Answer:
Determining how many ozone molecules are lost in the atmosphere.
Explanation:
According to Le Châtelier's principle, an increase in temperature favors the endothermic process. Since the reaction, which proceeds towards bromine chloride, is endothermic, the reaction would shift right, making more products.
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN
Explanation:
- A good buffer system contains a weak acid and its salt or a weak base and its salt.
- In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
- HCN is a weak acid, while LiCN is a salt of the weak acid, that is, CN- conjugate of the acid.
Answer: 2.58 days
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = age of sample = 6 days
a = initial amount of the reactant = 1 g
a - x = amount left after decay process
= 0.2 g
a) to find the rate constant
b) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
The half life is 2.58 days
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
