PbSO₄ partially dissociates in water. the balanced equation is;
PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X
Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm
Answer:
Percentage yield = 0.49 × 10² %
Explanation:
Given data:
Actual yield of SO₂ = 4.309 ×10² g
Theoretical yield of SO₂ = 8.78 ×10² g
Percentage yield = ?
Solution:
Chemical equation:
2NiS₂ + 5O₂ → 2NiO + 4SO₂
Percentage yield:
Percentage yield = actual yield / theoretical yield × 100
Percentage yield = 4.309 ×10² g / 8.78 ×10² g × 100
Percentage yield = 0.491 × 100
Percentage yield = 49.1%
In scientific notation:
Percentage yield = 0.49 × 10² %
Answer:
0.0102 mole
Explanation:
Use and solve proportions for these with x as the unknown value.
The production of manganese peroxidase (MnP) by Irpex lacteus, purified to electrophoretic homogeneity by acetone precipitation, HiPrep Q and HiPrep Sephacryl S-200 chromatography, was shown to correlate with the decolorization of textile industry wastewater. The MnP was purified 11.0-fold, with an overall yield of 24.3%. The molecular mass of the native enzyme, as determined by gel filtration chromatography, was about 53 kDa. The enzyme was shown to have a molecular mass of 53.2 and 38.3 kDa on SDS-PAGE and MALDI-TOF mass spectrometry, respectively, and an isoelectric point of about 3.7. The enzyme was optimally active at pH 6.0 and between 30 and 40 degrees C. The enzyme efficiently catalyzed the decolorization of various artificial dyes and oxidized Mn (II) to Mn (III) in the presence of H(2)O(2). The absorption spectrum of the enzyme exhibited maxima at 407, 500, and 640 nm. The amino acid sequence of the three tryptic peptides was analyzed by ESI Q-TOF MS/MS spectrometry, and showed low similarity to those of the extracellular peroxidases of other white-rot basidiomycetes.
Answer:
Redox reaction and single displacement
Explanation:
This reaction is first of all a redox reaction. A redox reaction is a reaction that involves both oxidation and reduction. Oxidation involves increase in oxidation number while reduction involves decrease in oxidation number.
Copper (Cu) had an oxidation number of "0" as a reactant but had an oxidation number of "2+" in the product [Cu(NO₃)₂] hence oxidation occurred.
Nitrogen (N) had an oxidation number of "5+" in the reactant (HNO₃) but had an oxidation number of "4+" in the product (NO₂) hence reduction also occurred.
Also, from the reaction, it can be deduced that copper (Cu) displaced hydrogen (H) from the nitric acid (HNO₃) solution to form copper (II) nitrate [Cu(NO₃)₂]. It should be noted that copper can displace hydrogen because it is higher than hydrogen in the electrochemical series. Hence, this reaction can also be called a single displacement reaction. A single displacement reaction is a reaction in which an atom of an element replaces another atom in a compound (as seen in the equation given in the question).
