Answer:
3.18 mol
Explanation:
n(CO2) = mass/ Mr.
= 25.5 / 16
= 1.59 mol
As per the equation above,
n(LiOH) : n(CO2)
2 : 1
∴ 3.18 : 1.59
Answer :
The time taken by the reaction is 19.2 seconds.
The order of reaction is, second order reaction.
Explanation :
The general formula to determine the unit of rate constant is:
Unit of rate constant Order of reaction
0
1
2
As the unit of rate constant is 
. So, the order of reaction is second order.
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = 
t = time = ?
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
= final concentration = 0.97 M
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 2.48 M
Now put all the given values in the above expression, we get:
Therefore, the time taken by the reaction is 19.2 seconds.
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of 
indicates that each of these ions carries a charge of 
. That corresponds to the shortage of one electron for each 
ion.
Similarly, the superscript 
on each 
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is 
, and that the coefficient of (among the reactants) is 
.
.
There would thus be silver (
) atoms and aluminum (
) atoms on either side of the equation. Hence, the coefficient for 
and 
would be 
and 
, respectively.
.
The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the 
ions on the right-hand side of this equation would correspond to the shortage of 
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, 
.
The smallest integer and that could satisfy this relation are 
and 
. The equation becomes:
.
Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7
