Answer:
T<span>he gaseous product of this reaction is water (Option-A).
Explanation:
This is a very interesting experiment. Take sugar in a beaker and add concentrated Sulfuric Acid into it. After a while an exothermic reaction will initiate with the formation of Carbon Black and Water vapors. You will observe the formation of hard and hot stem like body which is completely Black. This blackness is due to C and the water vapors will eliminate in the form of steam as the temperature has arised.</span>
A second-order extension of the Kohn-Sham total energy in density-functional theory (DFT) with respect to charge density fluctuations serves as the foundation for the density functional based tight binding (DFTB) approach.
What is DFTB method?
- The density functional based tight binding (DFTB) electronic structure method was used to study the clusters of bare TiO2 and TiO2 with linked organic ligands modeling polyorganic composites used as photocatalytic materials.
- The results were compared to those obtained from B3LYP/6-31G(d,p) calculations, semiempirical methods PM6 and PM7, and available experimental data.
- It was discovered that the highly scalable DFTB approach produces outcomes that are nearly on the level of theory B3LYP/6-31G(d,p).
- The trans3d set more accurately reproduces the energies of the composite material production in polycondensation processes, but the corrected version of the tiorg DFTB parameter set (tiorg-smooth) performs better for structural parameter estimations.
- The tiorg-smooth and trans3d settings perform better than the matsci set in some way. Studies of adsorption complexes of bare TiO2 clusters can be conducted using the tiorg-smooth and matsci sets.
Learn more about the Density with the help of the given link:
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Answer:
168°C is the melting point of your impure sample.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= ?
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample = 0.275 mol/kg
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( non electrolyte)
168°C is the melting point of your impure sample.
Answer:
Energy was released when the sodium and hydroxide ions formed new bonds with the water.
Explanation:
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
