Answer:
The statement that best describes the trend in first ionization enery of elements on the periodic table is:
It generally decreases down a group because valence electrons are farther from the nucleus.
The first ionization energy measures how difficult is to release an electron from the outermost shell. The higher the ionization energy the more difficult it is to release an electron, the lower the ionication energy the easier to release an electron.
As the atomic number of the atom increases (which is what happens when you go down a group) the furthest the outermost shell of electrons will be (the size of the atoms increases) and so those electrons require less energy to be released, which means that the ionization energy decreases.
Hope it helps!
Answer:
The total pressure after one half is 6.375 atm.
Explanation:
The initial pressure of product is increases while the pressure of reactant would decrease.
Balanced chemical equation:
2N₂O → 2N₂ + O₂
The pressure of N₂O is 5.10 atm. The change in pressure would be,
N₂O = -2x
N₂ = +2x
O₂ = +x
The total pressure will be
P(total) = P(N₂O) + P(N₂) + P(O₂)
P(total) = ( 5.10 - 2x) + (2x) + (x)
P(total) = 5.10 + x
After one half life:
P(N₂O) = 1/2(5.10) = 5.10 - 2x
x = 5.10 - 1/2(5.10) /2
x = 5.10 - 0.5 (5.10) /2
x = 5.10 - 2.55 / 2
x = 2.55 /2 = 1.275 atm
Thus the total pressure will be,
P(total) = 5.10 + x
P(total) = 5.10 + 1.275
P(total) = 6.375 atm
Answer:
1.274 moles
Explanation:
The equation for the reaction can be represented as follows:
⇄
+ 
K = 0.060
K = ![\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Concentration of
= 
Concentration of
= 
Concentration of
= 0.398 moles
If we construct an ICE table for the above equation; we have:
⇄
+ 
Initial 0.398 0 0
Change - x + x + x
Equilibrium (0.398 - x) x x
K = ![\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
K = ![\frac{[x][x]}{[0.398-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.398-x%5D%7D)
K = 
0.060 = 
0.06(0.398-x) = x²
0.02388 - 0.060x = x²
x² + 0.060x - 0.02388 = 0 (quadratic equation)
a = 1; b= 0.06; c= -0.02388
Using quadratic formula;
= 
= 
= 
= 
= 
=
or 
=
or 
= 0.1274 or -0.1874
We go by the positive value which says:
[x] = 0.1274 M
number of moles = 0.1274 × 10.0
= 1.274 moles
∴ the number of moles of Cl₂ produced at equilibrium = 1.274 moles
A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale. For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.
Answer:
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Explanation: