What is the concentration of a solution made by diluting 95 mL of 6.0 M HCl to a final volume of 750 mL

I just did this on my Homework

C1V1=C2V2
6.0*95=C2*750
6*95=570
570=C2*750
Balance the equation using algrebra
570/750=750/750
0.76/1= 0.76
0.76 M

A 0.105-L sample of an unknown HNO3 solution required 36.9 mL of 0.200 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?

Ah, HCl, aka the "acidic delight." Let's do some clown math, shall we?

So, you have a 95 mL of 6.0 M HCl and you want to dilute it to a final volume of 750 mL. That's quite a dilution, my friend!

To find the new concentration, we can use the formula:
M1V1 = M2V2

M1 = initial concentration (6.0 M)
V1 = initial volume (95 mL)
M2 = final concentration (unknown, let's call it X)
V2 = final volume (750 mL)

Now, plugging in the numbers, we get:
(6.0 M) * (95 mL) = X * (750 mL)

Simplifying this equation of hilarity, we get:
570 = 750X

And solving for X, we have:
X = 570/750

Doing the clown division, we find:
X = 0.76 M

So, the concentration of your diluted solution is approximately 0.76 M HCl. Don't worry, even though it's less concentrated, it can still sting a bit!

To find the concentration of the solution made by diluting 95 mL of 6.0 M HCl to a final volume of 750 mL, we can use the formula for dilution:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

In our case, C1 = 6.0 M, V1 = 95 mL, C2 is the concentration we want to find, and V2 = 750 mL.

Plugging in the values:

(6.0 M)(95 mL) = C2(750 mL)

To find C2, rearrange the equation:

C2 = (6.0 M)(95 mL) / 750 mL

C2 = 570 / 750

C2 = 0.76 M

Therefore, the concentration of the solution made by diluting 95 mL of 6.0 M HCl to a final volume of 750 mL is 0.76 M.

c1v1 = c2v2

6.0*95 = c*750
c = ?