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Vaselesa [24]
3 years ago
5

A gas at constant temperature is enclosed in a container with a movable piston. It occupies a space of 10L at a pressure of 500

torr. If the pressure is increased to 1200 torr, what is the new volume?
Chemistry
1 answer:
Liula [17]3 years ago
8 0

Answer:

4.17L

Explanation:

V1 = 10L

V2 =?

P1 = 500torr

P2 = 1200torr

Boyle's law states that at constant temperature, the volume of a gas is inversely proportional to its pressure.

P1V1 = P2V2

V2 = ( P1 * V1 ) / P2

V2 = 4.17L

The new volume of the gas is 4.17L

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A solution was prepared by mixing 20.00 mL of 0.100 M and 120.00 mL of 0.200 M. Calculate the molarity of the final solution of
Marat540 [252]

Answer:

0.186M

Explanation:

First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:

<em>Moles Nitric acid:</em>

0.0200L * (0.100mol / L) = 0.00200 moles

0.120L * (0.200mol / L)= 0.02400 moles

Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid

Molarity: 0.026 moles / 0.140L

<h3>0.186M</h3>
6 0
3 years ago
2) 2KClO3 --&gt; 2KCl + 3O2
aleksandrvk [35]

2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2

a)

2 \text{ mols of KClO}_3 \equiv 3  \text{ mols of O}_2

19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5  \text{ mols of O}_2

\boxed{19 \text{ mols of KClO}_3 \equiv 28,5  \text{ mols of O}_2}

b)

2 \text{ mols of KClO}_3 \equiv 2  \text{ mols of KCl}

62 \text{ mol of KClO}_3 \equiv 62  \text{ mol of KCl}

Using the atomic mass given in the periodic table:

62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

62\cdot122,5 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

7595 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

\boxed{7,595 \text{ kg of KClO}_3 \equiv 62  \text{ mol of KCl}}

c)

2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3

3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}

Using the atomic mass given in the periodic table:

3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5)  \text{ g of KCl}

96\text{ g of O}_2 \equiv 149\text{ g of KCl}

\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}

\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}

This result is an aproximation.

8 0
3 years ago
A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

  • 47.2 mL * 150 mM = 24.0 mM * V₂
  • V₂ = 295 mL

And <u>converting into L </u>becomes:

  • 295 mL * \frac{1 L}{1000mL} = 0.295 L

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3 years ago
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<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />
7 0
3 years ago
What’s 13 in scientific notation
MatroZZZ [7]
1.3*10^1 would be the answer
6 0
3 years ago
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