Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol

Answer:

0.25 mm> 0.50 mm > 0.75 mm> 1mm> 1.25 mm

Explanation:

Capillary movement involves the movement of liquid up through a capillary tube.

Looking at the question, we can see all the various diameters of the capillary tube are involved.

Note that, that the narrower the diameter of the capillary tube, the better the capillary action of the liquid (the liquid rises more now than then).

Answer:

The chemical formula of the unknown ionic compound is Ag₂CO₃.

Explanation:

The method used to identify the ions present in a given solution of an unknown compound is known as qualitative analysis. Specific tests are carried based on the chemical reaction of these ions and the results they give in these reactions, the ions present innthe compound are identified. Some of the results or observations include formation of precipitates on addition of certain reagents, evolution of gases, colour changes, as well as the color changes that are observed in a flame test of a solutionmof these compounds.

In the given question, when testing for cations, a white precipitate is formed when HCl is added to the unknown compound. When the white precipitate is heated in hot water, it does not dissolve.

From the table of observations, both Ag+ and Pb²+ give a white precipitate on addition of HCl but Zn²+ does not. So the cation present is either Ag+ or Pb²+. However, when heated in hot water, the white precipitate does not dissolve as will be observed if Ag+ were the cation present. Thus, the cation is Ag+.

When testing for anions, a white precipitate is formed when Ca(NO3)2 is added to the unknown compound and when HNO3 is added to the precipitate, it dissolves and bubbles are formed.

Now comparing these results with that as observed with CO₃²- and PO₄³-, both form white precipitates on addition of Ca(NO₃)₂, and the precipitate dissolves on addition of HNO₃. However, only CO₃²- dissolves with the production of bubbles when HNO₃ is added. Thus, the anion is CO₃²-.

Therefore, the compound is Ag₂CO₃ which is formed from the cation Ag+ and CO₃²-.

Answer:

Lower than

Higher than

Explanation:

The vapour pressure and boiling point of liquids are inversely related. Thus, the higher the vapour pressure of a liquid, the lower it's boiling point. Lower vapour pressure implies that the liquid is easily converted into vapour phase.

If hexane has a higher vapour pressure than water then its boiling point is lower than that of water and its vapor pressure when it is boiling will be higher than water when water is boiling.

Answer:

89.35 hour

Explanation:

Recall :

Charge on 1 electron = 1.6 × 10^-19 C

1 mole contains = 6.023 × 10^23

Therefore, the charge on 1 mole of electron will be :

Charge per electron × 1 mole :

(1.6 × 10^-19) * (6.023 * 10^23) = 96500 C = 1 Farad

1 Farad = 96500 C

Using the formula :

Q = Current(I) * time(t)

Q = I*t

t = Q/I

Current, I = 0.3 A

t = 96500 / 0.3

t = 321666.66 second

t = 321666.66 / 3600 = 89.35 hour

Answer:

?what

Explanation:

i am having a hard time reading this

Answer:

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Explanation:

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Answer:

K should be large so the temperature change will be large enough to measure.

Explanation:

Let us recall that;

∆T = k m I

Where;

∆T= change in boiling point

K= freezing point constant

m= molality

i= Van't Hoff factor

Since the change in temperature depends on k, the larger the value of k the larger the temperature change(Ts - Tb). Hence; K should be large so the temperature change will be large enough to measure.

Ts= freezing temperature of pure solvent

Tb= freezing temperature of solution

i think .117018

now to get to the 20charaeactrs minimum.

Answer:

The statement is false

Explanation:

Number of moles of alkene = 2.31 × 85/1000 = 0.196 moles

Number of moles of Br2 = 3.55 × 7.5/1000 = 0.0266 moles

Given that the reaction is 1:1

1 mole of alkene reacts with 1 mole of bromine

0.196 moles of alkene should react with 0.196 moles of bromine

Hence, to achieve 100%yield, 0.196 moles of bromine and not 25mmols of elemental bromine

Answer:

CO2

Explanation:

Carbon dioxide is a chemical compound that consists of two oxygen atoms and one carbon atom

The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.

A potential energy diagram or a reaction profile shows us the energ change between the reactants and the products.

As we look at the reaction profile, we observe that the products have a greater energy than the reactants hence the reaction is endothermic. The enthalpy chamgeis obtained by subtracting the energy of the products from the energy of the reactants.

The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.

Learn more about energy profile: https://brainly.com/question/11256472

Answer:

Balance the following reactions and identify the species that have been oxidized and the species that have been reduced.

CL2 +I- -------------> I2 +CL-

WO2 + H2 ---------------> W + H2O

CA + H2O -----------------> H2 + CA(OH)2

AL+ O2 --------------------> AL2O3

Explanation:

Oxidation is the process where an atom loses electrons.

During oxidation, the oxidation number of the atom increases.

Reduction is the process where an atom gains electrons.

During reduction, the oxidation number of the atom decreases.

The balanced chemical equations for the given reactions are shown below:

[tex]Cl_2 +2I^- -> I_2 + 2Cl^-[/tex]

Iodine undergoes oxidation and chlorine undergoes reduction.

[tex]WO_2 +2 H_2 -> W+ 2 H_2O\\[/tex]

In this reaction, W (tungsten) undergoes reduction and hydrogen undergoes oxidation.

[tex]Ca+2 H_2O ->H_2+Ca(OH)_2\\[/tex]

Ca undergoes oxidation and hydrogen undergoes reduction.

[tex]4Al+3O_2->2Al_2O_3\\[/tex]

Al undergoes oxidation and oxygen undergoes reduction.

Answer:

About redox reaction which of the given statements are true?

Explanation:

Redox reaction is the one in which both oxidation and reduction reactions take place simultaneously.

For example:

[tex]C(s)+O_2(g)->CO_2(g)[/tex]

In this reaction, carbon undergoes oxidation and oxygen undergoes reduction simultaneously.

During this reaction, mutual exchange of electrosn take place between the oxidant and the reductant.

Among the given options,

Option B. electrons are transferred

and

option C.They include both oxidation and reduction takes place are the correct answers.

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

of MgCO3.3H2O in the mixture?

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

Mass water:

3.883g - 2.927g = 0.956g water

Moles water -18.01g/mol-

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

Moles MgCO3.3H2O:

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

Mass percent:

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

Different boiling point.

Because in distillation the substance with lower boiling point is evaporated, leaving the other material(s).

Distillation relies on different boiling points to separate out parts of a solution.

Answer:

Mg3N2

Explanation:

it would be magnesium as it would loss to electron so it would have 10 electron. you can see in the picture above .

hope this helps :)

Explanation:

For this question, we apply the equation: Q = mCp AT Where m is the mass of the substance, Cp

is its specific heat capacity and AT is the

temperature change. Q = 896 x 0.45 x (5-94)

Q = -35884.8 Joules

So about -36 kilojoules of heat is released.

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Answer:

The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Explanation:

We are given that

Volume of H2 at STP=52.8mL

Mass of magnesium metal ,M(Mg)=0.055g

We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.

Molar mass of Mg=24.305 g/mol

Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]

Using the formula

Number of moles of Mg=[tex]\frac{0.055}{24.305}[/tex]moles

Number of moles of Mg=0.00226moles

Number of moles of Mg=Number of moles of H2

Number of moles of H2=0.00226moles

Molar volume of Hydrogen gas (H2)=[tex]\frac{volume\;at\;STP}{No\;of\;moles\;H_2}[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}mL/mol[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}\times 10^{-3}L/mol[/tex]

[tex]1L=1000mL[/tex]

Molar volume of Hydrogen gas (H2)=23.36L/mol

Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Answer:

C₂H₃O

Explanation:

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

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

Answer:

D. Two scientists may have different underlying assumptions that lead them to different conclusions about the same thing.

Explanation:

It's all about how a person analyzes data. Some do it mathmatically while others do it logically. By doing it different ways, you may still come to the same conclusion, despite working at the problem in a different way.

Answer:

1.40x10^24 atoms of H

Explanation:

Answer:

HOPE IT helps much as you can

Answer:

vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.

Answer:

0.111 g

Explanation:

1 g = 1000 mg

Doctor ordered the following concentration of Claforan:

C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL

If we add 2 g iof Claforan, we obtain:

2 g Claforn ---- 180 mg/mL Claforan

To reach a concentration equal to C (10 mg/mL), we need:

10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn

Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.  

Two chromosomes

Explanation:

The chromosomes are XY

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Answer:

Two

Explanation:

Answer:

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]

Explanation:

For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:

[tex]\bar v = Z^2 R_H \Big(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\Big)[/tex]

where:

emitted radiation of the wavenumber  [tex]\bar v[/tex] = ???

atomic no of helium Z = 2

Rydberg's constant [tex]R_H = 1.097*10^7 \ m^{-1}[/tex]

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

Now, the emitted radiation of the wavenumber can be computed as:

[tex]\bar v = (2)^2 (1.097*10^7 \ m^{-1} ) \Big(\dfrac{1}{1^2}-\dfrac{1}{2^2}\Big)[/tex]

[tex]\bar v = 3.291 \times 10^ 7/m[/tex]

Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber  [tex]\bar v[/tex], which is:

[tex]\bar v = \dfrac{1}{\lambda}[/tex]

[tex]\lambda = \dfrac{1}{\bar v}[/tex]

[tex]\lambda = \dfrac{1}{3.291 \times 10^{7}}\times \dfrac{m}{1}[/tex]

[tex]\mathbf{\lambda =3.03859 \times 10^{-8} \ m}[/tex]

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]

Answer:

Explanation:

The structural formula of 2-methyl-3-hexene is drawn in the attached image below. After then, the three possible structures were drawn and the mechanism reaction.

From the mechanism shown, methanol deprotonates methoxide proton and neutralizes the positive charge to form methoxide. This is followed by the movement of electrons that takes place from the region of higher electron density to lower electron density.