Answer:
Explanation:
MM of I2 = 2 (127 g) = 254 g/mol
0.065 mol I2 x 254g I₂/ 1 mol I₂ = 16.5 g I2
Answer:
The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.
Explanation:
Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
Best regards!
The bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.
No of moles of calcium bromide is calculated as follows:
moles = 1.642 ÷ 199.89 = 8.215 × 10-³moles
Molarity = 8.215 × 10-³moles ÷ 469.1mL = 1.75 × 10-⁵M
Therefore, the bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
Learn more about concentration at: brainly.com/question/10725862
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