B. is ur answer I bieleve
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
For a good consert mix aggregate needed to the clean hard strong partical free of absorb chemicals or coating clay and other fine materials
Answer:
209.68
Explanation:
The only number that is relevant (though the rest are quite interesting) is the last one 1.98 * 10^24
1 mole of Barium Acetate Contains 6.02*10^23 particles.
There are 4 moles of carbon to every mole of Barium Acetate.
1.98 * 10^24 atoms / (4*6.02*10^23)
0.8223 moles of Ba(C2H3O2)2
Ba = 137
4C = 4*12 48
6H = 6*1 6
4O = 4*16 64
1mole 255 grams
0.8223 * 255 = 209.68 grams
I have used rounded masses for these elements depending on the periodic table you use. Go through the question with your masses to get a more accurate answer. My answer will not differ by much. It is a guide.
It is called a phenyl group
