In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol 
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is 
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration ![[A_{0} ]](https://tex.z-dn.net/?f=%5BA_%7B0%7D%20%5D)
is 100 then concentration at time t ![[A_{t} ]](https://tex.z-dn.net/?f=%5BA_%7Bt%7D%20%5D)
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.
Answer:
glucose is in large sugar molecules, while fructose is in simple fruits and vegetables
I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra
It’s “c. +1” because the Sodium is losing a proton, which has a positive chage
1. the Moon must be in line with the Earth and Sun - New Moon for a solar eclipse and Full Moon for a lunar eclipse.2. the Moon must be at one of its nodes. Most of the time the Moon at New or Full passes above or below the Sun because its orbit is tilted relative to the ecliptic. An eclipse can occur only when it is at a node, in other words crossing the ecliptic.
