WHAT IS THE ABILITY OF A SUBSTANCE TO BE DISSOLVED

The normal freezing point of a certain liquid

slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$

3 0

2 years ago

Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2

weqwewe [10]

Equation of reaction

NaHCO3 + HCl ---------> NaCl + H2O + CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

8 0

2 years ago

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A sample of sulfurous acid (H2SO3) has a mass of 1.31 g.

lakkis [162]

Answer:

a) 2 (H+) ions

b) 1 (SO3²-) ions

c) 1.36 × 10^-22 grams.

Explanation:

According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.

Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:

2H+ + SO3²- → H2SO3

Hence;

- there are 2 hydrogen ions (2H+) present in H2SO3.

- there is 1 sulfite ion (SO3²-) present in H2SO3.

c) The mass of one formula unit of H2SO3 is calculated thus:

= 1.008 (2) + 32.065 + 15.999(3)

= 2.016 + 32.065 + 47.997

= 82.08 a.m.u

Since, 1 gram is = 6.02 x 10^23 a.m.u

82.08 a.m.u = 82.08/6.02 × 10^-23

= 13.6 × 10^-23

= 1.36 × 10^-22 grams.

6 0

2 years ago

Help with this question plz

Triss [41]

Increase, because you need heat to melt a solid to a liquid, so the temperature will have to get greater.

6 0

3 years ago

What are some physical properties of a candle

Troyanec [42]

The wick and the wax

Sorry if that was useless, I'm not sure how generalized you were being

6 0

2 years ago

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