(0.80M)(5.00mL) = c2 (75.00mL)
c2 = 5.33e-2 M
Is the correct amount of sig figs?
No. If the 0.80 is correctly written then you may have only 2 s.f. in the answer.
Ok so the answer is just 5.33? For an equation like this (0.50M)(1.75mL)= C2(75.00mL) would the answer just be 1.16e-3 or 1.16?
You are confusing scientific notation and sigfigs. Review these concepts to help you answer your own question.
5.33E-2 is three s.f. You may have only 2. That is 5.3E-2 or 0.0053
To determine the correct number of significant figures, we need to follow the rules for multiplying and dividing with significant figures.
In this case, we have the multiplication of (0.80 M) and (5.00 mL), resulting in the value c2 (75.00 mL).
The rule for multiplication and division states that the result should have the same number of significant figures as the measurement with the fewest significant figures.
Among the given values, the number 0.80 M has only 2 significant figures. Therefore, the answer should also have 2 significant figures.
The value mentioned, c2 = 5.33e-2 M, has 4 significant figures. Since this is more than the original value, it has excessive significant figures and is not correctly rounded.
To correct it to have the correct number of significant figures, we need to round it to 2 significant figures.
Rounding the value 5.33e-2 M to 2 significant figures gives us c2 = 0.053 M.
Therefore, the correct amount of significant figures for c2 is 2, which makes the correct value c2 = 0.053 M.