Answer:
30.62 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 55 L
Initial pressure (P₁) = 3.2 atm
Initial temperature (T₁) = 520 K
Final temperature (T₂) = 760 K
Final pressure (P₂) = 8.4 atm
Final volume (V₂) =?
The final volume of the gas can be obtained as follow:
P₁V₁ / T₁ = P₂V₂ / T₂
3.2 × 55 / 520 = 8.4 × V₂ / 760
176 / 520 = 8.4 × V₂ / 760
Cross multiply
520 × 8.4 × V₂ = 176 × 760
4368 × V₂ = 133760
Divide both side by 4368
V₂ = 133760 / 4368
V₂ = 30.62 L
Therefore, the new volume of the gas is 30.62 L
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for 
with a rate constant of 
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
Answer:
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Explanation:
Equating coefficients, you get ...
aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃
For Ba: 3a = d
For P: 2a = 4c
For O: 8a +2b = 10c +3d
For Si: b = d
__
Expressing everything in terms of b and c, we get ...
d = b
a = b/3 = 2c
From the second, b = 6c, so we have ...
a = 2c
b = 6c
c = c
d = 6c
And we can write the equation with c=1 as ...
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
The correct answer is letter <span>C. mixture in which its components retain their identity. A heterogeneous mixture is a mixtures in which the component of the mixed are not uniform. You can see that there are localized regions that have different properties. The components have the capacity to retain their identity.</span>
<span>
</span>
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =
