answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zavuch27 [327]
2 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]2 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]2 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>
You might be interested in
In 1-2 sentences, explain why weather can be predicted only as probable, not definite. (2 points) Plaese help me ​
anyanavicka [17]

Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.

It is common observation that weather forecasts are often given as a probability and never in definite terms.

This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.

The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.

Learn more: brainly.com/question/21209813

8 0
2 years ago
In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
2 years ago
A gem has a mass of 4.50 g. When the gem is placed in a graduated cylinder 12.00 mL of water, the water level rises to 13.45 mL.
Mandarinka [93]
<span>Displaced volume :

</span>Final volume - <span>Initial volume

</span>13.45 mL - 12.00 mL => 1.45 mL

Mass =  4.50 g

Therefore:

density = mass / volume

D = 4.50 / 1.45

<span>D = 3.103 g/mL </span>
6 0
2 years ago
The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalp
wariber [46]

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

7 0
2 years ago
What vitamin a compounds bind with opsin to form rhodopsin??
irina [24]
The organic compound retinal binds with opsin and forms rhodopsin. Retinal is part of the molecule that is responsible for its color. This part is called chromophore. On the other hand, opsins are the proteins in photoreceptor cells. Retinal bounds with these opsins and forms rhodopsin: the basis of the human vision. Rhodopsin is also a protein.It is the pigment in the retinas of humans and animals.
7 0
2 years ago
Other questions:
  • When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the n2 gas cha
    11·2 answers
  • Question 1 (Matching Worth 3 points)
    12·1 answer
  • What is the specific heat (J/g°C) of a metal object whose temperature increases by 3.0°C when 17.5 g of metal was heated with 38
    13·1 answer
  • The diagram shows the movement of particles from one end of the container to the opposite end of the container. A cylindrical co
    11·2 answers
  • A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi
    15·1 answer
  • 7.47 Two atoms have the electron configurations 1s22s22p6 and 1s22s22p63s1. The first ionization energy of one is 2080 kJ/mol an
    6·1 answer
  • 7. If you fill a balloon with 5.2 moles of gas and it creates a balloon with a volume of 23.5 liters, how many moles are in a ba
    12·1 answer
  • Just Lemons Lemonade Recipe Equation:
    5·1 answer
  • A student is curious about the Ksp value for NaCl. The student looks up the value om the appendix of his textbook but cannot fin
    15·1 answer
  • Calculate the molar mass of a 2.89 g gas at 346 ml, a temperature of 28.3 degrees Celsius, and a pressure of 760 mmHg.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!