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:

Vanlency of hydrogen - 11

Electrons of hydrogen - 1

Answer:

The answer is: 1

Hope this helps :) <3

Explanation:

Answer:

F 64.6 percent of carbon may be

Answer:

[tex]\boxed {\boxed {\sf 0.566 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of atoms to moles.  

We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.

[tex]\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}[/tex]

[tex]3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}[/tex]

[tex]\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.5662570575\ mol \ Fe[/tex]

The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.5662570575) tells us to leave the 6 in the thousandths place.

[tex]0.566 \ mol \ Fe[/tex]

3.41 × 10²³ atoms of iron is equal to approximately 0.566 moles of iron.

Answer:

a measure of concentration equal to the gram equivalent weight per liter of solution.

Explanation:

Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

hope it helped

Answer:

The concentration of KOH is  0.06137 M

Explanation:

Step 1: Data given

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.4877 grams

Volume of water = 50 mL

Volume of KOH solution = 38.91 mL

Step 2: The balanced equation

C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O

Step 3: Calculate number of moles of KHP

Moles = Mass / molar mass

Moles KHP = 0.4877 grams / 204.22 g/mol

Moles KHP = 0.002388 moles

Step 4: Calculate moles of KOH

For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O

For 0.002388 moles KHP we need 0.002388 moles KOH

Step 5: Calculate the concentration of KOH

Concentration = moles / volume

Concentration of KOH = 0.002388 moles / 0.03891 L

Concentration of KOH = 0.06137 M

The concentration of KOH is  0.06137 M

Answer:

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

__________________________________________

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻]  = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

___________________________________________________

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹                                   Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl          

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

-----------------------------------------------------------

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

----------------------------------------------------------

From the 0.36M CH₃NH₃⁺

=>       CH₃NH₃⁺ + H₂O  ⇄ CH₃NH₄⁺ + OH⁻

C(eq)   0.36M        ----              x             x     (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

--------------------------------------------------------

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and

CH₃NH₃Cl and the mixture give the following approximate values;

A. The pH value of the 0.18 M CH₃NH₂ is 11.93

B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69

C. The pH value of the mixture is 10.644

A. 0.18 M CH₃NH₂

The solution is presented as follows;

CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻

Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we

have;

The number of moles of CH₃NH₂ remaining = 0.18 - x

Which gives;

[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]

[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴

Therefore;

[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]

4.167 × 10⁻⁴ × (0.18 - x) = x²

4.167 × 10⁻⁴ × (0.18 - x) - x² = 0

Which gives;

x = [OH⁻] = 8.455 × 10⁻³

pH = 14 + log[OH⁻]

Which gives;

pH = 14 + log(8.455 × 10⁻³) ≈ 11.93

B.  0.18 M CH₃NH₃Cl

The solution is presented as follows;

CH₃NH₃⁺ → CH₃NH₂ + H⁺

Let x represent the number of moles of CH₃NH₂ and H⁺ produced,

respectively, we have;

The number of moles of CH₃NH₃⁺ remaining = 0.18 - x

Which gives;

[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]

Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹

Therefore;

[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]

2.27 × 10⁻¹¹ × (0.18 - x) = x²

2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0

Which gives;

x = [H⁺] ≈ 2.02 × 10⁻⁶

pH = -log[H⁺]

Which gives;

pH = -log(2.02 × 10⁻⁶) ≈ 5.69

C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;

Based on the Henderson-Hasselbalch equation, we have;

[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]

Which gives;

[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]

Learn more about Henderson-Hasselbalch equation here:

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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:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]

[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]

[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]

= 66.67%

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

Answer:

26.0L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of CH4 reacts with 1 mole of H2O to produce 1 mole of CO and 3 moles of H2.

That is, 1 mole of each reactant produce 4 moles of gases

As in the reaction, 0.1500 moles of CH4 and 0.1500 moles of H2O are added, 0.1500 moles of CO and 0.4500 moles of H2 are produced.

Before the reaction, the moles of gas are 0.3000 moles and after the reaction the moles are 0.6000 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 13.0L

n1 = 0.3000 moles

V2 = ?

n2 = 0.6000 moles

13.0L*0.6000 moles / 0.3000 moles = V2

V2 = 26.0L is the volume of the sample after the reaction

Answer:

"[Temperature is a measurement of the average kinetic energy of the molecules in an object or a system. Kinetic energy is the energy that an object has because of its motion. The molecules in a substance have a range of kinetic energies because they don't all move at the same speed.]"

Answer:

Temperature is directly proportional to the average translational kinetic energy of molecules in an ideal gas

Explanation:

Answer:

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

Explanation:

The Oxidation no. is +3

Answer:

See explanation

Explanation:

The particular reactants in the Fischer esterification reaction were not stated.

Generally, a Fischer esterification is a reaction that proceeds as follows;

RCOOH + R'OH ⇄RCOOR' + H2O

This reaction occurs in the presence of an acid catalyst.

We can shift the equilibrium of this reaction towards the products side in two ways;

I) use of a large excess of either of the reactants

ii) removal of one of the products as it is formed.

Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer:

The order of reactivity towards electrophilic susbtitution is shown below:

a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene

b. p-cresol>p-xylene>toluene>benzene

c.Phenol>propylbenzene>benzene>benzoic acid

d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene

Explanation:

Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.

For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.

The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.

Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.

The question is incomplete, the complete question is;

When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.

First, the oxidant, like hypochlorite, oxidizes

Choose...

neutral iodine into iodide ion

iodide ion into neutral iodine

iodate polyatomic ion into iodide ion

When starch and iodine are both present, the solution is

Choose...

blue-black

brownish yellow

clear

During the titration, the titrant, like thiosulfate, reduces the

Choose...

iodide ion into iodate polyatomic ion

neutral iodine into iodide ion

iodide ion into neutral iodine

When the iodine has completely reacted at the endpoint of the titration, the solution should become

Choose...

clear

blue-black

brownish yellow

Answer:

1. iodide ion into neutral iodine

2. blue-black

3. neutral iodine into iodide ion

4. clear

Explanation:

Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;

ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)

When iodine is added, the colour of the starch solution immediately changes to blue-black.

A reduction reaction occurs when the titrant, thiosulfate is added as follows;

I2 + 2S2O32- → 2I- + S4O62-

The solution at end point is found to become clear again.

Answer:

a) 22 calories

b) 30 mg

Explanation:

Divide number of cal or mg of pot by 8 fl oz.

Answer:

Volume = 1222.5cm³

Explanation:

If the question is about the volume of the rectangle:

The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.

In the problem, the length of the rectangle is 0.162m = 16.2cm

The width is 7.7cm

And the height is 9.8cm

The volume is:

Volume = 16.2cm*7.7cm*9.8cm

Answer:

d. 60.8 L

Explanation:

Step 1: Given data

Step 2: Calculate the work (W) done by the system

We will use the following expression.

ΔU = Q + W

W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J ×  1 atm.L/101.325 J = -1.593 atm.L

Step 4: Calculate the initial volume

First, we will use the following expression.

W = - P × ΔV

ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

The initial volume is:

V2 = V1 + ΔV

V1 = V2 - ΔV

V1 = 63.2 L - 2.35 L = 60.8 L

Answer:

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Explanation:

Given the data in the question;

wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m

we determine the energy of the proton;

E = hc / λ

where h is plank constant ( 6.626 × 10⁻³⁴ JS )

and c is the speed of light ( 3 × 10⁸ m/s )

we substitute

E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]

E = 8.61 × 10⁻¹⁹ J

we know that, bond energy for H-I is 295 kJ/mol

so, H = 295 × 10³ J/mol

Now, energy to dissociate HI will be;

⇒ H / N

where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )

energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )

= 4.898 × 10⁻¹⁹ J

Therefore, Excess energy over dissociation will be;

⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )

= 3.712 × 10⁻¹⁹ J

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Answer:

5.0 × 10⁻⁶

Explanation:

Step 1: Write the balanced equation for the solution of chromium(III) iodate

Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

Step 2: Calculate the solubility product constant (Ksp)

To relate Ksp and the solubility (S), we will make an ICE chart.

        Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

I                               0                 0

C                             +S              +3S

E                                S               3S

The solubility product constant is:

Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶

1.42liters, which is equivalent to 3pints, of blood is required for the surgery

1 US pint = 0.473 liters

3 × 0.473

= 1.419 liters of blood for the surgery

Learn more at: https://brainly.com/question/24168664

Answer:

A. 1.9 × 103 g

(next one)

Which metric unit would be the best choice to report the result?  

A. kg

Answer:

1. 2

2. 1.9 × 10^3 g

3. kg

Explanation:

Answer:

3.89 ×10^-5

Explanation:

Since they are gaseous reactants, we obtain the equilibrium constant from the given partial pressures;

p(H2O) = 0.0900 bar

p(H2) = 0.00100 bar

p(O2) = 0.00350 bar

The equation of the reaction is;2H2O(g)⇄2H2(g)+O2(g)

Kp= p(H2) . p(O2)/p(H2O)

Kp= 0.00100 × 0.00350/0.0900

Kp= 3.89 ×10^-5

Answer:

12

Explanation:

2+8+2=12

atomic no is the No of protons

Answer:

Atomic number is 12.

Explanation:

Atomic number = electrons in filled shells + outermost electrons

= 2 + 8 +2

= 12

Answer:

d

Explanation:

Answer:

[H₃O⁺] = [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]  

Explanation:

By definition pH = -log[H⁺] => [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]

Using your calculator ... I am using a TI-30XA scientific calculator.

Depending on the calculator, the answer may show as 0.0138, or  1.38x10⁻², or 1.38E-2 (=1.38 x 10⁻²). It is the user's job to insert dimensional units into answer of interest =>  0.0138M, or  1.38 x 10⁻²M, or 1.38E-2M.

1.38E-2 which is 1.38 x 10⁻².

how many electrons does tin have?

Answer : 2, 8, 18, 18, 4

Therefore, total electrons = 50