Answer:
8.1×10^-8 mols-1
Explanation:
Now we have the mass of copper sulphate produced after three days. Recall that the rate of reaction is given as;
Rate= change in the concentration of product/time
At the beginning of the reaction, there was 0 moles of copper sulphate
After 72 hours or 259200 seconds, there was 3.4g/160gmol-1 = 0.021 moles of copper sulphate.
Note that 160gmol-1 is the molar mass of copper sulphate.
Hence;
Rate of reaction= 0.021 moles /259200 seconds
Hence, the rate of reaction is 8.1×10^-8 mols-1
Rate of reaction= 8.1×10^-8 mols-1
Answer:
To increase surface area of the platinum electrode which results in superior quality and action of the electrodes as opposed to normal platinum electrodes.
Explanation:
Platinization of Platinum is the process of covering platinum electrode with a layer of platinum black. Platinum black is a finally divided form of platinum, optimized for catalysing the addition of hydrogen to unsaturated organic compound. This increases the surface area of the platinum electrodes and therefore exhibits action superior to that of normal electrodes.
Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256
Explanation:
Molarity is defined as the number of moles dissolved per liter of the solution.
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of 
solution = 1.42 M
Volume of solution = 180.0 mL
Putting values in equation 1, we get:
Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.
Answer: The concentration of excess ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of 
give = 1 mole of 
Thus 0.019 moles of give = 0.019 mole of
2. moles of 
According to stoichiometry:
1 mole of 
gives = 2 moles of 
Thus 0.012 moles of give = 
moles of
As 1 mole of neutralize 1 mole of
0.019 mole of will neutralize 0.019 mole of
Thus (0.024-0.019)= 0.005 moles of will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of in solution is 0.017 M.
