Asked by Max
The following volumes of 0.000300 M SCN are diluted to 15.00mL determine the concentration of SCN in each sample after dilution.
Sample 0.000300 M SCN(ml) SCN (M)
1 1.50
2 3.50
Can someone please explain how to do this?
Sample 0.000300 M SCN(ml) SCN (M)
1 1.50
2 3.50
Can someone please explain how to do this?
Answers
Answered by
Max
Should have been in a table but i guess it could not fit there.
Sample 1 = 1.50 (0.000300M SCN (mL)), SCN (M) = ?
sample 2 = 3.50 (0.000300M SCN (mL)), SCN (M) = ?
Sample 1 = 1.50 (0.000300M SCN (mL)), SCN (M) = ?
sample 2 = 3.50 (0.000300M SCN (mL)), SCN (M) = ?
Answered by
DrBob222
You take 1.50 mL and dilute to 15.00; therefore, the new concn is
0.00030 x (1.50/15.00) = ??
If you prefer, you can use the dilution formula of c1v1 = c2v2 which is the same as mL1 x M1 = mL2 x M2
0.00030 x (1.50/15.00) = ??
If you prefer, you can use the dilution formula of c1v1 = c2v2 which is the same as mL1 x M1 = mL2 x M2
Answered by
Parsa Zarei
0.00003
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