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PilotLPTM [1.2K]
3 years ago
6

Calculate the volume the gas will occupy if the pressure is increased to 1.89 atmatm while the temperature is held constant. Exp

ress the answer in liters to three significant figures.
Chemistry
1 answer:
Y_Kistochka [10]3 years ago
6 0

Question is incomplete, complete question is :

A fixed quantity of gas at 21°C exhibits a pressure of 758 torr and occupies a volume of 5.42 L . Calculate the volume the gas will occupy if the pressure is increased to 1.89 atm while the temperature is held constant.

Answer:

The final volume of the gas will be 2.86 L.

Explanation:

initial pressure of the gas = P_1=758 Torr=\frac{758}{760} atm=0.997 atm

initial volume of the gas = V_1=5.42 L

Final pressure of the gas = P_2=1.89 atm

Final volume of the gas = V_1=?

Applying Boyle's law;

P_1V_1=P_2V_2 ( at constant temperature)

V_2=\frac{P_1V_1}{P_2}=\frac{0.997 atm\times 5.42 L}{1.89 atm}

V_2=2.86 L

The final volume of the gas will be 2.86 L.

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OverLord2011 [107]

Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

 Experiment

 In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

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To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

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In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.

Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

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Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.

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6 0
3 years ago
A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out
Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

                       = 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0
3 years ago
What size volumetric flask would you use to create a 1.00M solution using 166.00 g of KI?
mr Goodwill [35]

Answer:

A 1 liter volumetric flask should be used.

Explanation:

First we <u>convert 166.00 g of KI into moles</u>, using its <em>molar mass</em>:

Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol

  • 166.00 g ÷ 166 g/mol = 1 mol KI

Then we <u>calculate the required volume</u>, using the <em>definition of molarity</em>:

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Liters = moles / molarity

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3 0
2 years ago
Which best describe the isovolumetric contraction phase of the cardiac cycle?
cluponka [151]
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5 0
3 years ago
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Find the number of moles of barium iodide if you have 5.23 x 1024 formula units of Bal2
Oksi-84 [34.3K]

Answer:

8.68 moles of BaI₂

Explanation:

Given data:

Number of moles of BaI₂ = ?

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Solution:

By using Avogadro number,

1 mole of any  substance contain 6.022× 10²³ formula units.

5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units

0.868 × 10¹ moles of BaI₂

8.68 moles of BaI₂

Thus, 5.23× 10²⁴ formula units of BaI₂ contain 8.68 moles of BaI₂

7 0
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