Answer:
A)
,
, ![E_{a} = 50KJ/mol](https://tex.z-dn.net/?f=E_%7Ba%7D%20%3D%2050KJ%2Fmol)
A = 1.5×
, A = 1.9×
, A=1.5×![10^{-7} s^{-1}](https://tex.z-dn.net/?f=10%5E%7B-7%7D%20s%5E%7B-1%7D)
B) 4.469
Explanation:
From Arrhenius equation
![K=Ae^{\frac{E_{a} }{RT} }](https://tex.z-dn.net/?f=K%3DAe%5E%7B%5Cfrac%7BE_%7Ba%7D%20%7D%7BRT%7D%20%7D)
where; K = Rate of constant
A = Pre exponetial factor
= Activation Energy
R = Universal constant
T = Temperature in Kelvin
Given parameters:
![E_{a} =165KJ/mol](https://tex.z-dn.net/?f=E_%7Ba%7D%20%3D165KJ%2Fmol)
![T_{1}=505K](https://tex.z-dn.net/?f=T_%7B1%7D%3D505K)
![T_{2}=525K](https://tex.z-dn.net/?f=T_%7B2%7D%3D525K)
![R=8.314JK^{-1}mol^{-1}](https://tex.z-dn.net/?f=R%3D8.314JK%5E%7B-1%7Dmol%5E%7B-1%7D)
taking logarithm on both sides of the equation we have;
![InK=InA-\frac{E_{a} }{RT}](https://tex.z-dn.net/?f=InK%3DInA-%5Cfrac%7BE_%7Ba%7D%20%7D%7BRT%7D)
since we have the rate of two different temperature the equation can be derived as:
![In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )](https://tex.z-dn.net/?f=In%28%5Cfrac%7BK_%7B2%7D%20%7D%7BK_%7B1%7D%20%7D%20%29%3D%5Cfrac%7BE_%7Ba%7D%20%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%20%7D%20-%5Cfrac%7B1%7D%7BT_%7B2%7D%20%7D%20%29)
![In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )](https://tex.z-dn.net/?f=In%28%5Cfrac%7BK_%7B2%7D%20%7D%7BK_%7B1%7D%20%7D%20%29%3D%5Cfrac%7B165000J%2Fmol%7D%7B8.314JK%5E%7B-1%7Dmol%5E%7B-1%7D%20%20%7D.%28%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%29)
= 19846.04×7.544×
= 1.497
=
= 4.469
Answer:
2,75 mol of O2 it's 88 g of O2.
Explanation:
The weight of the diatomic molecule O2 is 32 g/mol. So considering that, you should multiply 2,75 mol · 32 = 88g :)
Answer:
orbital
Explanation:
electrons are found in an orbital
4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.
Solution:
We will start with the Molarity
![\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Molarity%20%7D%3D%5Ctext%20%7B%20Mole%20of%20solute%20%7D%20%5Cdiv%20%5Ctext%20%7B%20liter%20of%20solution%20%7D)
Also we know 1000 ml = 1 L
Therefore 37.5 ml by 1000ml we obtained 0.0375L
Equation for solving mole of solute
![\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Mole%20of%20solute%20%7D%3D%5Ctext%20%7B%20Molarity%20%7D%20%5Ctimes%20%5Ctext%20%7B%20Liters%20of%20solution%20%7D)
Now, multiply 0.750M by 0.0375
Substitute the known values in the above equation we get
![0.750 \times 0.0375=0.0281](https://tex.z-dn.net/?f=0.750%20%5Ctimes%200.0375%3D0.0281)
Also we know that Molar mass of KI is 166 g/mol
So divide the molar mass value to get the no of grams.
![0.028 \times 166=4.648](https://tex.z-dn.net/?f=0.028%20%5Ctimes%20166%3D4.648)
So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.