The right answer is D - mass.
4 infiltration percolation!! I think! Correct me if I’m wrong
Answer:
A) , ,
A = 1.5×, A = 1.9×, A=1.5×
B) 4.469
Explanation:
From Arrhenius equation
where; K = Rate of constant
A = Pre exponetial factor
= Activation Energy
R = Universal constant
T = Temperature in Kelvin
Given parameters:
taking logarithm on both sides of the equation we have;
since we have the rate of two different temperature the equation can be derived as:
= 19846.04×7.544× = 1.497
= = 4.469
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48