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

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 + O₂ —> CO₂

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

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.

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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Answer:

No it has not.

Explanation:

The actual chemical compound has not changed, the materials have mixed but no reaction has taken place. Blowing the bubbles causes a physical change. Hope this helped!

Mixing soap and water and forming bubbles is a physical change as there is no change in composition.

Physical changes are defined as changes which affect only the form of a substance but not it's chemical composition. They are used to separate mixtures in to chemical components but cannot be used to separate compounds to simpler compounds.

Physical changes are always reversible using physical means and involve a change in the physical properties.Examples of physical changes include melting,boiling , change in texture, size,color,volume and density.Magnetism, crystallization, formation of alloys are all reversible and hence physical changes.

They involve only rearrangement of atoms and are often characterized to be changes which are reversible.

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Answer:

32

Explanation:

Answer:

la primera va con la última

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

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The final volume of diluted solution is 214.39 mL

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

Molarity of diluted solution (M₂) = 0.7 M

Molarity of stock solution (M₁) = 6 M

Volume of stock solution (V₁) = 25 mL

M₁V₁ = M₂V₂

6 × 25 = 0.7 × V₂

150 = 0.7 × V₂

Divide both side by 0.7

V₂ = 150 / 0.7

Therefore, the volume of the diluted solution is 214.39 mL

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Answer:

I believe they deserve to be here.

Explanation:

The partial pressure of oxygen is equal to 10.2 atm.

Given the following data:

To determine the partial pressure of oxygen:

First of all, we would find the total number of moles of the elements:

[tex]n=3.25+2.75[/tex]

n = 6 moles

Next, we would determine the mole fraction of the oxygen by using this formula:

[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]

Substituting the values, we have:

[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]

For oxygen:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]

Partial pressure of oxygen = 10.2 atm.

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Answer:

12.1

Explanation:

3.25 + 2.75= 6 moles total

then take: 3.25/6 to get 0.542

multiply 0.542 by 22.4 to get 12.1

the other answer solved for nitrogen, not oxygen.

#let's study

(a) The number of molecules contained by the balloon at the given condition is 6,438.97 moles.

(b) The mass of air in the balloon is 186.73 kg.

The given parameters;

The number of molecules contained by the balloon at the given condition is calculated as follows;

[tex]PV = nRT[/tex]

where;

[tex]n = \frac{PV}{RT} \\\\n = \frac{10,100 \times 1500}{8.314 \times 283} \\\\n = 6,438.97 \ moles[/tex]

The mass of air in the balloon is calculated as;

[tex]n = \frac{m}{M} \\\\m = nM\\\\m = 6,438.97\ moles \ \times \ 29 \ g/mol\\\\m = 186.73 \ kg[/tex]

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Answer:

75% Cl-35 and 25% Cl-37

Explanation:

chlorine has two isotopes chlorine-35 and chlorine-37 with average atomic mass 35.5.find the percentage abundance of each isotopes​

assume Cl-35 has Y%  Then Cl-37 has (100-Y) %

The weighted average of 35.5 is given by

[(35 x Y) + ((100-Y) x 37)]/100 = 35.5

so

(35 x Y) + ((100-Y) x 37) =3550

35Y + 3700 -37Y =  3550

-2Y = -15075% Cl-35 and 25% Cl-37

Y = so average weight chlorine is

75% Cl-35 and 25% Cl-37

Electrolysis is the production of a chemical reaction by passing an electric current to liquid containing ions.

The presence of electron donating methoxy group in  p-methoxybenzaldehyde , makes it less reactive towards  nucleophilic additions.

Nucleophilic addition reactions that involve of p-Nitrobenzaldehyde and p-methoxybenzaldehyde both occur at the C=O functional group. The electron density around the carbonyl is a key factor that controls the reactivity of each compound.

In  p-Nitrobenzaldehyde, the electron withdrawing nitro group decreases the electron density around the carbonyl making it more vulnerable to nucleophilic attack.

In p-methoxybenzaldehyde, the methoxy group donates electrons by resonance towards the carbonyl thereby increasing the electron density around it so that it is less reactive towards nucleophillic addition reaction.

The nucleophilic attack on p-Nitrobenzaldehyde leads to more resonance structures than nucleophilic attack on p-methoxybenzaldehyde.

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Answer:  a.1.37  b.4.11 c.6.00  d.12.3

Explanation:

Answer:

Ideal Gas Law Calculator. Easily calculate the pressure, volume, temperature or quantity in moles of a gas using this combined gas law calculator (Boyle's law calculator, Charles's law calculator, Avogadro's law calculator and Gay Lussac's law calculator in one).Supports a variety of input metrics such as Celsius, Fahrenheit, Kelvin, Pascals, bars, atmospheres, and volume in both metric and

Answer:

5.39 X 10-4

Explanation:

Try Checking It Out If It Right Or Not

Explanation:

filling air inside an empty o2 medal containers, if air is added, the air inside, creates volume inside.

due to heat, weather,... depending on what indoor or outdoor storage said containers r placed in,;

So, I can't give you an answer, due to lack of details..sorry...

an ionic compound is made of two or more elements held by a strong electrostatic force. metal atoms lose electrons to become positive ions, and non metals gain electrons to become negative ions. ionic bonding occurs between metals and non metals

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The Correct choice is ~ D

Solute is the substance that dissolves into another substance

This is the answer i hope it helps

A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?

Answer:

River esk the third one is righttt

Answer:

933.33 m/ min

Explanation:

1 mile = 1.6 km

So the car travels 1.6× 35 km in 1 hour

One hour has 60 minutes

So speed of car = 1.6×35 km / 60 minutes

1km = 1000 m

Speed = 1.6×35×1000 m/ 60 minutes

= 933.33 meters per minute

A car is traveling 35 miles per hour. The speed that is measured in meters per minute is 933.33 m/ min.

Speed is the dimension of distance. It is calculated in meters per second. It is the rate of change of an object in a direction. Speed is calculated by dividing the distance by time. It is a scalar quantity.

Given, that a car is traveling 35 miles per hour. The speed is given, in miles we just have to convert the miles into meters per minute. Miles if the English unit of measurement and meter is the SI unit of measurement.

1 mile = 1.6 km

So the car travels 1.6× 35 km in 1 hour

One hour has 60 minutes

So speed of car = 1.6×35 km / 60 minutes

1km = 1000 m

Speed = 1.6 × 35 × 1000 m/ 60 minutes

= 933.33 meters per minute

Thus, the speed of the car is 933.33 m/ min.

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The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

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

Original amount (N₀) = 0.240 mole

Amount remaining (N) = 1.88×10¯³ mole

N = 1/2ⁿ × N₀

1.88×10¯³ = 1/2ⁿ × 0.240

Cross multiply

1.88×10¯³ × 2ⁿ = 0.240

Divide both side by 1.88×10¯³

2ⁿ = 0.240 / 1.88×10¯³

2ⁿ = 128

2ⁿ = 2⁷

Thus, 7 half-lives has elapsed

Number of half-lives (n) = 7

Half-life (t½) = 11.4 days

t = n × t½

t = 7 × 11.4

Therefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

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ok done. Thank to me :>

Answer: Mass percent of Li: total mass Li = 6.94 g/mol. % Li = mass Li mass LiClO 2 × 100 % Li = 6 . ...

Mass percent of Cl: total mass Cl = 35.45 g/mol. % Cl = mass Cl mass LiClO 2 × 100 % Cl = 35 . ...

Mass percent of O: total mass O = 32.00 g/mol.

Explanation:

Answer:

1. A/Thermal Energy

2. B/The particles within the system will have greater motion, and the temperature will increase.

3. C/Attractions occur due to electrostatic forces. When particles move fast enough, these forces can no longer keep particles together.

4. D/ For a phase change from solid to liquid, the bonds do not break completely and particles can still slide past each other.

5. D/ a sample of liquid benzene at 80ºC and a sample of gaseous benzene at 80ºC

Explanation:

I took the Test.

1. Internal energy associated with systems and particles is thermal energy. Thus, option A is correct.

The internal energy has been described as the energy that has been assessed by the molecules and bonds at the STP. Thermal energy has been the energy of particular molecules and the system comprising all the molecules have the summation of the thermal energy.

Thus, internal energy associated with systems and particles is thermal energy. Thus, option A is correct.

2. The addition of energy results in greater motion with an increase in temperature has been representation no phase change. Thus, option B is correct.

The addition of energy to the substance results in the increased motion of the particles. The energy higher than the threshold energy has been responsible for the phase change of the substance.

The addition of energy results in greater motion with an increase in temperature has been representation no phase change. Thus, option B is correct.

3. The movement of the particles under the gravitational attraction is not enough to hold the particles together at a longer distance. Thus, option B is correct.

The attraction has been the force that has been responsible for holding the particles in a molecule. Different attractions have a different levels of stability and strength.

The movement of the particles under the gravitational attraction is not enough to hold the particles together at a longer distance. Thus, option B is correct.

4. The phase change of the molecules has been mediated by overcoming the force of attraction that allows the molecules to move freely. The solid to liquid phase change has been mediated with the breaking of the attraction between the bonds allowing free movement.

Thus, statement 4 is incorrect, and option D is correct.

5. The kinetic energy has been the energy possessed by the moving molecules in the medium. The molecules in liquid and gas have been possessing the same energy at the same temperature, Thus, option D is correct.

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Answer: the answer is 0.074Hz

Explanation:

Given, (In air)

Velocity V=340m/s

Frequency f=20,000Hz

Wavelegth λ=?

V=f.λ

λ=

F

V

​

=

20,000

340

​

=0.017Hz

Also, Given (in Water)

Velocity, V=1480m/s

Frequencyf=20,000Hz

wavelength, λ=?

V=F.λ

λ=

F

V

​

=

20,000

1480

=​0.074Hz

Answer:

frequency

Explanation:

frequency is velocity/ wavelength

340/1.36

250

Fire extinguishment is referred to the stage in which flames are not visible

and the combustible item no longer generates heat.

This stage refers to when the fire has been put out through a fire

extinguisher or other compounds.   When fire is put out, there is no longer

flames present.

The absence of flames also means that there is no heat generation or

combustion products due to the absence of a heat source.

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Answer: O 2, 1,4

Explanation:

Answer:

Model C has more double the oxygen than A. I think it is B

Explanation:

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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