Answer:
1.76 * 6.02*10^23 = 1.05952*10^24
1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms
1 mol --------- 6.02x10²³ atoms
6.3 moles ---- ??
atoms = 6.3 * 6.02x10²³
= 3.7926x10²⁴ atoms
hope this helps!
Answer:
The volume of the liquid should be measured before it is heated.
Explanation:
Answer:
0.124 M
Explanation:
The reaction obeys second-order kinetics:
![r = k[BrO^-]^2](https://tex.z-dn.net/?f=r%20%3D%20k%5BBrO%5E-%5D%5E2)
According to the integrated second-order rate law, we may rewrite the rate law in terms of:
![\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_t%7D%20%3D%20kt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D)
Here:
is a rate constant,
is the molarity of the reactant at time t,
is the initial molarity of the reactant.
Converting the time into seconds (since the rate constant has seconds in its units), we obtain:

Rearranging the integrated equation for the amount at time t:
![[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7Bkt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D%7D)
We may now substitute the data:
![[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7B0.056%20M%5E%7B-1%7Ds%5E%7B-1%7D%5Ccdot%2060.0%20s%20%2B%20%5Cdfrac%7B1%7D%7B0.212%20M%7D%7D%20%3D%200.124%20M)