This ether can, in principle, be synthesized by two different combinations of haloalkane and metal alkoxide. Draw the combination of alkyl chloride and potassium alkoxide that forms the higher yield of ether. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Separate multiple reactants using the sign from the drop-down menu.

Answer:

Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield  

Explanation:

Answer:

t = 0.049 mins or 2.94 secs

Explanation:

For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;

Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.

Using the integrated rate law:

I/[NO2]t - 1/[NO2]o = Kt

Where K = 3.40 L/mol/min

[NO2]t = 1.5 mol/L

[N02]o = 2.0 mol/L

t = ?

Making t subject of formula;

t = (1/[NO2]t - 1/[NO2]o) / K

t = (1/1.5 - 1/2.0)/3.40

t = 0.049 mins or 2.94 secs

Answer:

2.0 g Na

Explanation:

Stoichiometry.

8.4g sodium hydrogen citrate x (1 mol sodium hydrogen citrate / 192 g sodium hydrogen citrate) x (2 mol Na/1 mol sodium hydrogen citrate) x (23g Na/1 mol Na)

^write it out it makes more sense that way

Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium

Explanation:

I hope it will help you

Answer: B. Potassium(K)

Explanation:

Answer:

HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)

Explanation:

The equation to distinguish between cation and anion hydrolysis is given below :  

HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)

The important thing to remember is their origin. The anions can react with water and can produce hydroxide ions while hydroxide ions make a solution basic.

Answer:

Explanation:

1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.

Yes, there are new substances created from this mixture.

Answer:

filtering

Explanation:

you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.

Answer:

a. 2,3-dimethylbutane < 2-methylpentane < n-hexane

Explanation:

The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.

The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.

n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

Mass of the Vanadium, number of O2 atoms present, or the mole ratio of   1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.

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Answer:

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

Explanation:

Palladium is a chemical element with the symbol Pd and atomic number 46.

The electronic configuration is;

[Kr] 4d¹⁰

The full electronic configuration observed for palladium is given as;

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

The reason for for the anomlaous electron configuration is beacuse;

1. Full d orbitals are more stable than partially filled ones.

2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.

Answer:

         H

         ° *         . .

H  ° *   C  * ·    Cl :

           * ·           . .

       :  Cl :

          . .

Explanation:

Carbon has 4 valent electrons

  *

 *C*

   *

Hydrogen has 1 electron

H°

Cl has 7 electrons on the last level.

 . .

: Cl·

 . .

         H

         ° *         . .

H  ° *   C  * ·    Cl :

           * ·           . .

       :  Cl :

          . .

The Lewis dot structure for [tex]H_2CCl_2[/tex] is explained in the explanation part below.

A Lewis structure is a symbolic depiction of a molecule or ion that depicts the arrangement of atoms and valence electrons.

It is also known as a Lewis dot structure or electron dot structure. Gilbert N. Lewis, an American chemist, invented it.

The total number of valence electrons in the molecule must be determined before writing the Lewis dot structure for H2CCl2 (dichloroethylene).

H has one valence electron, while C has four.

Cl has seven valence electrons.

Thus, the Lewis dot structure for the given compound is attached below as image.

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Answer:

1.) zinc and aluminum

2.)

a.) The old coating was made of tin.

b.) Aluminum would be a good choice for repairing the fracture.

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed  by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

Answer:

All ionic substances are written as separate ions in solution

All ionic substances are written as separate ions in solution in a complete ionic equation. Therefore, option (C) is correct.

A complete ionic equation can be described as a particular chemical equation where charged atoms such as ions are expressed in a given solution. The complete ionic equations always contain all ions that are formed or act during a particular chemical reaction.

The net ionic equation can be described as an equation that provides information about ions that exists in an aqueous medium. Salts get dissolved in polar solvents such as water which are present as cations and anions in their dissolved state.

The ionic equation shows the chemical species that undergo a chemical change. The ions which are present on both sides of the equation are considered to be spectator ions. Therefore, in order to obtain the net ionic equation we can eliminate them.

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Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Answer:

B. Infrared.

Explanation:

Referring to the electromagnetic spectrum, ultraviolet rays can be measured with a frequency of 10‐⁸, infrared has a frequency of 10‐⁵, visible radiation has a frequency of 0.5 x 10‐⁶ meanwhile X-rays show a frequency of 10‐¹⁰.

Hence, the largest magnitude among the rest goes to infrared rays, which makes B the correct answer.

Answer:

i have no clue imma go with c

Explanation:

Strong acids are completely ionised and weak acids are partly ionised

Answer:

Como forman los iones en soluciión

Explanation:

Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.

Answer: True

Explanation:

The independent variable is the one which can be changed or manipulated in an experiment. The independent variable exerts its influence on the dependent variable. The dependent variable is the result of the experiment.

The amount of sunlight, can be regulated or changed in an experiment, thus it is an independent variable. The effect of sunlight on different plants is the dependent variable.

Answer: The minimum [tex][Ag^{+}][/tex] concentrations required to precipitate out the anions is [tex]9 \times 10^{-9}[/tex] M.

Explanation:

We know that,

  [tex]K_{sp}[/tex] for AgCl is [tex]1.8 \times 10^{-10}[/tex]

and,  [tex]K_{sp}[/tex] for [tex]Ag_{2}CrO_{4}[/tex] is [tex]9 \times 10^{-12}[/tex]

Now, we will calculate the concentration of  at which these ions precipitate out are as follows.

For AgCl :

[tex][Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]}[/tex]

             = [tex]\frac{1.8 \times 10^{-10}}{0.02}[/tex]

             = [tex]9 \times 10^{-9}[/tex] M

For  [tex]Ag_{2}CrO_{4}[/tex] :

[tex][Ag^{+}]^{2} = \frac{K_{sp}}{CrO^{2-}_{4}}[/tex]

              = [tex]\frac{9 \times 10^{-12}}{0.01}[/tex]

              = [tex]9 \times 10^{-10}[/tex]

[tex][Ag^{+}] = \sqrt{(9 \times 10^{-9})}[/tex]

            = [tex]3 \times 10^{-5}[/tex] M

This shows that concentration of  ions in AgCl is less than the concentration of AgCl will precipitate first.

Answer:

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Explanation:

Yo know the following balanced reaction:

CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)

You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:

Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?

[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]

mass of Zn= 185.49 grams

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Answer:

Group 1 or akali metals have the greatest metallic property.

Group 17 has the lowest metallic character.

C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.

Explanation:

Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.

Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.

Answer:

4–octene.

Explanation:

To name the compound given in the question, we must determine the following:

1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.

2. Locate the longest continuous carbon. This gives the parent name of the compound.

3. Combine the above to obtain the name of the compound.

Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:

1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.

2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.

3. Therefore, the name of the compound is:

4–octene.

Answer:

0.125 mg

Explanation:

The correct answer would be 0.125 mg

According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.

milligram = [tex]10^{-3}[/tex]

microgram = [tex]10^{-6}[/tex]

Hence,

1 milligram = 1000 micrograms or 1 microgram = [tex]10^{-3}[/tex] milligram

Therefore, 125 micrograms will be:

  125/1000 = 0.125 milligram

Answer:

density = 8.824g/mL

Explanation:

given

mass = 75g

volume = 8.5mL

density = mass/volume

density = 75g/8.5mL

density = 8.824g/mL

Answer:0.088g/ml

Explanation:

Density=mass/volume

d=0.75g/8.5ml

d=0.088g/ml

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

[tex]2*n_{acid}=n_{base}[/tex]

And in terms of concentrations and volumes:

[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, we solve for the volume of acid:

[tex]V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL[/tex]

Best regards.

Answer:

B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS

Explanation:

Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.

Answer:

B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.

Explanation:

Answer:

Rate of diffusion is same .

Explanation:

As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula:  ∆Go ’= -RTln K’eq

where,  

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

=  - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

=  - 8.315 x 298 x 6.79

=  - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

=  -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

Answer:

Percent yield = 22.8 %

Explanation:

Step 1: Data given

Numbers of moles  copper = 0.0970 moles

Mass of copper(I) sulfide = 1.76 grams

Step 2: The balanced equation

2Cu + S ⇒ Cu2S

Step 3:  Calculate moles of Cu2S

For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S

For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles

Step 4: Calculate mass of Cu2S

Mass Cu2s = moles Cu2S * molar mass Cu2S

Mass Cu2S = 0.0485 moles * 159.16 g/mol

Mass Cu2S = 7.72 grams

Step 5: Calculate percent yield

Percent yield = (actual yield/ theoretical mass) * 100%

Percent yield = (1.76 grams / 7.72 grams)*100%

Percent yield = 22.8 %

The percentage yield of the experiment obtained by the reaction of 0.0970 mole of copper with excess sulfurs is 22.8%

We'll begin by calculating the number of mole of Cu₂S produced from the reaction. This can be obtained as follow:

2Cu + S —> Cu₂S

From the balanced equation above,

2 moles of Cu reacted to produce 1 mole of Cu₂S.

Therefore,

0.0970 mole of Cu will react to produce = [tex]\frac{0.0970}{2}[/tex] = 0.0485 mole of Cu₂S.

Next, we shall determine the theoretical yield by calculating the mass of 0.0485 mole of Cu₂S.

Molar mass of Cu₂S = (63.5×2) + 32 = 159 g/mol

Mole of Cu₂S = 0.0485 mole

Mass = mole × molar mass

Mass of Cu₂S = 0.0485 × 159

Thus, the theoretical yield of Cu₂S is 7.7115 g

Finally, we shall determine the percentage yield of Cu₂S.

Actual yield = 1.76 g

Theoretical yield = 7.7115 g

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{1.76}{7.7115} * 100\\\\[/tex]

Therefore, the percentage yield of the experiment is 22.8%

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