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

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:

joib

Explanation:

Answer:

The student completed the experiment correctly and there were no errors in the experiment.

Explanation:

When a pea size amount of Ag is dropped into a solution of HCl, the following reaction occurs;

Ag(s) + 2HCl(aq) ----> AgCl(aq) + H2(g)

The gas H2 burns with a pop sound. The gas obtained does not react to litmus. The products of the reaction are expected.

Hence, the experiment was properly conducted and the student completed the experiment correctly and there were no errors in the experiment.

Answer is 9

pKb=−logK

b=−log10^-5=5

pOH=pKb+log[acid]*[salt]

Substitute values in the above expression.

pOH=5+log0.1*0.1=5

Hence, the pH of the solution is pH=14−pOH=14−5=9

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Answer:

hydrogen, oxygen, chlorine, ammonia

Explanation:

Air is a mixture of gases. When we say "clean" air here, we are referring to air that does not contain pollutant gases.

Some components of air such as water vapour, methane, CO2, and N2O are greenhouse gases. They are known to contribute towards global warming.

Some gases such as SO2 and NO2 contribute towards acid rain. The oxides of nitrogen are particularly involved in the formation of photochemical smog.

The halogens are known to lead to the depletion of the ozone layer and radon is a radioactive gas.

Hence, hydrogen, oxygen, chlorine, ammonia have no negative environmental impact hence they are found in clean air.

Answer:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And 0.119M is the concentration of the citric acid.

Explanation:

In an acid-base reaction, the proton H+ and the hydroxil ion OH- reacts producing water. The ions of the acid and base (C6H5O7³⁻ and Ca²⁺ ions produce the respective salt) as follows:

Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the Calcium ions:

3Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the C6H5O7³⁻ ions:

3Ca(OH)2 + 2C6H8O7 → H2O + Ca3(C6H5O7)2

And to balance the oxygens of water:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And this is the balanced reaction.

The moles of Ca(OH)2 that reacts are:

41.27mL = 0.04127L * (0.108mol/L) = 0.004457 moles Ca(OH)2

Moles of citric acid:

0.004457 moles Ca(OH)2 * (2mol C6H8O7 / 3mol Ca(OH)2) = 0.002971 moles C6H8O7

In 25.00mL = 0.02500L:

0.002971 moles C6H8O7 / 0.0250L =

Answer:

4.33 L

Explanation:

Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:

Where in this case:

We input the given data:

And solve for V₂:

Answer:

3.65 g of cyclohexene

Explanation:

Cyclohexanol + phosphoric acid ----> cyclohexene

The reaction is 1:1 hence the limiting reactant is phosphoric acid.

Hence,

1 mole of phosphoric acid yields 1 mole of cyclohexene

0.0444 moles of phosphoric acid yields 0.0444 moles of cyclohexene

Theoretical yield = number of moles of cyclohexene × molar mass of cyclohexene

Theoretical yield = 0.0444 moles of cyclohexene × 82.143 g/mol

Theoretical yield = 3.65 g of cyclohexene

Answer:

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Answer:

Certain ketones such as fructose can be oxidized by Benedict's reagent under basic conditions to form what type of compound

Explanation:

Benedict's solution is a mixture of copper sulfate, anhydrous sodium carbonate, and sodium citrate.

In presence of mild reducing agents, Cu(II) ion in Benedict's solution becomes the Cu(I) ion.

Fructose has an alpha-hydroxy ketone group.

It undergoes tautomerism and forms alpha-hydroxy aldehyde which gives a positive test with Benedict's reagent.

During this test, aldehydes will be converted into carboxylic acids.

The reaction of fructose with Benedict's reagent is shown below:

Answer:

26.9 g

81%

Explanation:

The equation of the reaction is;

4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)

Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles

4 moles of KO2 yields 2 moles of K2CO3

0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3

Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles

2 moles of CO2 yields 2 moles of K2CO3

1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3

Hence the limiting reactant is KO2

Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g

Percent yield = actual yield/theoretical yield × 100

Percent yield = 21.8/26.9 × 100

Percent yield = 81%

Answer:

SN1 substitution reaction

Explanation:

cyclohexanol is a secondary alkanol. The mechanism of acid catalysed dehydration of cyclohexanol involves the protonation of the -OH group.

This is followed by loss of water which is a good leaving group. At this stage, a proton could be abstracted to yield cyclohexene by E1 mechanism or an SN1 substitution reaction may occur to yield Cyclohexyl chloride.

The both reactions are equally likely.

Answer:

cell

chloroplast and cell wall

nucleus

life processes

cell membrane

shape and size

vacuole

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Answer:

Write a chemical equation for KOH(aq) showing how it is an acid or a base according to the Arrhenius definition.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Write a chemical equation for NH4+(aq) showing how it is an acid or a base according to the Arrhenius definition.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Explanation:

According to this theory, an acid is a substance that is a proton donor when the substance is dissolved in water.

The base is the one that gives OH- ions when dissolved in water.

KOH(aq) is a base.

Because it releases OH- ions when dissolved in water.

[tex]KOH(aq)->K^+(aq)+OH^-(aq)[/tex]

The ammonium ion is an acid.

It can donate a proton when dissolved in water.

[tex]NH_4^+(aq)->NH_3(aq)+H^+(aq)[/tex]

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

Moles E-stilbene:

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

Mass Product = Theoretical yield:

0.00277 moles * (340.058g/mol) = 0.9433g

Answer:

D

Explanation:

PV =nRT

So V and T are inversely proportional

Answer:

1) Ra

K

Na

Li

2) Na

Mg

Al and Si

Answer:

0.1 L

Explanation:

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

Concentration of stock solution (C₁) = 1.25 M

Volume of diluted solution (V₂) = 0.5 L

Concentration of diluted solution (C₂) = 0.25 M

Volume of stock solution needed solution (V₁) =?

The volume of the stock solution needed can be obtained as follow:

C₁V₁ = C₂V₂

1.25 × V₁ = 0.25 × 0.5

1.25 × V₁ = 0.125

Divide both side by 1.25

V₁ = 0.125 / 1.25

V₁ = 0.1 L

Therefore, the volume of the stock solution needed is 0.1 L

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

[tex]n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol[/tex]

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

Answer:

10.5 × 10^5 M

Explanation:

E°cell = E°cathode - E°anode

E°cell = 0.85 - (-0.74) = 1.59 V

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

1.499 = 1.59 - 0.0592/6 log [Cr^3+]/6.35x10^-4

1.499 - 1.59 = - 0.0592/6 log [Cr^3+]/6.35x10^-4

-0.091 = -0.00987 log [Cr^3+]/6.35x10^-4

-0.091/ -0.00987 = log [Cr^3+]/6.35x10^-4

9.22 = log [Cr^3+]/6.35x10^-4

Antilog (9.22) = [Cr^3+]/6.35x10^-4

1.66 × 10^9 = [Cr^3+]/6.35x10^-4

[Cr^3+] = 1.66 × 10^9 × 6.35x10^-4

[Cr^3+] = 10.5 × 10^5 M

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

[tex]1 \ mg = \frac{1}{1000} \ g[/tex]

then,

⇒ [tex]43 \ mg = \frac{43}{1000} \ g[/tex]

              [tex]= 0.043 \ g[/tex]

Thus, the above is the correct alternative.

Answer:

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Answer:

41264 g of CO₂

Explanation:

Combustion reaction is:

C₃H₈ + 5O₂ →  3CO₂  +  4H₂O

1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.

In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.

We have the volume of propane but we need moles of it, so we need to apply density.

Density = mass / volume so mass = density . volume.

Density of propane is: 493 g/L

Mass of propane is 493 g/L . 27.9L = 13754.7 g

We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles

According to reaction, 1 mol of propane can produce 3 moles of CO₂

Our 312.6 moles will produce 312.6 . 3 = 937.8 moles

We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g

Answer:

The answer is d because the light is energy of wave

The frequency of a light wave can be determined

So, option D is correct one.

What is frequency?

Energy equation,

[tex]E= hv[/tex]

where,

[tex]E[/tex] = Energy of light

[tex]h[/tex]= Planks constant

[tex]v[/tex] = Frequency of wave.

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Answer:

a) The mechanism of the reaction is the Elimination Bimolecular or E2.

b) Steps for the mechanism of this reaction is given as follows,

c) Reaction rate = K[Organic compound][[tex]NaoCH_{2} CH_{3}[/tex]].  

Explanation:

a) The mechanism of the reaction is the Elimination Bimolecular or E2.

c) This is an E2 reaction, so it depends on the concentration of both substrate and reactant. If we increase the concentration of [tex]NaoCH_{2} CH_{3}[/tex], the reaction rate will be increased.

Reaction rate = K[Organic compound][[tex]NaoCH_{2} CH_{3}[/tex]].  

b) Steps for the mechanism of this reaction is given as follows,

Answer:

d. the conjugate base of the weak acid

Explanation:

The strong base (BOH) is completely dissociated in water:

BOH → B⁺ + OH⁻

The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.

By other hand, the weak acid (HA) is only slightly dissociated in water:

HA ⇄ H⁺ + A⁻

The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.

Therefore, the resulting solution will have a pH > 7 (basic).

Answer:

Percent yield = 48.3%

Explanation:

The reaction is:

CCl₄  +  2HF → CF₂Cl₂ + 2HCl

1 mol of CCl₄ reacts with 2 moles of hydrofluoric acid in order to produce 1 mol of CF₂Cl₂ and 2 moles of hydrogen chloride.

HF is in excess, so the limiting reagent is the CCl₄.

We convert mass to moles:

32.9 g . 1mol / 153.8g = 0.214 moles

Ratio is 1:1. In conclussion: 0.0813 moles of CCl₄ can produce 0.0813 moles of CF₂Cl₂. We convert moles to mass, to determine the theoretical yield:

0.214 mol . 120.91g /mol = 25.8 g

Percent yield = (Yield produced /Theoretical yield) . 100

Percent yield = (12.5 g/ 25.8g) . 100 = 48.3%

Answer:

A molecule of sucrose (C12H22O11) has 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.

Explanation:

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There are   1.4376  × 10²⁶ carbon atoms in 15 lbs of sugar (C12H22O11).

From the given information,

Using the standard conversion method;

1 lbs = 453.592 gram

∴

15 lbs = (453.592 gram × 15 lbs/1 lbs)

= 6803.88 grams

Now, we will need to determine the molar mass of the sugar compound C12H22O11.

Molar mass of C12H22O11 =  (12 × 12) +(1 × 22) + (16 × 11)

Molar mass of C12H22O11 = 144 + 22 + 176

Molar mass of C12H22O11 = 342 g/mol

Using the relation:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

Number of moles = [tex]\dfrac{6803.88 }{ 342}[/tex]

Number of moles of C12H22O11   = 19.894 moles

Since we've known the number of moles present in C12H22O11, the next thing to do is determine the number of molecules of sugar by using Avogadro's constant:

i.e.

number of moles of sugar = [tex]19.894 moles \times \dfrac{6.023 \times 10^{23}}{mol}[/tex]

= 1.198 × 10²⁵ molecules of C12H22O11

Now to determine the number of carbon atoms in 15 lbs, we have:

= number of carbon atoms × amount of molecules

= 12 × 1.198 × 10²⁵ carbon atoms

= 1.4376  × 10²⁶ carbon atoms

Therefore, we can conclude that there are  1.4376  × 10²⁶ carbon atoms present in 15 lbs of sugar, C12H22O11

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