2.0 g 1.0g 0.5g 0.25g
3 and a half lives = 42 days
Answer:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant. If a number ends in zeros to the right of the decimal point, those zeros are significant.
Explanation:
1.138 has 4 significant figures, which are 1, 1, 3 and 8. The numbers after the decimal point are decimals and are significant figures.
Given reactions:
(A) 6CO2(g) + 6H2O(l) + sunlight → C6H12O6(aq) + 6O2(g)
(B) 2H2(g) + O2(g) → 2H2O(g) + energy
Exothermic reactions are those which proceed with the release of heat/energy. In contrast, endothermic reactions proceed with the absorption of energy in the form of heat or light.
Since reaction A required sunlight, it is endothermic. Reaction B releases energy, hence exothermic
Ans: (B)
A is endothermic
B is exothermic
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
