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igomit [66]
2 years ago
6

Calculate the number of grams of ethanol nevessary to prepare 268g of a 9.76m solution of ethanol in water

Chemistry
1 answer:
stiv31 [10]2 years ago
7 0

The amount, in grams, of ethanol needed to prepare 268 grams of a 9.76 M solution of ethanol in water will be 120.50 grams

<h3>Solution preparation</h3>

In order to prepare 268 grams of solution, 268 mL of water would be needed because 1 mL of water weighs 1 gram.

Since we now know the volume of solution that we want to prepare, the amount of solute (ethanol) that would be required in order to make a solution of 9.76 M will be:

Mole required = 9.76 x 268/1000 = 2.62 moles

Mass of 2.62 moles ethanol = 2.62 x 46.07 = 120.50 grams

More on solution preparation can be found here: brainly.com/question/14667249

#SPJ3

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4 0
3 years ago
A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of t
erik [133]
<span>Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

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3 years ago
Read 2 more answers
What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?
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The balanced equation for the above reaction is as follows;
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according to molar ratio 
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since we know the number of moles we can find volume using ideal gas law equation 
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where
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R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation 

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K 
V = 4.07 L 
volume of SO₂ formed is 4.07 L


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