From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent.
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K

7 0

3 years ago

what is matter that can not be divided into two or more diffrent types of matter by any physical or chemical called/

VikaD [51]

Answer an atom

Explanation:

6 0

3 years ago

Using the correct answer from Part B, calculate the volume of a rectangular prism with a length of 5.6 cm, a width of 2.1 cm, an

Alenkinab [10]

5.6 x 2.1 x 6.6 = 77.616 cm3

5 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago