Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
= 207+56+192 = 455 g/mol
Formula:
Solving:
Answer:
<span>
B. 0.119 M</span>
Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.
Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.
Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.
Let the molar solubility be S upon dissociation.
PbBr₂ or Lead Bromide dissociates in pure water as follows:
PbBr₂ ----------> Pb⁺² + Br⁻
S 2S
Ksp = [Pb⁺²] [ Br⁻]
Ksp = (S) (2S)²
Ksp = 4S³
6.60 × 10⁻⁶ = 4S³
S = 0.0118M
Hence, the Molar solubility S is 0.0118M.
Learn more about Molar solubility here, brainly.com/question/16243859
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Solar Radiation(Basically Radiation)
Answer:
Your question is half unfinished, regarding the chromium III oxide the correct option that expresses the inorganic formula of said compound is "A"
Explanation:
In the reaction an initial salt reacts giving as product water vapor, nitrogen gas and an oxide that is chromium oxide.
Chromium oxide is an oxide that adopts the structure of corundum, compact hexagonal. It consists of an anion oxide matrix with 2/3 of the octahedral holes occupied by chromium. Like corundum, Cr2O3 is a tough, brittle material.
It is used as a pigment, green in color.
Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = ![\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
