Answer:
The correct statements are:
The rate of disappearance of B is twice the rate of appearance of C.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
3A + 2B → C + 2D
Rate of the reaction:
![R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
The rate of disappearance of B is twice the rate of appearance of C.
![\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Answer:
151.63 g
Explanation:
We first get the number of moles;
Moles = Molarity × volume
= 2.0 × 0.475
= 0.95 moles
1 mole of CuSO4 = 159.609 g/mol
Therefore;
Mass = moles × molar mas
= 0.95 moles × 159.609 g/mol
= 151.63 g
True. Nuclear fusion of hydrogen to form helium occurs naturallyin the sun and other stars. It takes place only at extremely high temperatures.
