Answer:
34.23 g.
M = (no. of moles of solute)/(V of the solution (L)).
Answer:
0.120M is the concentration of the solution
Explanation:
<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>
<em />
Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.
To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:
<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>
2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3
<em>Molar concentration:</em>
0.0300 moles NaNO3 / 0.250L =
<h3>0.120M is the concentration of the solution</h3>
Hmm... interesting topic you're writing about here!
Well, for this essay, there must be specific criteria, correct? I'll make some suggestions, but of course you don't have to go by them if you don't like 'em. So... here they are!:
- List the physical and chemical properties of glue
- List the physical and chemical properties of liquid corn starch
- Compare and contrast the physical and chemical properties of glue with the physical and chemical properties of liquid corn starch
- You can conduct some experiments with the two substances also! You can mix the two together and see how they react with each other, and incorporate the results of the experiment into your essay.
- You can mix glue with some other liquid substances too, and incorporate your results into your essay. You can see whether each substance the glue mixes with creates a homogeneous or heterogeneous mixture, etc.
- You can mix liquid corn starch with some other liquid substances too, and incorporate your results into your essay. You can see whether each substance the liquid corn starch mixes with creates a homogeneous or heterogeneous mixture, etc.
That's all I can come up with at the moment. Maybe you'll take some of these suggestions into consideration...? Anyhow, I hope I helped somehow! :)
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
