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Tems11 [23]
3 years ago
15

Determine the number of moles and mass of the sample of argon occupying 37.8L at STP​

Chemistry
1 answer:
Delvig [45]3 years ago
8 0

Answer:

The number of moles and mass of the sample of argon occupying 37.8L at STP​ are 1.6875 moles and 67.41 grams respectively.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then you can apply the following rule of three: if by definition of STP 22.4 L are occupied by 1 mole of Ar, 37.8 L of Ar are occupied by how many moles?

amount of moles=\frac{37.8 L* 1 mole}{22.4 L}

amount of moles= 1.6875 moles

The atomic weight of Ar is 39.948 g/mol. So the mass of the sample of argon can be calculated as:

1.6875 moles* 39.948 \frac{g}{mol} = 67.41 grams

<u><em>The number of moles and mass of the sample of argon occupying 37.8L at STP​ are 1.6875 moles and 67.41 grams respectively.</em></u>

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Answer:

144cm²

Explanation:

Given dimensions:

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The widest part of this figure will be the face containing the length and the breadth.

The breadth is the width of the figure;

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --&gt; __ AlCl3 + __
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Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
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<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

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