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mario62 [17]
3 years ago
9

Calculate the volume of a tin block that has a mass of 95.04 grams at STP. Your response must include both a numerical setup and

the calculated result.
Element Density at STP (g/cm3)C 3.51Si 2.33Ge 5.32Sn 7.31Pb 11.35
Chemistry
1 answer:
Ahat [919]3 years ago
3 0

Answer:

Explanation:

As the problem states that we have STP, these conditions are 1 atm of pressure and 273 K of temperature.

Now, the equation we must use to solve this:

PV = nRT

Solving for V:

<em>V = nRT/P</em>

<em>Where:</em>

<em>V: Volume in Liters</em>

<em>n: moles of the tin block</em>

<em>T: temperature in K</em>

<em>P: Pressure in atm</em>

<em>R: gas constant which is 0.082 L atm / K mol</em>

But also the problem is giving us the density data for all elements. In the case of Tin it is 7.31 g/cm³ or 7.31 g/mL, so, with the formula of density:

<em>d = m/V  ----> V = m/d</em>

From the above formula, we can calculate the volume of tin so:

V = 95.04 / 7.31

<em>V = 13 mL</em>

This would be the volume of the tin block, but, we have this block at STP so we need to calculate the volume with the ideal gas equation above. We need the molecular mass of Tin which is 118.71 g/mol, so let's calculate the moles:

n = m/MM

n = 95.04 / 118.71 = 0.8 moles

Now, solving for V:

V = 0.8 * 0.082 * 273 / 1

<em>V = 17.91 L</em>

<em>And this would be the volume of the tin block at STP conditions.</em>

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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

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The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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