Answer:
[Ar] 3d10 4s2 4p4
Explanation:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
Answer:
a. 300 kg of Fertilizer
b. 225 kg of fertilizer
c.400 Kg of fertilizer
d.600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.
The percentages can be expressed as fractions as follows:
For nitrogen; 40/100 = 0.4
For phosphorus; 15/100 = 0.15
For potassium; 10/100 = 0.1
To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
Answer:
Answer: V=67.2 L
Explanation:
Ideal Gas Law: PV=nRT
P=1.00 atm (STP)
V=?
n=3.00 mol
R=0.08206Latm/Kmol
T=273.15 K (STP)
To find V, we would manipulate the equation to V=nRT/P
With significan figures, our answer is V=67.2 L.
This will be classified as light on the API scale due to the large percentage of lighter fractions such as paraffins and naphthenes.
Answer:
a. Kp=1.4


b.Kp=2.0 * 10^-4


c.Kp=2.0 * 10^5


Explanation:
For the reaction
A(g)⇌2B(g)
Kp is defined as:

The conditions in the system are:
A B
initial 0 1 atm
equilibrium x 1atm-2x
At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.
Replacing these values in the expression for Kp we get:

Working with this equation:

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1
The general expression to solve these kinds of equations is:
(equation 1)
We just take the positive values from the solution since negative partial pressures don´t make physical sense.
Kp = 1.4


With x1 we get a partial pressure of:


Since negative partial pressure don´t make physical sense x1 is not the solution for the system.
With x2 we get:


These partial pressures make sense so x2 is the solution for the equation.
We follow the same analysis for the other values of Kp.
Kp=2*10^-4
X1=0.505
X2=0.495
With x1


Not sense.
With x2


X2 is the solution for this equation.
Kp=2*10^5
X1=50001

With x1


Not sense.
With x2


X2 is the solution for this equation.