The final temperature :T₂=411.125 K = 138.125 °C
<h3>Further explanation</h3>
Given
T₁=-20 °C+273 = 253 K
V₁=40 ml
V₂=65 ml
Required
Final temperature
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature

Input the value :
T₂=(V₂.T₁)/V₁
T₂=(65.253)/40
T₂=411.125 K = 138.125 °C
Answer:
![r = k . [CO] .[Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%20.%5BCl_%7B2%7D%5D)
Explanation:
Let´s consider the following reaction:
CO + Cl₂ ⇒ COCl₂
The general rate law is:
![r = k . [CO]^{m}. [Cl_{2}]^{n}](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%5E%7Bm%7D.%20%5BCl_%7B2%7D%5D%5E%7Bn%7D)
where,
r is the rate of the reaction
k is the rate constant
[CO] and [Cl₂] are the molar concentrations of each reactant
m and n are the reaction orders for each reactant
Since the reaction is first order in CO, m = 1. The overall order is the sum of all the individual orders. In this case, the overall order m + n = 2. Then,
m + n = 2
n = 2 - m = 2 - 1 = 1
The reaction is first order in Cl₂.
The rate law is:
![r = k . [CO]. [Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D.%20%5BCl_%7B2%7D%5D)
Answer:
Volume of solution = 80.5 mL
Explanation:
Given data:
Molarity of solution = 4.50 mol/L
Mass of ethanol = 16.7 g
Volume of solution = ?
Solution:
Volume will be calculated from molarity formula.
Molarity = number of moles / volume in L
Number of moles:
Number of moles = mass/molar mass
Number of moles = 16.7 g/ 46.07 g/mol
Number of moles = 0.3625 mol
Volume of solution:
Molarity = number of moles / volume in L
4.50 mol/L = 0.3625 mol / volume in L
Volume in L = 0.3625 mol /4.50 mol/L
Volume in L = 0.0805 L
Volume in mL:
0.0805 L ×1000 mL/1 L
80.5 mL
Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-
Explanation:
The resonance compounds are compounds that have the same position of the atoms (same quantity and elements) but the position of the electrons arround them is different in each resonance compound.
In reality, the compound switches between all the resonance structures it has.
One example of resonance Lewis structures is the Ozone's as can be seen in the figure.