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mr Goodwill [35]
3 years ago
14

Consider the temperature versus time graph below. I which region is the substance in both the solid phase and the liquid phase?

1 2 3 4

Chemistry
2 answers:
SVEN [57.7K]3 years ago
8 0

Answer:

2

Explanation:

The change of state occurs at a constant temperture and pressure. In the grahp we can see while the time passes, the temperature doesn't change.

The rect number 4 correspond to a liquid-gas phase

Rasek [7]3 years ago
4 0

Answer : The stage 2 shows the substance in both the solid phase and the liquid phase.

Explanation :

In the given phase diagram,

Stage 1 : It shows the solid state of a substance.

Stage 2 : It shows the melting and freezing process of a substance. It means that when the phase changes from solid state to liquid state at constant temperature is know as melting process and when the phase changes from liquid state to solid state at constant temperature is know as freezing process.

Stage 3 : It shows the liquid state of a substance.

Stage 4 : It shows the evaporation and condensation process of a substance. It means that when the phase changes from liquid state to gaseous state at constant temperature is know as evaporation process and when the phase changes from gaseous state to liquid state at constant temperature is know as condensation process.

Stage 5 : It shows the gaseous state of a substance.

Hence, from the above information we conclude that the stage 2 shows the substance in both the solid phase and the liquid phase.

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
Annette [7]

Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

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

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  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

4 0
3 years ago
A. 1
Lubov Fominskaja [6]
I think it’s B. I’m not 100% sure but I believe it is B! Srry if this is late btw.
8 0
3 years ago
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