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zavuch27 [327]
3 years ago
14

Matt did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers la

beled A and B. He put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown.
Experimental ObservationsSubstance Mass of Undissolved Substance at Different Temperatures (gram)
20 °C 40 °C 60 °C 80 °C
A 50 50 50 50
B 10 8 5 2



Part 1: Which, if any, substance is soluble in water?
Part 2: Explain how the data helped you determine solubility for both substances for temperatures 20 °C to 80 °C.
Chemistry
2 answers:
Black_prince [1.1K]3 years ago
4 0

Part 1 : Answer is only B substance is soluble in water.

In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.

Part 2 : Substance A is not soluble in water and substance B is soluble in water.

According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.

But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.


Reptile [31]3 years ago
4 0
<span>Both of the substances are water soluble as the experiment only measured the weigh that was left over. Therefore if the substance was not soluble the substances weight would not of changed. The data shows the substances become more soluble as the temperature raises as the weight decreases as temperature rises.</span>
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What is the maximum radiation pressure exerted by sunlight in space ( s = 1350 w/m 2) on a flat black surface? a. 2. 25 × 10−5 p
mel-nik [20]

The maximum radiation pressure exerted by sunlight in space on a flat black surface is  4.5 × 10^{-6}  P a. So, the correct option is (b).

Radiation pressure is the name for the force electromagnetic wave particles exert on a surface. It is inversely related to the wave's speed. Given data

Solar constant  ( S )  =  1350W / m ^2

Now, the radiaton pressure is given by

P = 2 S /c

where c is the speed of the light

P = 2 × 1350 /3 × 10 ^8

P = 9 × 10^{-6}  P a

For a black surface, P = 4.5× 10^{-6}  P a

Therefore, maximum radiation pressure exerted by sunlight in space on a flat black surface is  4.5 × 10^{-6}  P a

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8 0
1 year ago
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
3 years ago
Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?
Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

7 0
3 years ago
Read 2 more answers
PLEASE ANSWER I AM BEGGING
TiliK225 [7]

Taking into account the definition of dilution:

  • you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
  • If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

<h3>Volume of stock solution</h3>

In this case, you know:

  • Ci= 7 M
  • Vi= ?
  • Cf= 0.480 M
  • Vf= 0.120 L

Replacing in the definition of dilution:

7 M× Vi= 0.480 M× 0.120 L

Solving:

Vi= (0.480 M× 0.120 L)÷ 7 M

<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)

Finally, you will need 8.23 mL of the stock solution.

<h3>Concentration of the diluted solution</h3>

In this case, you know:

  • Ci= 7 M assuming the stock solution is 7.00 M HNO₃
  • Vi= 20 mL= 0.02 L
  • Cf= ?
  • Vf= 0.270 L

Replacing in the definition of dilution:

7 M× 0.02 L= Cf× 0.270 L

Solving:

(7 M× 0.02 L)÷ 0.270 L= Cf

<u><em>0.518 M= Cf</em></u>

Finally, the concentration is 0.518 M.

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1 year ago
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