V ( HCl ) = 16.4 mL / 1000 => 0.0164 L
M( HCl) = ?
V( KOH) = 12.7 mL / 1000 => 0.0127 L
M(KOH) = 0.620 M
Number of moles KOH:
n = M x V
n = 0.620 x 0.0127
n = 0.007874 moles of KOH
number of moles HCl :
<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
</span>
moles HCl = 0.007874 * 1 / 1
= 0.007874 moles of HCl
M = n / V
M = 0.007874 / <span>0.0164
</span>= 0.480 M
Answer (2)
hope this helps!
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
Explanation:
According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.
Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.
The possible transition for the electron given in the question is :
n = 2, 3 and 4
The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .
A cell is a microscopic living organism which is the smallest part of an organism such as a plant or animal.
