Identify the major product that is obtained when 1-hexyne is treated with H2 and Pd. cis-2-hexene 1-hexene hexane trans-2-hexene

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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The balanced molecular chemical equation for the reaction is (NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq).

The rule for writing a balanced chemical reaction equation is that the number of atom of each element on the right hand side must be the same as the number of atoms of the same element on the left hand side of the reaction equation. This is achieved by performing an atom count.

The balanced molecular chemical equation for the reaction in aqueous solution for ammonium sulfate and iron(II) chloride is;

(NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq)

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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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Answer:

D  BLESSS Y

Explanation:

Explanation:

[tex]\tiny\implies Molarity = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{volume \: of \: the \: solution \: (in \: ml)} [/tex]

[tex]\tiny\implies 0.50 = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{15}[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 0.50 \times 15[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 7.5[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute = \dfrac{7.5}{1000} [/tex]

[tex]\tiny\implies \bf no. \: of \: moles \: of \: solute = 0.075 \: mol[/tex]

Answer:

49.979

Explanation:

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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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:

engey

Explanation:

If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.

Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.

Hence, if  the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.

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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.

Answer:

i can't understand the question

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Answer:

A-Solid

Explanation:

Waves travel faster in solids & slowest in gases

In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.

Let's consider the following balanced redox reaction.

2 Sc + 3 Br₂ ⟶ 2 ScBr₃

We can identify both half-reactions.

Oxidation: 2 Sc ⟶ 2 Sc⁺³ + 6 e⁻

Reduction: 6 e⁻ + 3 Br₂ ⟶ 6 Br⁻

As we can see, 6 electrons are involved in the formation of 2 formula units of ScBr₃. Thus, 3 electrons are involved in the formation of 1 formula unit of ScBr₃.

In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.

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When the oil is added to the heated copper, the energy in the system is

conserved.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The mass of oil in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

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Possible part of the question obtained from a similar question online, are;

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating = 65.17°C

As the solar nebula cooled METAL compounds are the first to condense from a gas to a solid. The solar nebula gave birth to the Solar system.

A solar nebula is a disc-shaped cloud of gases and grain dust, which gave birth to the Sun and planets of the Solar system, approximately 4.6 billion years ago.

The solar nebula is at the beginning a mixture of interstellar gases (hydrogen and helium) and dust grains.

As the solar nebula cools, heavy elements such as metals in the disk condensate into planetesimals.

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Answer:

The answer would be 4.

The agencies are:

Department of Agriculture's Forest Service (AFS)

Department of the Interior's Bureau of Land Management (BLM)

Fish and Wildlife Service (FWS)

National Park Service (NPS)

These four agencies cover around 95% of the land and are the only federal agencies.

I learned about these agencies in school.

Answer:

I believe there are 30

Explanation:

Answer:

[tex]11 \times 6.022 \times {10}^{23} \\ = 66.242\times {10}^{23} \: of \: \\ water \: molecules[/tex]

Answer:

247 ml

Explanation:

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml

0.000150 x ? = 0.0370 moles HF

? = 0.0370/0.000150 = 247 ml

check

247 ml = 247/1000 L = 0.247

(0.247) x (0.150) =0.370 check

Answer:

dog dog im not gon hold u we kinda screwed

Answer:

´protons: 16

neutrons: 16

name: sulfur

Explanation:

From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.

The correct responses are;

Reasons:

The given parameters are;

The concentration of the KCl solution = 4% m/v

Taking the solution as solution of KCl

The volume of the solution in container 1 = Two times less than the volume of KCl solution.

[tex]V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}[/tex]

The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.

[tex]V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}[/tex]

The volume of the solution in container 3 = Two times larger than the container two solution volume.

[tex]V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}[/tex]

Therefore;

[tex]V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}[/tex]

Part A Required:

a. To select the container that represent the dilution of the 4% solution to 2%

Solution:

The dilution formula is; C₁·V₁ = C₂·V₂

Therefore;

[tex]\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}[/tex]

C₁ =4%, C₂ = 2%, we get;

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1[/tex]

The volume of the container that represents a 2% dilution is container 2

which is filled with a solution of a volume two times larger compared to the

KCl solution.

Part B:

Required:

The container diluted to a 1% m/v KCl solution.

Solution;

Using the dilution formula, we have;

C₁ = 4%, C₂ = 1%

Therefore;

[tex]\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}[/tex]

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}[/tex]

The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.

Possible parts of the question are;

Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;

Part A: a 2% (m/v) KCl solution

Part B: a solution that is a 1% (m/v) KCl solution

Please see attached drawings

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Answer:

I believe they deserve to be here.

Explanation: