Answer:
Pb oxidation number is +2
C in CH4 is -4. H is +1. 4H + 1C = 0. ; 4(+1) + C = 0 ; C = -4
O is usually -2. 3O = 3(-2) = -6. ; 2Fe =. +6 ; each Fe is +3
2Ag = +2 since O = -2 ; each Ag = +1
Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
Answer:
Hi
Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).
Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).
As an example (figure 3).
In the attached file are each of the steps of Williamson's synthesis.
Explanation:
Answer:
1. Hubble
2.Aristarchus
3.Galileo
4.Aristotle
5.Kepler
6.Jupiter
7.Mercury
8.Uranus
9.Venus
10.Earth
11. Jupiter and Mars
12. sun
13.D
14. Saturn
15.Gas giants
16.Kuiper Belt and Ceres,Pluto,eris and haumea.
17.pluto
18.Ptolemy's model
19.meteor- a small body of matter from the outer surface that enters the earth's atmosphere.
meteoroid- a small body moving in the solar system that would become a meteor if it enters the earth's atmosphere.
meteorite- a piece of rock or metal that has fallen to the earth's surface from outer space as a meteor.
the relationship is the diagram I drew at the top.
hope this was helpful
Answer: 36.53g
Explanation:
First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl
Molarity = 5M
MM of NaCl = 58.45
Molarity = Mass conc (g/L) / MM
Mass conc. (g/L) of NaCl = Molarity x MM
= 5 x 58.45 = 292.25g
Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)
From the calculations above,
292.25g of NaCl dissolved in 1L
Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e
Xg of NaCl = 292.25 x 0.125 = 36.53g.
Therefore 36.53g of NaCl will dissolve in 125mL of the solution
