Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
Answer:
Explanation:
We are given that 25 mL of 0.10 M 
is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of 
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of 
=0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+=
=0.0025 moles
Number of moles of 
=0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28
]=-log4.28+2 log 10=-0.631+2
The correct Answer is A.
The Flow chart of the Carbon Cycle, will look as follows:
1. Carbon dioxide in the atmosphere.
2. Producers undergo photosynthesis (in this process plants among other producers like algae, use CO2 to produce energy).
3. Consumers eat producers (a primary level of consumer like a rabbit eats the producers or plants).
4. Decomposers return Carbon to the soil and release waste.
And the cycle continues again, by going back to step 1.
Crushing a can, breaking glass, cutting paper, boiling water, chopping wood, and mixing water and sand
