Answer:
The essence including its given problem is outlined in the following segment on the context..
Explanation:
The given values are:
Moles of CO₂,
x = 0.01962
Moles of water,



Compound's mass,
= 0.4647 g
Let the compound's formula will be:

Combustion's general equation will be:
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
Now,
x : y : z = 
= 
= 
= 
So that the empirical formula seems to be "C₃H₆O₂".
1400 centimeters in 1.4 millimeters
Answer:
rate of recrystallization = 4.99 × 10⁻³ min⁻¹
Explanation:
For Avrami equation:

To calculate the value of k which is a dependent variable for the above equation ; we have:


The time needed for 50% transformation can be determined as follows:
![y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ t =[ \dfrac{-In(1-y)}{k}]^{^{1/n}}](https://tex.z-dn.net/?f=y%20%3D%201-e%20%5E%7B%28-kt%5En%29%7D%20%5C%5C%20%5C%5C%20e%5E%7B%28-kt%5En%29%7D%20%3D%201-y%5C%5C%20%5C%5C%20-kt%5En%20%3D%20In%281-y%29%20%5C%5C%20%5C%5C%20t%20%3D%5B%20%5Cdfrac%7B-In%281-y%29%7D%7Bk%7D%5D%5E%7B%5E%7B1%2Fn%7D%7D)
![t_{0.5} =[ \dfrac{-In(1-0.4)}{9.030 \times 10^{-7}}]^{^{1/2.5}}](https://tex.z-dn.net/?f=t_%7B0.5%7D%20%3D%5B%20%5Cdfrac%7B-In%281-0.4%29%7D%7B9.030%20%5Ctimes%2010%5E%7B-7%7D%7D%5D%5E%7B%5E%7B1%2F2.5%7D%7D)
= 200.00183 min
The rate of reaction for Avrami equation is:


rate = 0.00499 / min
rate of recrystallization = 4.99 × 10⁻³ min⁻¹
Your answer would be (D), Noble Gases, or Inert Gases - The far right on the periodic table is also known as Group(0)., or Group 18 on the periodic table. Elements in this group includes, Neon(NE), Argon(AR), Krypton(KR), Xenon(XE), and Radon(RN).
Hope that helps!!!
Answer:
the balanced equation to the combustion of gasoline is C8H18 + 12.5 O2 → 8 CO2 + 9 H2O (1)