Answer:
4Al203 + 9Fe -> 3Fe3O4 + 8Al
Amount of Al on reactant side: 8
Amount of Al on product side: 8
Amount of O on reactant side: 12
Amount of O on product side: 12
Amount of Fe on reactant side: 9
Amount of Fe on product side: 9
Answer:
for the reaction is -198.762 J/K
Explanation:

Standard change in entropy for the system (
) is given by-
![\Delta S^{0}=[2moles\times S^{0}(NH_{3})_{g}]-[1mole\times S^{0}(N_{2})_{g}]-[3\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B2moles%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B1mole%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B3%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where
represents standard entropy.
Here
,
and 
So, ![\Delta S^{0}=[2\times 192.45]-[1\times 191.61]-[3\times 130.684]J/K=-198.762J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B2%5Ctimes%20192.45%5D-%5B1%5Ctimes%20191.61%5D-%5B3%5Ctimes%20130.684%5DJ%2FK%3D-198.762J%2FK)
Answer:
Classification of elements in groups provide us with a fixed pattern in which the elements change their properties periodically.
Explanation:
<u>Answer:</u> The correct answer is Option d.
<u>Explanation:</u>
Electronegativity is defined as the tendency of an element to attract the shared pair of electron towards itself in a compound.
When electron is transferred from less electronegative atom to more electronegative atom.
The atom which is more electronegative gains electron and undergoes reduction reaction.
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

During the transfer of electrons, energy is released when an electron looses is potential energy during the transfer.
Hence, the correct answer is Option d.
Radioactive decay is a pseudo-first order reaction. When you know the half-life of the material, you could use this equation.
A= A₀(1/2)^t/h
where
A is the final activity
A₀ is the initial activity
t is the time
h is the half-life
A = (0.64)(1/2)^88/44 = <em>0.16 mbq</em>