The S and P blocks are known as orbitals
Answer:
b. 0.47 moles Ca.
Explanation:
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In this case, since 1 mole of any element contains 6.022x10²³ atoms of the same, it is possible for us to compute the moles in 2.8x10²³ atoms of calcium via the Avogadro's number:
Therefore, the answer would be b. 0.47 moles Ca.
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Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Answer: Complete ionic equations dissociate all aqueous solutions into ions. Net ionic equations show the change that occurs in chemical reactions and do not show spectator ions that are the same in reactants and products. (b) If no spectator ions were present then complete and net ionic equations would be identical.
Explanation: