Answer:
Balanced equation have equal number of atoms of different elements in the side of reactants and products.
Answer:
A) M = 100X
B) M = 36X
C) M = 178.88X
Explanation:
Given data:
ASTM grain size number 7
a) total grain per inch^2 - 64 grain/inch^2
we know that number of grain per square inch is given as

where M is magnification, n is grain size
therefore we have

solving for M we get
M = 100 X
B) total grain per inch^2 = 500 grain/inch^2
we know that number of grain per square inch is given as

where M is magnification, n is grain size
therefore we have
solving for M we get
M = 36 X
C) Total grain per inch^2 = 20 grain/inch^2
we know that number of grain per square inch is given as

where M is magnification, n is grain size
therefore we have
solving for M we get
M = 178.88 X
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is

KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
1.
The balanced chemical reaction is:
N2 +3 I2 = 2NI3
We are given the amount of product formed.
This will be the starting point of our calculations.
3.58 g NI3 ( 1 mol NI3 / 394.71 g NI3 ) ( 3
mol I2 / 2 mol NI3 ) = 0.014 mol I2.
Thus, 0.014 mol of I2 is needed to form the
given amount of NI3.
Answer:
The activation energy is 164.02 kJ/mol
Explanation:
Log (k2/k1) = Ea/2.303R × [1/T1 - 1/T2]
k1 = 8.9×10^-4 s^-1
k2 = 9.83×10^-3 s^-1
R = 8.314 J/mol.K
T1 = 540 K
T2 = 578 K
Log (9.83×10^-3/8.9×10^-4) = Ea/2.303×8.314 × [1/540 - 1/578]
1.043 = 6.359×10^-6Ea
Ea = 1.043/6.359×10^-6 = 164020 J/mol = 164020/1000 = 164.02 kJ/mol