To develop the chemical formula of a substance or compound in this regard, one only needs to write the chemical symbol for each of the element present and write as subscript to each symbol the number of atoms of that element. For nitrobenzene, as described above, the chemical formula should be,
C₆H₅O₂N
If 0.380 mol of barium nitrate is allowed to react with an excess of phosphoric acid, 0.127 moles of barium phosphate could be formed.
<h3>How to calculate number of moles?</h3>
The number of moles of a compound can be calculated stoichiometrically as follows:
Based on this question, the following chemical equation is given:
Ba (NO3)2+ H3PO4 → Ba3(PO4)2+ HNO3
The balanced equation is as follows:
3Ba(NO3)2 + 2H3PO4 → Ba3(PO4)2 + 6HNO3
3 moles of barium nitrate produces 1 mole of barium phosphate
Therefore, 0.380 moles of barium nitrate will produce 0.380/3 = 0.127moles of barium phosphate.
Learn more about stoichiometry at: brainly.com/question/9743981
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
Answer: A)H+
Explanation:my teacher just told me
Volume:
a x b x c
Therefore:
5.54 x 10.6 cm x 199 => 11,686.076 cm³
