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:
CH4 - Methane
B2Si - Diboron monosilicide
N2O5 - Dinitrogen pentoxide
CO2 - Carbon dioxide
Explanation:
When it comes to naming covalent compounds, there are several rules.
The name is derived based on the formula. For example, N2O5. The first element is nitrogen. To the name of the element, you add the prefix that tells us how many of its atoms are in the compound. In this case, there are two atoms, which means that the prefix will be <em>di</em>- (dinitrogen). The second element is oxygen. You are supposed to take only the root of the second element's name and then add the prefix denoting the number of its atoms and the suffix <em>-ide</em> (pentoxide). This is how we'll get dinitrogen pentoxide.
The only exception is methane (CH4), which is an organic compound. Organic compounds are named using the IUPAC nomenclature.
Answer:
A Semipermeable membrane is a type of biological or synthetic, polymeric membrane that will allow certain molecules or ions to pass through it by diffusion or occasionally by more specialized processes of facilitated diffusion, passive transport or active transport.
Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J
Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418
Answer: monki says d
Explanation: monki big brain oof
