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xxTIMURxx [149]
3 years ago
8

A gas is contained in a thick-walled

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

                   T₁  =  194 K

Solution:

Data Given:

                 Initial Temperature  =  T₁  =  <u>??</u>

                 Final Temperature  =  T₂  =  465 K

                 Initial Pressure  =  P₁  =  18.7 psi

                 Final Pressure  =  P₂  =  20.2 psi

                 Initial Volume  =  V₁  =  0.475 L

                 Final Volume  =  V₂  =  1.054 L

Formula Used:

Let's assume that the hydrogen gas in balloon is acting as an Ideal gas, the according to Ideal Gas Equation,

                 P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹,

Taking number of moles and R constant we can have following formula for initial and final states,

                P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for T₁,

                T₁  =  P₁ V₁ T₂ / P₂ V₂

Putting values,

                T₁  =  (18.7 psi × 0.475 L × 465 K) / (20.2 psi × 1.054 L)

               T₁  =  194 K

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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
ankoles [38]

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.

Density of the ethanol = 0.785 g/cm³

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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a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.

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hope this helps! :)
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