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How many moles of magnesium is 1.25 x 1023 atoms of magnesium?

andrew-mc [135]

The moles of magnesium in 1.25 x 1023 are 20.8 mol.

4 0

3 years ago

Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

MAXImum [283]

Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Explanation:

Given that:

the equilibrium number of vacancies at 800 °C

i.e T = 800°C is 3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

where ;

$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

$\rho _{Ag}$ = Density of silver = 9.5 g/cm³

$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

$Q_v = (2.38 \ * 1.602176565 * 10^{-19} ) J/atom }$

$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

8 0

3 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago

What is the molecular structure of water? What are the physical and chemical properties of water?

Slav-nsk [51]

Water (H
2O) is a polar inorganic compound that is at room temperature a tasteless and odorless liquid, which is nearly colorless apart from an inherent hint of blue. It is by far the most studied chemical compound and is described as the "universal solvent" [18][19] and the "solvent of life".[20] It is the most abundant substance on Earth[21] and the only common substance to exist as a solid, liquid, and gas on Earth's surface.[22] It is also the third most abundant molecule in the universe.[21]

Water (H
2O)

NamesIUPAC name

water, oxidane

Other names

Hydrogen hydroxide (HH or HOH), hydrogen oxide, dihydrogen monoxide (DHMO) (systematic name[1]), hydrogen monoxide, dihydrogen oxide, hydric acid, hydrohydroxic acid, hydroxic acid, hydrol,[2] μ-oxido dihydrogen

Identifiers

CAS Number

7732-18-5

3D model (JSmol)

Interactive image

Beilstein Reference

3587155ChEBI

CHEBI:15377

ChEMBL

ChEMBL1098659

ChemSpider

937

Gmelin Reference

117

PubChem CID

962

RTECS numberZC0110000UNII

059QF0KO0R

InChI

InChI=1S/H2O/h1H2

Key: XLYOFNOQVPJJNP-UHFFFAOYSA-N

SMILES

O

Properties

Chemical formula

H
2OMolar mass18.01528(33) g/molAppearanceWhite crystalline solid, almost colorless liquid with a hint of blue, colorless gas[3]OdorNoneDensityLiquid:[4]
0.9998396 g/mL at 0 °C
0.9970474 g/mL at 25 °C
0.961893 g/mL at 95 °C
Solid:[5]
0.9167 g/ml at 0 °CMelting point0.00 °C (32.00 °F; 273.15 K) [a]Boiling point99.98 °C (211.96 °F; 373.13 K) [6][a]SolubilityPoorly soluble in haloalkanes, aliphaticand aromatic hydrocarbons, ethers.[7]Improved solubility in carboxylates, alcohols, ketones, amines. Miscible with methanol, ethanol, propanol, isopropanol, acetone, glycerol, 1,4-dioxane, tetrahydrofuran, sulfolane, acetaldehyde, dimethylformamide, dimethoxyethane, dimethyl sulfoxide, acetonitrile. Partially miscible with Diethyl ether, Methyl Ethyl Ketone, Dichloromethane, Ethyl Acetate, Bromine.Vapor pressure3.1690 kilopascals or 0.031276 atm[8]Acidity (pKa)13.995[9][10][b]Basicity (pKb)13.995Conjugate acidHydroniumConjugate baseHydroxideThermal conductivity0.6065 W/(m·K)[13]

Refractive index (nD)

1.3330 (20 °C)[14]Viscosity0.890 cP[15]Structure

Crystal structure

Hexagonal

Point group

C2v

Molecular shape

Bent

Dipole moment

1.8546 D[16]Thermochemistry

Heat capacity (C)

75.375 ± 0.05 J/(mol·K)[17]

Std molar
entropy (So298)

69.95 ± 0.03 J/(mol·K)[17]

Std enthalpy of
formation (ΔfHo298)

−285.83 ± 0.04 kJ/mol[7][17]

Gibbs free energy (ΔfG˚)

−237.24 kJ/mol[7]

6 0

3 years ago

Michelle is trying to find the average atomic mass of a sample of an unknown

GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

Abundance of isotope A (A%) = 59.34%
Mass of isotope A = 113.6459 amu
Mass of isotope B = 115.8488 amu
Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

$Average \: atomic \: mass \: = \frac{mass \: of \: A \times A\%}{100} + \frac{mass \: of \: B \times B\%}{100} \\ \\ Average \: atomic \: mass \: = \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\ \\ Average \: atomic \: mass \: = 114.54 \: amu \\ \\$

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0

2 years ago

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