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Effectus [21]
3 years ago
11

How many moles of gas occupy 56.3 L at 0.899 atm and 20.0°C? ​

Chemistry
1 answer:
zzz [600]3 years ago
3 0

Answer:

n = 2.1 mol

Explanation:

Given data:

Number of moles of gas = ?

Volume of gas = 56.3 L

Pressure of gas = 0.899 atm

Temperature of gas = 20°C (20+273 = 293 k)

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

0.899 atm × 56.3 L  = n × 0.0821 atm.L/ mol.K  × 293 k

50.614 atm.L = n × 24.055 atm.L/ mol

n = 50.614 atm.L / 24.055 atm.L/ mol

n = 2.1 mol

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2 years ago
You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the
Makovka662 [10]

The empirical formula of this compound is Fe_2O_3

<h3>Empirical formula </h3>

To calculate the empirical formula of a compound, the value of moles of each element is needed.

As we have the information of the mass value, we will use the molar mass expression, which corresponds to:

MM_O = 16g/mol\\MM_Fe = 55.8g/mol

                                              MM = \frac{m}{mol}

  • O

                                                   16 = \frac{4.8}{x}

                                                   x = 0.3mol

  • Fe

                                                    55.8=\frac{11.2}{x}\\x = 0.2

As the value of the empirical formula must be an integer, simply multiply the two values ​​by a common factor:

                                                O = 0.3 \times 10 = 3\\Fe = 0.2 \times 10 = 2

                                                       Fe_2O_3

So, the empirical formula of this compound is Fe_2O_3.

Learn more about empirical formula: brainly.com/question/1247523

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2 years ago
Where on the table would a group's<br> ionization energy be the greatest?
Ymorist [56]

Answer:

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3 years ago
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A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled
marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so  that the concept could be applied

Initial volume of the AICI3 solution =23.1 \mathrm{mL}

Initial Molarity of the solution =833 \mathrm{mM}

Final molarity of the solution =75.0 \mathrm{mM}

Final volume of the solution =?

From Law of Dilution, M_{f} V_{f}=M_{i} V_{i}

\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L

Final Volume of the solution =0.257

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