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jeka57 [31]
3 years ago
12

How many grams are in 0.02 moles of beryllium iodide

Chemistry
1 answer:
raketka [301]3 years ago
6 0
Beryllium iodide is BeI2. The molecular weight of BeI2 is 2(9.01) + 126.90 = 144.92g/mol. So, in 0.02 moles of BeI2, you have 0.02 * 144.92 = 2.9 grams (3 g if you want to stick to significant figures)
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A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak
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Answer:

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Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>
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