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myrzilka [38]
3 years ago
7

Which of the following techniques is most appropriate for the recovery of solid KNO3 from an aqueous solution of KNO3?A. Paper c

hromatographyB. FiltrationC. TitrationD. ElectrolysisE. Evaporation to dryness
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

E. Evaporation to dryness

Explanation:

E. - Evaporation to dryness is the best method for the recovery of solid KNO3 from an aqueous solution of KNO3.

(KNO3 is very soluble, and will violently decompose if overheated.)

Paper chromatography is for separation of different weight molecules in solution.

B. Filtration  won't work on a solution

C.Titration would contaminate the salt with something else and is used  

to determine concentrations

D. Electrolysis would destroy the salt

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A solution is prepared by mixing 25.0 g H2O and 25.0 g C2H5OH. Determine the mole fractions of each substance.
Veronika [31]

Answer:

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

Mole fraction H₂O = 1.39 / 1.933 → 0.72

Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

8 0
3 years ago
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Greeley [361]
B, the height above or below sea level
3 0
3 years ago
A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
Llana [10]

<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
                                    = </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
3 years ago
6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?
Darina [25.2K]
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
 </span>
4 0
3 years ago
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nydimaria [60]
It’s( A)Condensation
6 0
4 years ago
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