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arsen [322]
4 years ago
15

How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Chemistry
1 answer:
Lunna [17]4 years ago
6 0

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated  12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

n_1=n_2

Which can also be written in terms of concentrations and volumes:

M_1V_1=M_2V_2

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.

Best regards.

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A chemist requires 0.802 mol Na2CO3 for a reaction. How many grams does this correspond to?
Komok [63]

Answer:

Ok:

Explanation:

So grams = mols*MolarMass. Here, MolarMass (MM) = 105.99g which can be found using the periodic table. mols is given to be 0.802. We can then plug in to get that it corresponds to 85.0g.

7 0
3 years ago
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and                                Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is =  5.451 X 10³ kg.

 

5 0
3 years ago
Score! You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some o
mamaluj [8]

Answer:

You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some of the first unlabeled solution and some indicator.

The color changes to yellow.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

What are the identities of each unlabeled solution?

Explanation:

Bromothymol blue is a dye and it is used as an indicator.

It is used as a pH indicator.

In acids, it becomes yellow n in color.

In bases, it turns blue.

You take one of the empty beakers and add some of the first unlabeled solution and some indicator. The color changes to yellow.

That means the unlabeled solution is an acid.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

It is a basic solution.

6 0
3 years ago
In this experiment you are directed to add a limited amount of NaOH (aq) and then an excess amount. A similar strategy will be u
Verizon [17]

The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.

In qualitative analysis is a common to add the base in drops and then in excess. When added in drops, the metal hydroxide is formed. This metal hydroxide is often insoluble.

After this metal hydroxide is formed, the base could be added in excess such that the metal hydroxide dissolves in the excess base by forming a complex.

For instance;

CuCl2(aq) + 2NaOH(aq)  -------> Cu(OH)2(s) + 2NaCl(aq)

Cu(OH)2(s) + 2OH^-(aq) -------> [Cu(OH)4]^2+(aq)

Learn more: brainly.com/question/1527403

7 0
3 years ago
A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of Oxygen.calculate the percentage compos
oksano4ka [1.4K]

Answer:

Mass of compound = 0.24 g and, mass of boron = 0.096 g percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% mass of oxygen = 0.144 g again, mass of compound = 0.24 g percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60%

<h2>Hope it's Helpful!!✌️</h2>
5 0
2 years ago
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