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dusya [7]
2 years ago
10

Please Help Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potass

ium sulfate to 225g water, carefully heating the solution, and cooling it to 40 Celsius. A homogeneous solution is obtained. Is this solution saturated, unsaturated, or supersaturated? The beaker is shaken and precipitation occurs. How many grams of potassium sulfate would you except to crystallize out?
Chemistry
1 answer:
Svetradugi [14.3K]2 years ago
5 0
To determine the state of saturation of the solution, we calculate the mass of solute per mass of water for the given amounts and compare this value to the solubility. If the value is less than the solubility, then the solution is unsaturated. If it is greater than solubility, then it is supersaturated. If it is equal to the solubility, then it is saturated.

mass solute / mass water  = 39.0 grams K2SO4 / 225 grams H2O = 0.173 g K2SO4/ g H2O
solubility = 15 g /100 g = .15 g/g

Therefore, the solution is supersaturated. When it is shaken, some of the solute would precipitate out. 

mass of solute soluble to water = .15 g K2SO4/ g water ( 225 g water ) = 33.75 g K2SO4
mass of K2SO4 that would crystallize = 39.0 - 33.75 = 5.25 g K2SO4
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The statement that percent yield can never be greater than theoretical yield is another example of the ________.
Gnom [1K]
We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.
3 0
2 years ago
How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
2 years ago
A mining company is interested in monazite, a mineral substance that forms in igneous rock in Earth’s interior. They think they
VARVARA [1.3K]
Sedimentary rock can be minazute by adding more heat n pressure
7 0
2 years ago
A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
Summarize the process a scientist goes through to come up with a<br> satisfactory solution.
maria [59]
Guess and check, test, trial and error, completion.
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