Answer:
Esterification reaction
Explanation:
When we have to go from an acid to an ester we can use the <u>esterification reaction</u>. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).
In this case, we need the <u>methyl ester</u>, therefore we have to choose the <u>appropriate alcohol</u>, so we have to use the <u>methanol</u> as reactive if we have to produce the methyl ester.
The phase change in which the water molecules become most orderly is the freezing. This is the process of changing water as liquid to its solidified form. The process of freezing is an exothermic which means that for this to occur, heat should be removed from the system.
Answer:
60 V
Explanation:
From;
Vs/Vp = Ns/Np
Where;
Vs = voltage in the secondary coil = 6V
Vp = voltage in the primary coil= ??
Ns = number of turns in the secondary coil = 9
Np= number of turns in the primary coil = 90
6/Vp = 9/90
Vp= 90 * 6/9
Vp= 60 V
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2
