Eugenol is a molecule that contains the phenolic functional group. Which option properly identifies the phenol in eugenol

Answer: Thus the cell potential of an electrochemical cell is +0.28 V

Explanation:

The calculation of cell potential is done by :

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]

[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]

As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.

[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]

[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]

Thus the cell potential of an electrochemical cell is +0.28 V

Answer:

B

Explanation:

The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.

The element you hit is Argon.

The answer is B

Answer:

B ar

Explanation:

pen foster answer

Answer:

The correct answer is -2878 kJ/mol.

Explanation:

The reaction that takes place at the time of the oxidation of glucose is,  

C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)

The standard free energy change for the oxidation of glucose can be determined by using the formula,  

ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)

The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.  

Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)

ΔG°rxn = -2878 kJ/mol

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of Al produce 2 moles of Al₂O₃

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

Answer and Explanation:

Given the following chemical equation:

Cl₂(g) + F₂(g) ⇒ 2ClF(g)

The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:

On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).

On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).

Answer: C

Explanation:

Right on edge 2020

The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.

Land management refers to the activities which are done in order to protect and preserve the land as well the resources found on land.

The Bureau of Land Management was created to manage land in the US.

The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.

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Answer:

92.93 g

Explanation:

Number of half lives that have elapsed in eight days =8/14.3 = 0.559

Fraction of the radioactive nuclide that remains after 0.559 half lives is given by

N/No=(1/2)^0.559

Where N= mass of radioactive nuclides remaining after a time t

No= mass of radioactive nuclides originally present

N/No=(1/2)^0.559= 0.679

Mass of nuclides present eight days before= 63.1g/0.679

Mass of nuclides present eight days before=92.93 g

Answer:B

Explanation: A P E X

Answer:

1.7 × 10⁹ L

Explanation:

Step 1: Write the balanced equation

CO(g) + 2 H₂(g) → CH₃OH(l)

Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]

Step 3: Calculate the theoretical yield of CH₃OH

The real yield of CH₃OH is 3.1 × 10⁷ mol  and the percent yield is 40%. The theoretical yield is:

[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]

Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH

The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol

Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP

The volume of 1 mole of CO at STP is 22.4 L.

[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]

Answer:

A. Electron clouds or energy levels

Explanation:

Electrons can exist in two states:

Electrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.

Hence, the correct answer is A.

Answer:

Up and DOWN spin

Explanation:

Answer:

[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q  = It  

[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.  

(a) Calculate q

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]

Note: The answer can have only one significant figure because that is all you gave for the time.

Answer: stay away from doors and windows.

Explanation:

to aviod geting hit by glass

Answer:

Stay away from doors and windows.

Explanation:

Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.

Answer:

Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.

Answer:

wavelength, λ =  486.6 nm

frequency, f = 6.16 * 10¹⁴ Hz

Explanation:

a. Wavelength

Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)

Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s

1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)

1/λ = 2.055 * 10⁶ m

λ = 4.866 * 10⁻⁷ m

wavelength, λ =  486.6 nm

b.  Frequency

Using f = c/λ

f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m

frequency, f = 6.16 * 10¹⁴ Hz

Answer:

Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include

B) Gas condensing below its condensation point

C) Liquid vaporizing above its boiling point

D) Liquid freezing below its freezing point

F) Solid melting above its melting point

Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include

A) Solid melting below its melting point

E) Liquid freezing above its freezing point

H) Gas condensing above its condensation point

Equilibrium system

G) Liquid and gas together at boiling point with no net condensation or vaporization

I) Solid and liquid together at the melting point with no net freezing or melting

A) Solid melting below its melting point - nonspontaneous process

B) Gas condensing below its condensation point - spontaneous process

C) Liquid vaporizing above its boiling point - spontaneous process

D) Liquid freezing below its freezing point - spontaneous process

E) Liquid freezing above its freezing point - nonspontaneous process

F) Solid melting above its melting point - spontaneous process

G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system

H) Gas condensing above its condensation point - nonspontaneous process

I) Solid and liquid together at the melting point with no net freezing or melting -  Equilibrium system

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Answer:

T2 = 500K

Explanation:

Given data:

P1 = 1atm

V1 = 300ml

T1= 27 + 273 = 300K

T2 = ?

V2 = 1.00ml

P2 = 500atm

Apply combined law:

P1xV1//T1 = P2xV2/T2 ...eq1

Substituting values into eq1:

1 x 300/300 = 500 x 1/T2

Solve for T2:

300T2 = 500 x 300

300T2 = 150000

Divide both sides by the coefficient of T2:

300T2/300 = 150000/300

T2 = 500K

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

[tex]p^{OH}[/tex]=14-56.17

      =8.823

The volume of the [tex]NH_{3}[/tex] = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated [tex]NH_{3}[/tex] =0.10 M

The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M

The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml

convert into the liter= 0.030L  

The value of concentrated [tex]H_2So_4[/tex]=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]

calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:

[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]

create the ICE table:    

[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

[tex]\frac{V}{T}=k[/tex]

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

[tex]\frac{P}{T}=k[/tex]

Combinado las mencionadas tres leyes se obtiene:

[tex]\frac{P*V}{T} =k[/tex]

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

Reemplazando:

[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]

Resolviendo:

[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]

V2= 12.95 L

El volumen del gas era 12.95 L

Answer:

negative charge electrode

Explanation:

In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.

Answer:

The electrode that removes ions from the solution :) a p e x

Answer:

a) [tex]\Delta G=2.6kJ[/tex]

b) [tex]\Delta G=-979.57kJ[/tex]

c) [tex]\Delta G=264.21kJ[/tex]

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

[tex]\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol[/tex]

So we proceed as follows:

a)

[tex]\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ[/tex]

b)

[tex]\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ[/tex]

c)

[tex]\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ[/tex]

Regards.

Answer:

a) 1.5 x 10^-3 mol/L

b) 1.35×10^-8

c) decrease

Explanation:

The solubility of lead II iodide is given by the equation;

PbI2(s) -----> Pb^2+(aq) + 2I^-

By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L

Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L

Ksp= [Pb^2+] [2I^-]^2 =

Let the molar solubility of each ion be x, therefore;

Ksp= 4x^3

Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8

Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.

a) The molar solubility of PbI₂ is  [tex]1.5 * 10^{-3} mol/L[/tex]

b) The solubility constant is [tex]1.35*10^{-8}[/tex]

c) The molar solubility of lead (II) will decrease.

The solubility of lead II iodide is given by the equation;

[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]

By looking at the ICE table,

[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex]  [tex]1.5 * 10^{-3} mol/L[/tex]

Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]

[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]

Let the molar solubility of each ion be x, therefore;

[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]

The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.

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Answer:

E. O2

Explanation:

All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.

In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.

Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.

Answer:

-second

Explanation:

6-ethyl-2-octyne is an unsaturated compound with a triple bond.

6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number  "2" before suffix refers to the position of triple bond that is second carbon.

Hence, the correct option is  "-second ".

Answer:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Explanation:

To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.

Nitrite ion (NO₂⁻)

1 × N + 2 × O = -1

1 × N + 2 × (-2) = -1

N = +3

Nitrous oxide (NO)

1 × N + 1 × O = 0

1 × N + 1 × (-2) = 0

N = +2

Nitrate ion (NO₃⁻)

1 × N + 3 × O = -1

1 × N + 3 × (-2) = -1

N = +5

Ammonia (NH₃)

1 × N + 3 × H = 0

1 × N + 3 × (+1) = 0

N = -3

Nitrogen gas (N₂)

2 × N = 0

N = 0

The order of increasing nitrogen oxidation state is:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Answer:

sodium

Explanation:

(Na) atom is paramagnetic and sodium is a na atom.

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

Answer:

B and C. Just finished my lesson on Edge.

Answer:

1)is more electronegative

2)

3) is more electronegative

Explanation:

1) for the first question, iodine is more electronegative than tellurium hence we naturally expect that HI should be more acidic than H2Te since electronegativities play a role in the acidity of chemical species.

2) the correct option is not listed because the H2Te bond is weaker than the H2S bond. This makes it easier for H2Te to dissociate releasing H^+ , thereby being more acidic than H2S.

3) Hydrogen is more electronegative than sodium hence it cannot be ionized thus NaH is not acidic.

Answer:

0.2 moles, assuming weight of dried salt

Explanation:

In order to determine the number of moles, we need to be aware of the mass of the substance in question.

Assuming the mass of the dehydrated [tex]KAl(SO_{4} )_{2}.H_{2} O[/tex] is 50g.

No. of moles = mass of substance/ molar mass of the substance.

= [tex]\frac{50g}{39+27+32*2+16*4*2\\)g/mol}[/tex]

= 0.2 moles moles.

Answer:

The  energy change is  [tex]E = 9.0 *10^{9}\ J[/tex]

Explanation:

   From the question we are told that

          Mass loss  is  [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]

  Generally the energy change that  would accompany this loss  is mathematically represented as

     [tex]E = m * c^2[/tex]

Where  c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]

     [tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]

     [tex]E = 9.0 *10^{9}\ J[/tex]

Answer:

1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻

2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻

3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃

Hope this helps.

The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).

The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻

Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.

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