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vlabodo [156]
3 years ago
13

A mixture of chalk powder and water can be

Chemistry
1 answer:
grin007 [14]3 years ago
4 0

Answer:

<h3>Right answer is: ( a) chalk powder remains suspended in water.</h3>

Explanation:

Filtration is the technique used to separate suspended solute particles from a solution . The chalk powder remains suspended in the solution and can easily be filtered through a filter paper , the chalk powder can be collected on the filter paper and clear solvent is collected as the filtrate.

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Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for
Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

<h2>[OH⁻] = 14.29 </h2>

Hope this helps

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3. Use the chart to answer the following question. Amanda found a stone
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Answer:

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Explanation:

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3 years ago
Which solution has a molality of 0.25m nacl?
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Molality can be expressed by moles of solute over kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl is equal to 0.25 moles of NaCl over 1 kg of water.

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I just need help with question 10.??? Please ASAP
Alex

Answer:

4 hydrogen / 1 Carbon

Just look at the diagram

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