Answer: When using 645 L /s of O2 in a temperature and pressure of 195°C, 0.88 atm respectively, we will get 0.355Kg /s NO
Explanation:
- First we review the equation that represents the oxidation process of the NH3 to NO.
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
- Second we gather the information what we are going to use in our calculations.
O2 Volume Rate = 645 L /s
Pressure = 0.88 atm
Temperature = 195°C + 273 = 468K
NO molecular weight = 30.01 g/mol
- Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT
Let´s keep in mind that using this equation our constant R is 0.08205Lxatm/Kxmol
PV =n RT
n= PV / RT
n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]
n= 14.781 moles /s of O2
-
Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .
14.781 moles /s of O2 x 4moles NO / 5 moles O2 x 30.01g NO / 1 mol NO x 1Kg NO /1000g NO = 0.355Kg /s NO
A hypothesis is how you think the experiment is going to end
the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)
Answer:
tide
Explanation:
A tide is when the moon reactes to the moon is caused gravity to make the tide moves
To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:
HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-
Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
HC2H3OO H+ C2H3OO-
I 0.195 0 0.125
C -x +x +x
------------------------------------------------------------------
E 0.195-x x 0.125 + x
Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x
x = 2.81x10^-5 M = [H+]
pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55
Therefore, the pH of the resulting solution would be 4.55.</span>
