The table with the data is in the picture attached.
Answer:
Explanation:
The reaction equation suggests that the law could have this form:
Then, the work is to find the values of the exponents that satisfy the initial rate data.
A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.
Then, so far you can say:
When you use trials 1 and 2, you get:
Now, you can use trials 1 and 3 to determine the other exponent:
Thus, you have the rate law:
Now, you just use any trial to obtain k. Using trail 1:
Which yields:
2. D) A solution that has more solute in it than it can hold is called a supersaturated solution. An easy way to know this is because super means that there is more.
3. I'm not sure about this one, but all I can tell you is that the gram formula mass of H2SO4 is 98 grams.
4. D) C2H6SO is the empirical formula. We know this because all of the subscripts are simplified as much as possible. You might try to simplify the C2 and H6 but take note that the S and O have no subscripts, so elements cannot be simplified anymore.
5. D) ethanoic acid (C2H4O2) has the same empirical formula as glucose. The empirical formula of glucose is CH2O. This is because the subscripts of 6/12/6 have a common factor of 6, so we divide them all by 6 to get subscripts of 1/2/1. Ethanoic acid has subscripts of 2/4/2. They have a common factor of 2, so when we divide them by 2, they will become 1/2/1.