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Fittoniya [83]
1 year ago
14

If a solution contains less than the maximum amount of solute that it could dissolve at that temperature, the solution would be

described as......Group of answer choicessaturatedunsaturatedsupersaturatedunfinished
Chemistry
1 answer:
tamaranim1 [39]1 year ago
4 0

The solubility of a substance tells us the amount of solute that is capable of dissolving a given amount of solvent at a given temperature. We speak that a solution is.

Now, if the amount is less than the statement says, it will be an unsaturated solution.

When the amount is greater, the solution is supersaturated and a precipitate of solute will form in the solution.

According to what has been explained, the solution described by the statement is an unsaturated solution.

Answer: Unsaturated

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vazorg [7]
It would be B, so yeah hope that helps ya ;)

5 0
3 years ago
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How many moles of calcium chloride are contained in a 333 gram sample?
Bogdan [553]

Answer & Explanation:

The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..

\frac{333 g(CaCl2)}{} *\frac{1mol(CaCl2)}{110.98g(CaCl2)}

Multiply straight across on the top and straight across on the bottom.

\frac{333}{110.98}

Now divide.

\frac{333}{110.98}=3.00

So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.

6 0
3 years ago
Which element has two valence electrons?
Colt1911 [192]

Answer:

It is A) Calcium

Explanation:

Calcium has an electronic configuration of 2,8,8,2

6 0
2 years ago
Read 2 more answers
100 POINTS!!!
mixas84 [53]

1) Area of one side of each cube

The area of one side of each cube can be calculated by multiplying the length times the width:

A=L\cdot W

where

L = length

W = width

For cube A: L=1 cm, W=1 cm

So the area is: A_A = (1)(1)=1 cm^2

For cube B: L=2 cm, W=2 cm

So the area is: A_B=(2)(2)=4 cm^2

For cube C: L=3 cm, W=3 cm

So the area is: A_C=(3)(3)=9 cm^2

2) Surface area of each cube

The surface area of a cube is given by the sum of the areas of all its faces. Since a cube has 6 identical faces, this means that the total surface area is equal to 6 times the area of one face:

A=6 A_1

where

A_1 is the area of one face of the cube

For cube A, the area of one face is A_A=1 cm^2

So the surface area of cube A is

A'_A = 6A_A=6(1)=6 cm^2

For cube B, the area of one face is A_B=4 cm^2

So the surface area of cube B is

A'_B = 6A_B=6(4)=24 cm^2

For cube C, the area of one face is A_C=9 cm^2

So the surface area of cube C is

A'_C = 6A_C=6(9)=54 cm^2

3) Volume of each cube

The volume of a cube is obtained by multiplying its length, its width and its height:

V=L\cdot W \cdot H

Where:

L = length

W = width

H = height

Moreover for a cube, all the sides have equal length, so L=W=H

So the volume can be rewritten as

V=L^3

For cube A: L = 1 cm

So the volume is V_A=(1 cm)^3 = 1 cm^3

For cube B: L = 2 cm

So the volume is V_B=(2 cm)^3 = 8 cm^3

For cube C: L = 3 cm

So the volume is V_C = (3 cm)^3 = 27 cm^3

4) Ratio surface area/volume for each cube

In this part, we have to calculate the ratio between surface area and volume of each cube:

r=\frac{A}{V}

where

A is the surface area

V is the volume

For cube A, we have:

A_A = 6 cm ^2 (surface area)

V_A=1 cm^3 (volume)

So the ratio for cube A is:

r=\frac{6}{1}=6

For cube B, we have:

A_B = 24 cm ^2 (surface area)

V_B=8 cm^3 (volume)

So the ratio for cube B is:

r=\frac{24}{8}=3

For cube C, we have:

A_C = 54 cm ^2 (surface area)

V_C=27 cm^3 (volume)

So the ratio for cube C is:

r=\frac{54}{27}=2

4 0
3 years ago
A scholar measures 20 mL of room temperature water (22ºC) and adds 80 mL of 75°C water to it, following the
Leni [432]

Answer:

Data is not valid

Explanation:

When two liquids having different temperatures are mixed, regardless of the volumes, the final mix temperature will ALWAYS be between the initial temperature values.

1st Law Thermo => Law of Conservation of Energy => Energy can not be created nor destroyed, only changed in form. Mixing 22°C with 75°C will NOT result in a mix having a final temperature of 80°C.

∑ΔE = 0 => (mcΔT)₁ + (mcΔT)₂ = 0

[(20g)(1cal/g·°C)(Tₓ - 22°C)] + [(80g)(1cal/g·°C)(Tₓ - 75°C)] = 0

=> 20(Tₓ - 22) + 80(Tₓ - 75) = 0

=> 20Tₓ - 440 + 80Tₓ - 75 = 0

=> 100Tₓ = 440 + 75 = 515

=> Tₓ = (515/100)°C = 51.5°C final mix temperature

4 0
3 years ago
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