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mina [271]
3 years ago
10

A chemist prepares a solution of aluminum chloride (AlCl3) by measuring out 11 gr of aluminum chloride (AlCl3) into a 50 ml volu

metric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's aluminum chloride solution. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

1.6 M

Explanation:

To answer this question first we <u>calculate the molar mass of AlCl₃</u>, using the atomic weights of Al and Cl:

  • AlCl₃ ⇒ 27 g/mol + (35.45 g/mol)*3 = 133.35 g/mol

Now with the molar mass, and the mass weighed, we can <u>calculate mol AlCl₃</u>:

  • 11 g AlCl₃ ÷ 133.35 g/mol = 0.082 mol AlCl₃

Finally we <u>calculate concentration in mol/L</u>, keeping in mind that 50mL = 0.05 L:

  • 0.082 mol AlCl₃ / 0.05 L = 1.6 M
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Two gas samples have the same number of molecules, the same volume, and the same pressure. Which of the following could be true?
bulgar [2K]
Lets name one gas sample as A and other gas sample as B.
we can apply ideal gas law equation for both samples
PV = nRT
P - Pressure of A = Pressure of B
V - volume of A = volume of B
n  - number of molecules of both A and B being equal is equivalent to number of moles of A = number of moles of B
R - universal gas constant
Tᵃ - temperature of A
Tᵇ - temperature of B
for gas A
PV = nRTᵃ  --1)
for gas B
PV = nRTᵇ  ---2)
when we divide both equations 
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Tᵃ = Tᵇ
both temperatures are equal 
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therefore 0 °C = 273 K 
the correct answer is 
A)
<span>The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 0 </span>°<span>C</span>
3 0
3 years ago
Read 2 more answers
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