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.
Answer:
1. 7.256g of NaCl
2. 47.33g of Cl2
Explanation:
2 moles of Na reacts to produce 2 moles of NaCl
8 moles of Na will still produce 8 moles of NaCl
Mass of NaCl = molar mass of Nacl/moles of Nacl
=58.5/8
=7.256g of NaCl
From the equation, 2 moles of Na reacts with 1 mole of Cl2
3/2 moles of Cl2 will react with 3 moles of Na
Mass of Cl2 = 71/1.5
=47.33g of Cl2
Atomic number and the number of protons are the same...
Neutrons = Mass number - number of protons
Electrons are same # unless there is a charge
The whole number you see on the periodic table is the atomic number of the element which is also same as the number of protons
1) carbon - 14 ; Mass number = 14 , Protons = 6 , Neutrons = 14 - 6 = 8
Electrons = 6
2) Lead - 208 ; Mass # = 208 , Protons = 82 , Neutrons = 208 - 82 = 126
Electrons = 82
3) Uranium - 239 ; Mass # = 239 , Protons = 92,Neutrons = 239 - 92 = 147
Electrons = 92
4) Uranium - 238 ; Mass # = 238 , Protons = 92 , Neutrons = 238 - 92 = 146
Electrons = 92
5) Tin - 118 ; Mass # = 118 , Protons = 50 , Neutrons = 118 - 50 = 68
Electrons = 50
Answer:
I don't know if this is right but try it. The amount of water vapor in the air is called absolute humidity. The amount of water vapor in the air as compared with the amount of water that the air could hold is called relative humidity. This amount of space in air that can hold water changes depending on the temperature and pressure.
Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
