1)For the reaction, A = B + C, the rate law is k(A). If it takes 80.0 seconds for 70.0% of a 10.0 gram sample of A to be transformed into products, what is the value of the rate constant?

I was going to do 80/10 = 8 and then 8 times .70 but im not sure this is correct.

Haven't I already done this for you? I did for someone.

ln(No/N) = kt
ln(10/3) = k*80
Solve for k.
Note: No = 10; that's the starting amount.
N = amount left. When 70% has been used then 30% is left and 30%*10 = 3.

To solve this problem, you need to use the integrated rate law for a first-order reaction and the given data.

The integrated rate law for a first-order reaction is: ln([A]t/[A]0) = -kt

Where:
- [A]t is the concentration of A at a given time t
- [A]0 is the initial concentration of A
- k is the rate constant
- t is the time

In this case, we are given that 70.0% (0.70) of a 10.0-gram sample of A is transformed into products after 80.0 seconds.

We can convert the given mass of A to its concentration by dividing the mass by the molar mass of A. Let's assume the molar mass of A is M grams/mol.

Concentration of A at t=0 ([A]0) = (10.0 g / M g/mol)
Concentration of A at t=80.0 s ([A]t) = (0.70 * 10.0 g / M g/mol)

Plugging these values into the integrated rate law, we have:
ln([A]t/[A]0) = -kt

ln((0.70 * 10.0 g / M g/mol) / (10.0 g / M g/mol)) = -k * (80.0 s)

Simplifying, we get:
ln(0.70) = -k * (80.0 s)

Now, solve the equation for the rate constant k by rearranging the equation:
k = -ln(0.70) / (80.0 s)

Using a calculator, you can find the value of ln(0.70) and divide it by 80.0 s to obtain the rate constant k.