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motikmotik
2 years ago
8

Questions & Conclusions:

Chemistry
1 answer:
dem82 [27]2 years ago
3 0

The amount of KNO₃ precipitated out of solution when you cooled the solution from 100°C to -22°C is obtained from the solubility curve

<h3>What is a solubility curve?</h3>

A solubility curve is a curve of the solubility of a solute against temperature.

The solubility curve shows that the solubility of different solute at different temperatures.

The solubility curve of KNO₃ is as shown. Solubility at 100 °C and  -22 °C is not shown in the curve.

However, the amount of KNO₃ precipitated out of solution when cooled the solution from 100 °C to -22 °C can be determined by subtracting the amount of solute dissolve at -22 °C from that dissolved at 100 °C.

In conclusion, the solubility curve is used to determined the amount of solute dissolved in a given volume of solvent at different temperatures.

Learn more about solubility curve at: brainly.com/question/928930

#SPJ1

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Alex Ar [27]

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5 0
3 years ago
PLZ HELP (SCIENCE) Match the following items.
FromTheMoon [43]
Oxidation: a chemical change resulting form a reaction with oxygen
Endothermic:a reaction that takes in energy
Exothermic: a reaction that releases energy
Decomposition:chemical change whereby a molecule breaks down into simpler moles clues or elements
Subscript:written underneath or below
6 0
2 years ago
Read 2 more answers
Describe the results of Ernest Rutherford's gold-foil experiment and explain how his results changed ideas about the distributio
sashaice [31]
Physicist Ernest Rutherford<span> established the nuclear theory of the atom with his </span>gold-foil experiment<span>. When he shot a beam of alpha particles at a sheet of </span>gold foil<span>, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.</span>
4 0
4 years ago
Read 2 more answers
A 300-g aluminum cup contains 700 g of water in thermal equilibrium with the cup at 60°C. The combination of cup and water is co
Rus_ich [418]

Answer:

117.3 W is being removed.

Explanation:

The heat removed can be calculated as:

Q = m*c*ΔT

Where m is the mass, c is the specific heat and ΔT is the temperature variation. Because there're two components:

Q = mwater*cwater*ΔT + maluminum*caluminum*ΔT

Q = (mwater*cwater + maluminum*caluminum)*ΔT

Searching in a thermodynamic table:

cwater = 4.184 J/g°C

caluminium = 0.9 J/g°C

In 1 minute, the temperature decreases 2.2°C, so ΔT =  -2.2°C

Q = (700*4.184 + 300*0.9) * (-2.2)

Q = -7037.36 J

The rate of energy is the potency (P), which is the heat divided by the time. So, for 1 minute (60 s):

P = -7037.36/60

P = -117.3 J/s

P = -117.3 W

The minus signal indicates that the energy is being removed.

3 0
4 years ago
Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume
lbvjy [14]

Answer:

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

Molar mass of Glucose(C6H12O6) = 180 g/mol

We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

4 0
4 years ago
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