Answer:
5.4 tonnes.
Explanation:
The first step is to find the molar mass of Al2O3. Aluminum has a molar mass of about 27 and oxygen has a molar mass of about 16, so 2(27)+3(16)= 102g/mol=0.102kg/mol. 10200kg/0.102kg/mol=100,000 moles of Al2O3 in 10.2 tonnes. Multiplying this by the molar mass of the two aluminums, you get a total of 54*100,000=5400000g=5400kg=5.4 tonnes. Hope this helps!
Balanced chemical reaction:
PbO₂<span>(s) + Sn(s)+ 4H</span>⁺(aq) → Pb²⁺(aq) + Sn²⁺(aq) + 2H₂O<span>(l).
Oxidation half-reaction: Sn </span>→ Sn²⁺ + 2e⁻.<span>
Reduction half-reaction: PbO</span>₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O.
Net reaction: Sn + PbO₂ + 4H⁺ + 2e⁻ → Sn²⁺ + 2e⁻ + Pb²⁺ + 2H₂O.
Oxidation is increase of oxidation number, reduction is decrease of oxidation number.
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
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
4 NH₃ + 3O₂ --> 2N₂ + 6H₂O
First, make sure that this is a balanced equation.
There are 4 moles of nitrogen on the left side, and 4 moles of nitrogen on the right side.
There are 12 moles of hydrogen on the left side, and 12 moles of hydrogen on the right side.
There are 6 moles of oxygen on the left side, and 6 moles of oxygen on the right side.
The equation is therefore balanced, and we may proceed.
a) the mole ratio for NH₃ to N₂ is 4 to 2, which can be simplified to 2:1 or 2/1.
b) the mole ratio for H₂O to O₂ is 6 to 3, which can be simplified to 2:1 or 2/1.
