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3 years ago

What is the molarity of 1 mole of HCl in 5 liters of solution?

2 answers:

5 0

Molarity = moles of solute(HCl)
------------------------------------
volume of the solution

= 1
------
5

= 0.2M.

Hence option B is correct.
Hope this helps!!

prohojiy [21]3 years ago

4 0

Answer:

B. 0.2 M

Explanation:

Hello,

In this case, by considering that the molarity is defined as:

$M=\frac{mol_{solute}}{V_{solution}}$

We replace the given 1 mole of HCl as the moles of the solute and 5 liters as the volume of the solution to obtain the required molarity as shown below:

$M=\frac{1mol}{5L}=0.2M$

Therefore, the answer is B. 0.2 M.

Best regards.

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What do we mean by atmospheric pressure

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Answer:

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Atmospheric Pressure is the force per unit area exerted on a surface due to the weight of air molecules above it. It is caused by the collisions of the air molecules with that surface.:

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3 years ago

Which one of the following substances would be the most soluble in CCl4?

maria [59]

Answer: Option (d) is the correct answer.

Explanation:

Carbon tetrachloride $(CCl_{4})$ is a non-polar solvent. Whereas out of the given options, $Na_{2}SO_{4}$ , $H_{2}O$ , $CH_{3}CH_{2}CH_{2}CH_{2}OH$ , and HI are all polar molecules.

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3 years ago

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

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$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

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3 years ago

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3 years ago

Use the Henderson-Hasselbalch equation and your knowledge of ionization to help you answer this question. Aspirin is a weak acid

Leya [2.2K]

Answer:

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Explanation:

Aspirin is more readily absorbed when it is protonated, that is when pH is lower than pKa (<em>more H⁺ available in the medium</em>). We can confirm this using Henderson-Hasselbalch equation for pH = 1.5:

$pH = pKa +log(\frac{conjugate base}{weak acid} )\\1.5=3.5 +log(\frac{conjugate base}{weak acid} )\\-2.0=log(\frac{conjugate base}{weak acid} )\\10^{-2.0} =\frac{conjugate base}{weak acid}\\\frac{conjugate base}{weak acid}=0.01=\frac{1}{100}$

When aspirin is absorbed more readily the ratio of conjugate base to acid is 1 to 100, being the acid the <em>predominant</em> form.

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3 years ago