Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Answer:
The pH is equal to 4.41
Explanation:
Since HClO is a weak acid, its dissociation in aqueous medium is:
HClO ⇄ ClO- + H+
start: 0.05 0 0
change -x +x +x
balance 0.05-x x x
As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.
the acidity constant when equilibrium is reached is equal to:
![Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BClO-%5D%2A%5BH%2B%5D%7D%7B%5BHClO%5D%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.05-x%7D%3D3x10%5E%7B-8%7D)
The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:
clearing the x and calculating its value we have:
![x=3.87x10^{-5}=[H+]=[ClO-]](https://tex.z-dn.net/?f=x%3D3.87x10%5E%7B-5%7D%3D%5BH%2B%5D%3D%5BClO-%5D)
the pH can be calculated by:
![pH=-log[H+]=-log[3.87x10^{-5}]=4.41](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D%3D-log%5B3.87x10%5E%7B-5%7D%5D%3D4.41)
"a quantity representing the amount of water vapor in the atmosphere or a gas."
Answer:
32.25
Explanation:
8NH3 + 3Cl2 → 6NH4Cl + N2
NH3 = 17 g/mol
number of moles = 1
NH4Cl= 43 g/mol
number of moles = 3/4
mass= 43 × ¾ ≈ 32.25 g
<span>The correct answer is "Sugar is the solute and water is the solvent". Why? Sugar can be dissolved by water itself, especially in hot water, which is why water is considered to be a solvent, while sugar is considered to be a solute because it can be dissolved.</span>
