for the reac
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Answer:

0.040M

Explanation:

The following data were obtained from the question:

Volume of acid (Va) = 150mL

Volume of base (Vb) = 60mL

Molarity of base (Mb) = 0.1M

Molarity of acid (Ma) =..?

Next, the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Finally, we shall determine the concentration of the acid, HCl as follow:

MaVa / MbVb = nA/nB

Ma x 150 / 0.1 x 60 = 1/1

Cross multiply to express in linear form

Ma x 150 = 0.1 x 60

Divide both side by 150

Ma = 0.1 x 60 / 150

Ma = 0.04M

Therefore, the concentration of the acid is 0.04M

5 0
3 years ago
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3 years ago
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

a) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

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3 years ago
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a pure compound because an element is untouched and is just itself

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