The question is incomplete, the complete question is:
Standard reduction potentials for zinc(II) and copper(II)
The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:
Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V
Part B
What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Express your answer to three decimal places and include the appropriate units.
Answer:
1.100 V
Explanation:
E∘cell= E∘cathode - E∘anode
E∘cathode= +0.337 V
E∘anode= −0.763 V
E∘cell= 0.337-(-0.763)
E∘cell= 1.1V
Answer:
Answer) The first box represent the budding type of asexual reproduction . In budding due to the cell divison at the particular place in the organism , new individual will produced.
Second picture represent the budding in hydra. Budding is also a asexual reproduction.
Third type of asexual reproduction is found in plants it is vegetative reproduction.
Fourth is Binary fission . This is also a type of asexual reproduction, in which nucleus dicied into two new daughter cell via mitosis.Explanation:
Answer:
3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O
Explanation:
KOH + HBr ---> KBr + H2O
0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr
according to the reaction: 1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH
0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH
answer: 0,5 mol/dm³ KOH (molarity)
