Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,
Answer:The visible light contains 7 colours : Violet, indigo, Blue, Green, Yellow, Orange and Red.
White colour is formed by the interference of all 7 colours, thus white light has the combination of all frequencies of visible light.
Explanation:
Answer:
V = 12.5 L
Explanation:
Given data:
Volume of NO = 15.0 L
Temperature and pressure = standard
Volume of nitrogen gas produced = ?
Solution:
Chemical equation:
6NO + 4NH₃ → 5N₂ + 6 H₂O
Number of moles of NO:
PV = nRT
n = PV/RT
n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K
n = 15.0 atm.L / 22.43 atm.L /mol
n = 0.67 mol
now we will compare the moles of No and nitrogen gas.
NO : N₂
6 : 5
0.67 : 5/6×0.67 = 0.56
Volume of nitrogen gas:
PV = nRT
1 atm × V = 0.56 mol × 0.0821 atm.L /mol.K × 273.15 K
V = 12.5 atm.L / 1 atm
V = 12.5 L
23. The correct answer would be a (AlPO4 + H2O).
This is an example of a neutralization reaction (an acid and a base react to form water and a salt).
24. a, combination reaction. There are two reactants and one product.
Answer:
Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.
an azide ion
An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,
imide linkage
You should note the commonly used trivial names of the following compounds.
phthalic acid, phthalic anhydride, and phthalimide
The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.
If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).
Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).
The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.
Explanation:
