Answer:
The sestira structure makes the poem feel like a playful puzzle, but aciels nothing to the meaning of the poem
The repetition of the last word in each line creates a reflective rhythm and tone in the poem while also making significant words
easily recognizable
Explanation:
That was a really good one, I think you may want to go with that answer.
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
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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
Answer:
Ideal Solution: An ideal solution is a solution where interactions between molecules are identical between all the molecules in the solution. Non ideal Solution: A non ideal solution is a solution that has differences in the interactions between molecules of different components in the solution.
Explanation:
Given data:
Hydrogen (H) = 3.730 % by mass
Carbon (C) = 44.44%
Nitrogen (N) = 51.83 %
This means that if the sample weighs 100 g then:
Mass of H = 3.730 g
Mass of C = 44.44 g
Mass of N = 51.83 g
Now, calculate the # moles of each element:
# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles
# moles of C = 44.44/12 = 3.703 moles
# moles of N = 51.83/14 = 3.702 moles
Divide by the lowest # moles:
H = 3.730/3.702 = 1
C = 3.703/3.702 = 1
N = 3.702/3.702 = 1
Empirical Formula = HCN