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sweet [91]
3 years ago
12

a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of

HCL? what is the ph after the addition of. 0050 mol of NaOH?
Chemistry
1 answer:
maw [93]3 years ago
4 0
Hello there.

<span>A 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?

Part 2 answer: </span><span>pH = 4.67 </span>
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What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O
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<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

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We are required to determine the other product produced in the combustion of butane apart from water.

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  • Therefore, combustion of butane will yield carbon dioxide and water.
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3 years ago
What is the chemical formula and net ionic equations for all three solutions.
timofeeve [1]

Answer:

See answer below

Explanation:

As you are asking for chemical formula and ionic equation, then, I will assume that after the station #4 below, are the solutions you are requiring.

You are also not specifing if you want for example, result of solution 1 + solution 3. If you need that, please post that on another question.

Now for the chemical formula, you need to identify the elements in all 3 solutions, and also the type of compound.

<u>1. Solution 2 Potassium Chloride: </u>

In this case we have Potassium on one side, and Chlorine on the other side, the symbol for those are K and Cl. As Potassium have the +1 oxidation state, cause is the only one that it can have, when it's next to an halide like chlorine or bromine, it will form a binary salt. The halides, usually work with the lowest oxydation state. In the case of Chlorine it will be -1, so, the formula will be:

KCl

And the net ionic equation will be the chemical equation that shows how the charges and atoms are balanced. In this case it would be:

K⁺ + Cl⁻ ------> KCl

<u>2. Solution 1 Copper(II) sulfate: </u>

In this case we have a tertiary salt, The copper's symbol is Cu, and is working with it oxydation state +2. Sulfate is an anion and it's formula is SO₄ and works with oxydation state -2 instead.

The chemical formula and ionic equation will be:  

Copper(II) sulfate: CuSO₄

And the net equation:

Copper sulfate: Cu²⁺ + SO₄²⁻ -------> CuSO₄

<u>3. Solution 3 Sodium hydroxide:</u>

In this case, we have a compound that it's usually used in acid base reactions. This is a strong base or hydroxide, and we have the element of Sodium (Na) with the oxydation state +1, is the only one it can have, and for the other side we have the oxydrile anion OH, and together is working with the oxydation state -1. So the chemical formula will be:

NaOH

And the net ionic equation:

Na⁺ + OH⁻ -------> NaOH

Hope this helps

4 0
3 years ago
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