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Reika [66]
3 years ago
7

What is a precipitate?

Chemistry
1 answer:
Mariulka [41]3 years ago
6 0
Answer: 1) A solid product of a chemical reaction that is in aqueous form.
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Which change of phase is most likely occurring in the beaker
solniwko [45]

Answer:

Evaporation

Explanation:

Heat makes molecules move and eventually evaporate.

5 0
3 years ago
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.

5 0
3 years ago
Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c
DochEvi [55]
Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1% 

b)
1.1 volume percent required for LFL

1.1% x 1
= 0.0011 m³ of n-hexane required
6 0
3 years ago
Read 2 more answers
Which represents the correct equilibrium constant expression for the reaction below?
zubka84 [21]
K(eq) = concentration of products/concentration of reactant
         = [Cu+2] / [Ag+]^2 

Activity of pure solid and liquid is taken as 1.

Hence last option is correct.

Hope this helps, have a great day ahead!

6 0
3 years ago
Read 2 more answers
Calculate the number of moles of a gas if it occupies a volume of 4dm at 10atm and _200°c.
Rina8888 [55]
 The number  of moles of a gas if it occupy a volume of 4dm  at 10 atm and - 200 c   is 6.67  moles

  calculation
by use of ideal gas equation
that is PV =nRT
where  P(pressure) =  10 atm
           V( volume) =  4dm^3 = 4L
            n(number of moles) =?
             R(gas constant) = 0.0821 L.atm/mol.K
              T = -200  +273 =  73 K

make n the  formula  of  the subject  n = PV/RT

n is therefore =  (10 atm x 4l)/(0.0821 L.atm/mol.K  x 73 K)  =  6.67 moles
5 0
3 years ago
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