it wouldn't let me post after putting spaces in the url and everything, so i just copied and pasted the q instead :P sorry for all the messyness. i'll see if i can delete q's, in the mean time you can maybe look at this please?

Initial rate of rxn of H2O2(aq)->H2O+ 1/2 O2(g) is found at 1.7x10(-3)M/s and is constant for 2min. You start with 160ml of 16M H2O2.
How many mL of O2 in one minute? (Measured at 273K and 760mmHg).

I tried to find the new conc of H2O2 via NewConc= Old conc - kt
Then i multiplied the new conc by .16L, converted those moles to O2 and solved for volume via the PV=nRT, but ended up with 2.3ishL, which seems rather large considering it's asking for V in mL. Any idea if this is right or not?
(I doubt i can assume the volume is constant in my method, among all the other things most likely wrong with it XD)
Thanks

To solve this problem, you can use the given information to calculate the number of moles of H2O2 reacted in 2 minutes and then convert it into moles of O2. Finally, use the ideal gas law to calculate the volume of O2 at 273K and 760mmHg.

Here's a step-by-step solution:

Step 1: Calculate the number of moles of H2O2 reacted in 2 minutes:
Given initial rate of reaction = 1.7 x 10^(-3) M/s
Time = 2 minutes
Volume of H2O2 = 160 mL = 0.16 L
Initial concentration of H2O2 = 16 M

The rate of reaction can be represented as:
Rate = -d[H2O2]/dt

Since the initial rate of reaction is constant for 2 minutes, we can express it as:
Rate = Δ[H2O2]/Δt

Δ[H2O2] = (Final concentration - Initial concentration)
Δt = 2 minutes = 2 x 60 seconds

Hence,
1.7 x 10^(-3) M/s = Δ[H2O2]/(2 x 60 s)

Now, we can calculate the change in concentration of H2O2:
Δ[H2O2] = (1.7 x 10^(-3) M/s) x (2 x 60 s) = 0.204 M

Step 2: Calculate the number of moles of H2O2 reacted in 2 minutes:
Moles of H2O2 reacted = Δ[H2O2] x Volume of H2O2
= (0.204 M) x (0.16 L)
= 0.03264 moles H2O2

Step 3: Convert moles of H2O2 reacted to moles of O2:
According to the balanced equation, 1 mole of H2O2 produces 1/2 mole of O2.
Hence, 0.03264 moles of H2O2 will produce (1/2) x 0.03264 moles of O2.

Step 4: Calculate the volume of O2 at 273K and 760mmHg:
Now, we'll use the ideal gas law to find the volume of O2 (V) when measured at 273K and 760mmHg.
The ideal gas law equation is: PV = nRT

P = 760 mmHg
V = ?
n = (1/2) x 0.03264 moles
R = 0.0821 L·atm/(mol·K)
T = 273K

Rearranging the equation, we have: V = (nRT)/P
Substituting the values, we get:
V = [(1/2) x 0.03264 x 0.0821 x 273] / 760
= 0.00699 L

Finally, convert the volume from liters to milliliters:
V = 0.00699 L x 1000 mL/L
= 6.99 mL

Therefore, the volume of O2 produced in one minute is approximately 6.99 mL.