Answer: The rate constant for the reaction is 
Explanation:
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = age of sample = 559 min
a = let initial amount of the reactant = 
a - x = amount left after decay process = 



The rate constant for the reaction is 
<u>Answer:</u> The correct answer is 
<u>Explanation:</u>
We are given:

The substance having highest positive
potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.
Chromium will undergo oxidation reaction and will get oxidized.
The half reactions for the above cell is:
Oxidation half reaction: 
Reduction half reaction:
( × 3)
Net equation: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:

Putting values in above equation, we get:

Hence, the correct answer is 
Two things that store chemical energy are coal, and wood cause when you have a fire you have to get the chemicals from somewhere to keep your fire live.
Isotopes are substances that have the same number of protons but differ in the number of neutrons. Hence, the pair of isotopes above should be of the same element. In the given choices, 14C is not an isotope of 14N. 206Pb is an isotope of 208 Pb. O2 and O3 differ in molecular formula but still made up of same kind of atom, hence they are allotropes, while 32S and 32S2- are not isotopes.
Silicon is used in producing Pv celss because cells made from the crystallized polycrystalline silicon have efficiencies in the range of 14-18% and a moderate price.