Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer:
4600s
Explanation:
For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(B)]}{P(B)}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28B%29%5D%7D%7BP%28B%29%7D%3Dk%2Adt)
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)
Answer:
Ver las respuestas abajo.
Explanation:
Este problema se puede resolver conociendo la relacion entre horas y minutos, sabemos que:
1 hora [h] → 60 minutos [min]
De esta manera:
2 [min] = 2/60 = 0.033 [h]
15 [min] = 15/60 = 0.25 [h]
30 [min] = 30/60 = 0.5 [h]
10 [min] = 10/60 = 0.166 [h]
6 [min] = 6/60 = 0.1 [h]
20 [min] = 20/60 = 0.33 [h]
5 [min] = 5/60 = 0.0833 [h]
Answer:it’s C
Explanation:
A distant luminous object travels rapidly away from an observer.
Answer:
The Earth's magnetic field intensity is roughly between 25,000 - 65,000 north magnetic pole is 86.50°N and 164.04°E and the south magnetic pole is 64.07°S and 135.88°E
