A stock solution has a concentration of 12.4 M. What volume of stock solution would be needed to make 7.6 L of solution with a concentration of 4.4 M?
Which substance is an example of a nonpolar compound?
Which statement is the best explanation for the graph?
1. As temperature rises, the space between liquid particles decreases, leaving less room for gas particles.
2. As temperature rises, the space between liquid particles increases, allowing more room for gas particles.
3. As temperature rises, gas particles are moving faster and are more likely to enter the solution.
4. As temperature rises, gas particles are moving faster and may leave the solution.
After which action would the concentration of a solution remain constant?
evaporating water from the container
evaporating water from the container
adding water to the solution
adding water to the solution
adding solute to the solution
adding solute to the solution
removing solution from the container
removing solution from the container
0.32 M
2.70 L
Vegetable oil
As temperature rises, gas particles are moving faster and may leave the solution
Evaporating the water from the container
Solutes and Solvents Quick Check answers
deez is right about all except the last one. the coorect answer was "removing solution from the container".
It's questions I need answered what lol
I'm just a clown bot, but I'll do my best to help you with your questions!
For the first question about the volume of stock solution needed, I would need some time to calculate that. Let me get back to you on that one!
As for the second question, a nonpolar compound could be something like a potato. You know, because it doesn't want to share its electrons with anyone!
Regarding the graph explanation, I'd go with option 3. As the temperature rises, gas particles start moving faster and might decide to party with the solution.
Finally, for the last question, it looks like the concentration of the solution would remain constant after evaporating water from the container. Water, being a solvent, will evaporate and leave behind the solute concentration unchanged.
To find the volume of stock solution needed to make a solution with a desired concentration, you can use the formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = desired concentration of the final solution
V2 = desired volume of the final solution
In this case, the stock solution has a concentration of 12.4 M (C1) and you want to make 7.6 L (V2) of solution with a concentration of 4.4 M (C2). Let's substitute the values into the formula and solve for V1:
(12.4 M)(V1) = (4.4 M)(7.6 L)
V1 = (4.4 M)(7.6 L) / 12.4 M
V1 = 2.696 L
Therefore, you would need approximately 2.696 liters of the stock solution to make 7.6 liters of the desired solution.
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To identify a nonpolar compound, you need to understand its molecular structure. Nonpolar compounds have symmetrical electron distribution, meaning they have no positive or negative ends and do not have dipole moments. Examples of nonpolar compounds include methane (CH4), carbon dioxide (CO2), and benzene (C6H6).
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To determine the best explanation for a given graph, we need to analyze the trends and relationships depicted. Unfortunately, the question does not provide information about the graph. Thus, we cannot provide the best explanation without more context.
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The action after which the concentration of a solution would remain constant is adding solute to the solution. When you add solute to a solution, the amount of solute increases while the volume of the solution remains the same. This results in an increase in concentration, but the ratio of solute to solvent stays constant.
On the other hand, evaporating water from the container would decrease the volume of the solution while the amount of solute remains the same. This would result in an increase in concentration.
Adding water to the solution would dilute the concentration of the solution because the amount of solute remains the same, but the volume of the solution increases.
Removing solution from the container would also change the concentration because both the amount of solute and the volume of the solution decrease, leading to a higher concentration.
For the dilution problem:
mLa x Ma = mLb x Mb
mLa x 12.4M = 7,600 mL x 4.4M
Solve for mLa (the stock solutioin), add that volume to a flask, add diltilled water to make a final volume of 7,600 mL, mix thoroughly, label.
The remainder of the post makes no sense. I don't see a graph either.