Calculate the second ionization energy of the metal M (?Hion2� in kJ/mol) using the following data:


Lattice enthalpy of MO(s), ?Hl� = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

Enthalpy of sublimation of M = + 130 kJ/mol

First ionization energy of M = + 267 kJ/mol

Standard enthalpy of formation of MO(s), ?Hf� = -307 kJ/mol

Answers

Answer 1

From the information provided in the question, the second ionization energy of the metal is  578 kJ/mol.

From the question, we have the following information;

Lattice enthalpy of MO(s) = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

First ionization energy of M = + 267 kJ/mol

Heat of sublimation of M = + 130 kJ/mol

Standard enthalpy of formation of MO(s) = -307 kJ/mol

Using Hess law of constant heat summation;

ΔHf = ΔHs + BE + ∑IE + ∑EA + U

ΔHs  = Heat of sublimation of metal

ΔHf = Heat of formation MO

BE = Bond energy of O2

∑EA  = sum of electron affinities of Oxygen

∑IE  = Sum of the ionization energies of M

U = Lattice energy of MO

Let the second ionization energy be x

Substituting values;

(-307) = 130 + 498 + (267 + x) + 603 + (-2383)

(-307) = -885 + x

-x =   -885 + 307

-x = -578

x = 578 kJ/mol

The second ionization energy of the metal is  578 kJ/mol.

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Related Questions

10) Explain which substance(s) would you expect to be a gas at standard
temperature and pressure (STP).
NI3
BF3
PC13
A) BF3
Its molecules are nonpolar and, therefore, are subject only to
dispersion forces.
B) NI3
Its molecules are polar and possess dipole-dipole interactions
therefore allowing them to exist as a gas.
C) PCL3
Its molecules are slightly polar and, therefore, possess dipole-
dipole interactions. However, they are weak enough to allow it
to exist as a gas.
D) Nlz and BF3
Both molecules are very light in mass. Therefore, they have
weak dispersion forces which causes them to be attracted very
weakly and exist as a gas.

Answers

Answer:BF3

Explanation:USA test prep

Your skin is an important organ and has several functions. All of the functions below are performed by the skin EXCEPT
A)
makes vitamin D.
B
prevents-dehydration:
C)
maintains body temperature.
D)
works with bones to help you move

Answers

Answer:

D. works with bones to help you move

Explanation:

Your muscles and tendons are what help bones move.

A 4.0 L flask containing N2 at 15 atm is connected to a 4.0 L flask containing H2 at 7.0 atm and the gases are allowed to mix. What is the mole fraction of N2

Answers

The mole fraction of N₂ after the mixture of 4.0 L of N₂ at 15 atm with 4.0 L of H₂ at 7.0 atm is 0.68.

We can calculate the mole fraction of N₂ with the following equation:

[tex] X_{N_{2}} = \frac{n_{N_{2}}}{n_{t}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2}}} [/tex]   (1)

The number of moles of N₂ and H₂ can be found with the ideal gas law:

[tex] PV = nRT [/tex]

Where:

P: is the pressure

R: is the gas constant

T: is the temperature

V: is the volume

For nitrogen gas we have:

[tex] n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT} [/tex]   (2)

And for hydrogen:

[tex] n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} [/tex]   (3)

After entering equations (2) and (3) into (1), we get:

[tex] X_{N_{2}} = \frac{\frac{P_{N_{2}}V_{N_{2}}}{RT}}{\frac{P_{N_{2}}V_{N_{2}}}{RT} + \frac{P_{H_{2}}V_{H_{2}}}{RT}} [/tex]  

Since RT are constants, we have:

[tex] X_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{P_{N_{2}}V_{N_{2}} + P_{H_{2}}V_{H_{2}}} [/tex]                

We know that:

[tex] P_{N_{2}} = 15 atm[/tex]                

[tex] V_{N_{2}} = 4.0 L[/tex]                

[tex] P_{H_{2}} = 7.0 atm[/tex]                

[tex] V_{H_{2}} = 4.0 L[/tex]          

so:

[tex] X_{N_{2}} = \frac{15 atm*4.0 L}{15 atm*4.0 L + 7.0 amt*4.0 L} = 0.68 [/tex]                

Therefore, the mole fraction of N₂ is 0.68.

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which mass movement does this describe?

Answers

Answer:

landslide mudflows slump and creep

Explanation:

no explanation

please help asap!!!!!!!

Answers

Answer:

c) single bonds, double bonds, carbon atoms and hydrogen atoms

hydrogen bond is a strong dipole interaction for example water and other molecules true or false

Answers

Answer:

hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond it is a mere electrostatic attraction. These attractions can occur between molecules or within different parts of a single molecule. Intramolecular hydrogen bonds are those which occur within one single molecule.

Explanation:

CAN I GET BRAINLIEST

3.833 kJ of heat is required to convert a 36.8 g sample of ethyl
alcohol from the solid to liquid phase. What is the heat of
fusion of ethyl alcohol in J/g?

Answers

The heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.

The given parameters:

heat required to convert 36.8 g sample of ethyl alcohol, Q  = 3.3833 kJmass of the  ethyl alcohol, m = 36.8 g

The heat of fusion of the given sample of the ethyl alcohol converted from solid to liquid phase is calculated as follows;

[tex]H_f = \frac{Q}{m} \\\\H_f = \frac{3.833 \times 10^3\ J}{36.8 \ g} \\\\H_f = 104.16 \ J/g[/tex]

Thus, the heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.

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Can someone help me out with this question and explain me the answer too!!! Please

Answers

Answer:

sorry i dont know

Explanation:

How do we know flowing water is the geologic process that formed the channel on Mars?

Answers

Evidence that water was once present on a planet is evidence that the planet may once have had living organisms. When landforms on different rocky planets look similar, it is evidence that they may have been formed by the same geologic process. The channel on Mars may have been caused by flowing water or flowing lava.

Hopes this helps :)

Commercially prepared cloning vectors such as pUC18 are designed to contain several useful features. An example of one of these features is ________.

Answers

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

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You are given 1.091 grams of a white powder and told that it is a mixture of potassium carbonate and sodium carbonate. You are asked to determine the percent composition by mass of the sample. You add some of the sample to 10.00 mL of 0.8903 M nitric acid until you reach the equivalence point. When you have added enough carbonate to completely react with the acid, you reweigh your sample and find that the mass is 0.573 g. Calculate the mass of the sample that reacted with the nitric acid. Calculate the moles of nitric acid that reacted with the sample.

Answers

The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.

Molar mass of sodium carbonate = 106 g/mol

Molar mass of potassium carbonate = 138 g/mol

Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles

Mass of HNO3 =  0.008903 moles × 63 g/mol = 0.56 g

Mass of sample added = 1.091 g

Mass of sample left over = 0.573 g

Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g

The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.

Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O

106g of Na2CO3 reacts with 126g of HNO3

x g of Na2CO3 reacts with (126 × x/106)g of HNO3

K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O

138 g of K2CO3 reacts with 126 g of HNO3

(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g

Total mass of HNO3 used;

1.19x + 0.47 + 0.91x = 0.56

2.1x + 0.47 = 0.56

2.1x = 0.56 - 0.47

2.1x = 0.09

x =  0.09/2.1

x = 0.0043 g

Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g

Mass percent of K2CO3  = 0.5137 g/ 1.091 g × 100/1 = 47.1%

Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%

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Magnesium metal burns in air with an intense bright light according to the equation
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ

A.) What is the amount of energy in kJ produced when 4.5 mol of Mg is burned in the presence of excess oxygen?

Answers

The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.

A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;

2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ

From the reaction equation;

2 moles of magnesium produced 1200 kJ of heat

4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles

= 2700 kJ

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Doing Labs at home

I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!

Answers

Answer:

go get the stuff.

Explanation:

A particular reaction has a DH o value of -164 kJ and DS o of -185 J/K at 298 K. Calculate DG o at 617 K in kJ (with 3 significant digits), assuming that DH o and DS o hardly change with temperature

Answers

Answer:

e

Explanation:

A radioactive sample has a half life of 1 hour. If you start with 1.000 gram of it at noon, how much of it remains at 4pm

Answers

The amount of the sample remaining at 4pm is 0.0625 g

We'll begin by calculating the number of half-lives that has elapsed

Half-life (t½) = 1 hour

Time (t) = 4 hour

Number of half-lives (n) =?

n = t / t½

n = 4 / 1

n = 4

Finally, we shall determine the amount the sample remaining at 4pm

Number of half-lives (n) = 4

Initial amount (N₀) = 1 g

Amount remaining (N) =?

[tex]N = \frac{N_0}{ {2}^{n}} \\ \\ N = \frac{1}{ {2}^{4}} \\ \\ N = \frac{1}{16} \\ \\ N = 0.0625 \: g[/tex]

Thus, the amount remaining at 4pm is 0.0625 g

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he hybridization of carbon in diamond is _________. Enter your answer with no superscripts or subscripts, i.e., ab3.

Answers

Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.

Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.

Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.

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The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.

The correct answer is sp3.

The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.

The valence electrons are in p orbitals and p orbit mixed after the s orbitals.

Therefore, the correct answer is sp3

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What is the formula for sodium phosphate

Answers

Answer:

Explanation:

It''s chemical formula is Na3PO4

What identifies the number of protons in the nucleus of an atom?

Answers

Answer: Atomic number

Explanation:

I hope this helps you!

Please help asap chemistry worksheet doesn’t make sense

Answers

Answer:

dog dog im not gon hold u we kinda screwed

Which of the following is a way to increase pressure on a gas?

Answers

Answer:

increase the number of gas particles

Explanation:

How many moles of potassium iodide, KI, are required to precipitate all of the lead (II) ion from 25.0 mL of a 1.6 M Pb(NO3)2 solution? (First, write a balanced equation for the reaction.)

Answers

0.04 moles of iodide is required to precipitate all the lead II ions from  25.0 mL of a 1.6 M Pb(NO3)2 solution.

The reaction equation is;

2KI(aq) + Pb(NO3)2(aq) -------> PbI2(s) + 2KNO3(aq)

The net ionic equation is;

Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)

Number of moles of Pb(NO3)2  = 25/1000 L ×  1.6 M = 0.04 moles

Number of moles of Pb^2+ = 0.04 moles /2 = 0.02 moles

Since 2 moles of iodide reacts with 1 mole of Pb^2+

x moles of iodide reacts with 0.02 moles of Pb^2+

x =  2 moles × 0.02 moles/ 1 mole

x = 0.04 moles of iodide

Hence,  0.04 moles of iodide is required to precipitate all the lead II ions from  25.0 mL of a 1.6 M Pb(NO3)2 solution.

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true or false: When the value of secondary quantum no . (l) =1 , the magnetic quantum no . ( m ) = 3 .​

Answers

When the value of secondary quantum number (l) = 1, the magnetic quantum number can be from -1 to +1. The answer would be false.

What are Quantum numbers?

The angular quantum number (l) can take any integer between 0 and the value of the principal quantum number (n) less by 1.

In other words, l can be from 0 to n-1.

The magnetic quantum number (m), on the other hand, can take any value between -l to +l.

For example, if l = 1, m can be -1,0, and +1.

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#SPj1

which of the following are compounds (select all that apply)
a) Br2
b)NO2
c) KBr
d) Fe

Answers

Answer:

KNO2, KBr

Explanation:

Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.

Which of the following is true about a compound where there is a difference in
electronegativity of the two elements if the formula of the compound is greater than
1.7?
The compound is a binary covalent compound.
The compound is a covalent compound.
The compound is an ionic compound.
The compound is a polyatomic compound.

Answers

Answer:

D  BLESSS Y

Explanation:

helpp
What mass of carbon dioxide will be produced if
144 g of carbon react with 384 g of oxygen gas
according to the equation C+02 → CO2?

Answers

The mass of the carbon dioxide that will be produced when 144 g of carbon react with 384 g of oxygen gas is 528 g

We'll begin by calculating the masses of C and O₂ that reacted and the mass of CO₂ produced from the balanced equation.

C + O > CO

Molar mass of C = 12 g/mol

Mass of C from the balanced equation = 1 × 12 = 12 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 1 × 32 = 32 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of CO₂ from the balanced equation = 1 ×44 = 44 g

SUMMARY:

From the balanced equation above,

12 g of C reacted with 32 g of O₂ to produce 44 g of CO₂

Next, we shall determine the limiting reactant

From the balanced equation above,

, 12 g of C reacted with 32 g of O₂.

Therefore,

144 g of C will react with = (144 × 32)/12 = 384 g of O₂.

From the calculations made above, we can see that both C and O₂ are sufficient for the reaction. Thus, C and O₂ are both limiting reactants.

Finally, we shall determine the mass of CO₂ obtained from the reaction.

From the balanced equation above,

balanced equation above, 12 g of C reacted to produce 44 g of CO₂.

Therefore,

144 g of C will react to produce = (144 × 44)/12 = 528 g of CO₂.

Thus, the mass of CO₂ obtained from the reaction is 528 g

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What is the kinetic energy of a mole of Ar atoms moving with a speed of 650m/s

Answers

Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...

- BRAINLIEST answerer

Answer:

1.40×10-20

Explanation:

Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?

How many grams are 0.300 moles of glucose, C6H12O6?

Answers

Answer:

Explanation:

Firstly, let us calculate the molar mass of the glucose. To find the molar mass we need to add the masses of individual elements which constitute one glucose molecule.

Now, we know that

Molar mass of Carbon C=12gmol−1

Molar mass of Hydrogen H=1gmol−1

Molar mass of oxygen O=16gmol−1

Therefore, Molar Mass of glucose (C6H12O6) can be calculated as shown below

⇒ 6×12+12×1+6×16

⇒72+12+96⇒180gmol−1

Given mass of glucose = 300g

Now, we can calculate the number of moles in given mass of glucose, using the below formula,

Using the Formula numberofmoles=givenmassmolarmass we get

numberofmoles=300180 = 1.7 moles or 2 moles (approx.)

Hence the number of moles present in 300 g of glucose is 1.7 moles or 2 moles approximately.

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answers

Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation:

The structure of Disodium edta

Answers

Answer:

EDTA disodium salt | C10H14N2Na2O8

Explanation:

Which is larger? 12 milligrams or 12 kilograms


Answers

Answer:

12 kilograms is larger.

Explanation:

Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.

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