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zloy xaker [14]
3 years ago
14

 A chemist describes a particular experiment in this way: "0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol o

f O2." 
Chemistry
1 answer:
AleksandrR [38]3 years ago
4 0
Description:

<span>"0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol of O2." 

This means that a certain amount of H2O2 (0.0400 mol) decomposed or was broken down into two components, 0.04 mol of H2O and 0.02 mol of O2. To examine the system, we need a balanced equation:

H2O2 ---> H2O + 0.5O2

The final concentrations of the system indicates that the system is in equilibrium. </span>
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This family (ethane, propane, butane, etc) of materials is likely to have what set of properties?
Ilya [14]

This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.

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The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.

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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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Which statement about electrolytic cells is correct?
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30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?
USPshnik [31]
Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 &deg; C = 23+273.15 = 296.15 &deg; K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
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