Answer:
The formula to use is
VOLUME=LENGTH×BREADTH×HEIGHT
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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
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Mole number of 75 g is 0.4 mol. NO2 is 0.8 mol and is 36.8 g.
The chemical formula of ammonia is written as NH3 we can just use the ratio of the atoms since we are not given the reaction. We also make use the avogadro's number to convert from moles to molecules. We calculate as follows:
5.01 x 10^-4 g H2 ( 1 mol / 2.02 g H2 ) ( 2 mol NH3 / 3 mol H2 ) ( 6.022x10
^23 molecules / 1 mol ) = 9.96 x 10^19 molecules NH3
122 moles of Methane to liters is 2734.5