The target compound that you should synthesize is 3-chloro-1-butene. Again, this is an electrophilic alkene addition reaction.Examine the product to determine the location of the new functionality. Keep in mind the nature of the intermediate. The regioselectivity is controlled by the stability of this intermediate. Assume that only one equivalent of reagent is used.

Required:
State the starting agents, solvents, and products. What is the main reaction and mechanism? What are the TLC values?

Answer:

A. Reactants change into products

Answer:

A Standard Hydrogen Electrode is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode, just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram.

Answer:

The role of an electrode as cathode or anode depends on the nature and electrode potential of the other electrode with which it forms the complete electrochemical cell.

When a cell is to be made with zinc electrode and hydrogen electrode, the hydrogen electrode will behave as a cathode and the zinc electrode will behave as anode because zinc is present above hydrogen in the activity series. That is zinc is more electropositive than hydrogen.

If the cell is made with a copper electrode and hydrogen electrode, the hydrogen electrode will behave as anode and the copper electrode as a cathode. This is due to the fact that Cooper is present below hydrogen in the activity series. Copper is less electropositive than hydrogen.

Explanation:

Answer:

Explanation:

We are to match each land resource to what they are being used for.

Clay             →→→  used to make pottery and tiles

iron ore       →→→  used to make steel

Salt              →→→   used as a flavoring in food

aggregate   →→→   used in construction

graphite       →→→ used to make batteries

Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.

Iron ore:  The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.

Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.

Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,

Answer:

The options are incorrect.................

Explanation:

The Oxidation no. is +3

Answer:

c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.

Answer:

Vanlency of hydrogen - 11

Electrons of hydrogen - 1

Answer:

The answer is: 1

Hope this helps :) <3

Explanation:

Explanation:

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

Molarity of stock solution (M₁) = 1.2 M

Molarity of diluted solution (M₂) = 0.5 M

Volume of diluted solution (V₂) = 100 mL

The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:

1.2 × V₁ = 0.5 × 100

1.2 × V₁ = 50

Divide both side by 1.2

V₁ = 50 / 1.2

Thus, 42 mL of the stock solution is needed.

Learn more: https://brainly.com/question/24219233

Answer:

They need 41.7 mL of the original stock solution.

Explanation:

We can use the following equation for dilutions:

Cc x Vc = Cd x Vd

Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.

The concentrated solution is the original stock solution, and it has:

Cc = 1.2 M

The diluted solution must be:

Cd = 0.500 M

Vd = 100 mL

So, we have to calculate Vc. For this, we replace the data in the equation:

[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]

Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100  mL of a diluted solution with a concentration of 0.500 M.

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Answer:

a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.

Explanation:

Identify the correct descriptions of beta particles.

a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)

b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.

c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.

d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.

Answer:

λ = 150 nm

Explanation:

For C-O bond rupture:

The required energy to rupture C-O bond = bond energy of C-O bond

= 799 kJ/mol

[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]

[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]

[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]

Recall that the wavelength associated with energy and frequency is expressed as:

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]\lambda = \dfrac{hc}{E}[/tex]

[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]

[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]

λ = 150 nm

Answer:

molecules of real gases occupy no volume.

Explanation:

Answer:

See the answer below

Explanation:

From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.

Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.

Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.

In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.

Answer:

cis

Explanation:

Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.

Answer:

自分の仕事をする translate to english

Explanation:

Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Answer:

Alpha and beta particles originate from the nucleus, protection from radiation is important because if the radiation passes through the body it can damage cells. Exposure to radiation is often limited by increasing the distance from the radioactive source.

Explanation:  

Answer:

46.2%

Explanation:

Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles

Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.

Hence;

Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g

% yield = actual yield/theoretical yield × 100

% yield = 1.7 g/3.68 g × 100

% yield = 46.2%

Answer:

2.21% wt/v

Explanation:

The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL

The %wt/v of this solution is:

%wt /v = 1.77g / 80mL * 100

%wt/v = 2.21% wt/v

Answer:

d

Explanation:

Answer:

Specific gravity of the saturated solution is 2

Explanation:

The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:

Specific gravity  = Density

The density is the ratio between the mass of the solution and its volume.

In 100mL of water, the mass of KI that can be dissolved is:

100mL * (1g KI / 0.7mL) = 143g of KI

That means all the 100g of KI are dissolved (Mass solute)

As the volume of water is 100mL, the mass is 100g (Mass solvent)

The mass of the solution is 100g + 100g = 200g

In a volume of 100mL, the density of the solution is:

200g / 100mL = 2g/mL.

The specific gravity has no units, that means specific gravity of the saturated solution is 2

Answer:

d. End product is that product with a ketone and carboxylic acid.

Explanation:

[tex]{ \sf{NaBH_{4} : }}[/tex]

Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.

[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]

Then acidified water is an oxidising mixture which reverses the reduction reaction.

Explanation:

Option D is your answer

Hope it helps

Answer:

7 mol CuO

0.5 mol O₂

Explanation:

Step 1: Write the balanced equation

2 Cu + O₂ ⇒ 2 CuO

Step 2: Identify the limiting reactant

The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.

The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.

Since TMR > EMR, Cu is the limiting reactant

Step 3: Calculate the amount of CuO produced

7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO

Step 4: Calculate the excess of O₂ that remains

The amount of O₂ that reacts is:

7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂

The excess of O₂ that remains is:

4 mol - 3.5 mol = 0.5 mol

Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

Explanation:

The balanced chemical equation of the reaction is:

[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

Answer:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.

Explanation:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.

The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.

The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.

In the given reaction as shown below:

4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)

The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.

The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.

On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.

The oxidation numbers of hydrogen ion and sulfide ion do not change.

Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

Answer:

the metal become red hot