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poizon [28]
2 years ago
11

Structural, electronic, and thermodynamic properties of TiO$_{2}$/organic clusters: performance of DFTB method with different pa

rameter sets
Chemistry
1 answer:
Sladkaya [172]2 years ago
4 0

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:

brainly.com/question/23487480

#SPJ4

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Why would scientists use a spectrometer in space.
solniwko [45]

Answer:

A. To study the chemical composition of the surface of a planet

Explanation:

Strictly speaking, a spectrometer is any instrument used to view and analyze a range (or a spectrum) of a given characteristic for a substance (for example, a range of mass-to-charge values as in mass spectrometry), or a range of wavelengths as in absorption spectrometry like nuclear magnetic radiation spectroscopy

-hope this helps!

6 0
2 years ago
Read 2 more answers
A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0
4 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
Which is NOT a component of galaxies? * F. Universe G. Stars H. Dust J. Planets​
Afina-wow [57]

Answer:

F. Universe

Explanation:

Universe and Galaxies are two different things.

4 0
3 years ago
What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a tempe
garik1379 [7]
Answer:
             V₂  =  1070 mL or 1.07 L

Solution:

Data Given;
                  P₁  =  1170 mmHg

                  V₁  =  915 mL

                  T₁  =  24 °C  +  273 K  =  297 K

                  P₂  =  842 mmHg

                  V₂  =  ?

                  T₂  =  - 23 °C  +  273 K  =  250 K

According to Ideal gas equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂  =  P₁ V₁ T₂ / P₂ T₁

Putting Values,

                       V₂  = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

                       V₂  =  1070 mL or 1.07 L
5 0
3 years ago
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