For a spontaneous process, which of the following MUST be true?
A. TΔS>ΔH
B. ΔG>0
C. ΔSuniv>0
D. ΔSsys>ΔSsurr

Answer:

There are fifteen molecular orbitals in benzene filled with electrons.

Explanation:

Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;

There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.

There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons

The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.

All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.

Answer:

83914.52 grams

Explanation:

Given that,

Weight of a man is 185 lb

We need to find his weight in grams

For this, we must know the relation between lb and grams.

We know that,

1 lb = 453.592 grams

To find the mass of man in grams, the step is :

185 lb = (453.592 × 185) grams

= 83914.52 grams

So, the mass of a man is 83914.52 grams.

Answer:

The number of electrons transferred in the reaction

Explanation:

Answer:

A

Explanation:

Answer:

V= n/M

Explanation:

From;

π = nRT/V = MRT

Where;

n= number of moles

R= gas constant

T= absolute temperature

M= molar mass

V= volume of the solution

π= osmotic pressure

Thus;

nRT/V = MRT

nRT = VMRT

V= nRT/MRT

V= n/M

Answer:

b. 0°C and 1 bar

Explanation:

Hello,

In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.

Best regards.

Answer:

Explanation:

A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.

Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is

2-hydroxy-2,3-dimethylbutane

     H  OH   H    H

      |     |       |      |

H - C - C -   C  - C - H

      |     |       |      |

     H  CH₃  CH₃ H

From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain

Answer:

a. NCl3

Explanation:

All the elements stated are non metals. Generally the bond between non metals is a covalent bond. However depending on how the electrons are shared, this bond can either be polar or non polar.

A non polar covalent bond is formed when the electrons are shared equally between the atoms.

A way to determine if a bond is non polar or polar covalent is by comparing the electronegativity values.

If the electronegativity difference is less than 0.5, it is regarded as a non polar covalent bond.

Going back to the question;

Between N and Cl; the electronegativity difference is 3.0 - 3.0 = 0

Between Cl and O; the electronegativity difference is 3.5 - 3.0 = 0.5

Between Cl and P; the electronegativity difference is 3.5 - 2.1 = 1.4

From these we can tell that the correct option is option A.

Answer:

Electrons "surround"

Explanation:

Protons and neutrons "make up" the nucleus so they are contained "within" the nucleus meaning that electrons would "surround" the nucleus as they orbit around the nucleus

Answer:

Electrons

Explanation:

Protons and nuetrons are present inside the nucleus of an atom while the electron revolve around the nucleus in different energy levels.

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

I hope it helps you!

Answer:

e. The reaction is product favored and would be considered a voltaic (galvanic) cell

Explanation:

An electrochemical cell produces electrical energy from electrochemical reactions.

A voltaic cell is a type of electrochemical cell that produces electrical energy by spontaneous electrochemical reactions. In a voltaic cell, the cell potential is always positive unlike in an electrolytic cell.

Hence, given the fact that the cell potential is positive, it is a product favoured voltaic cell.

Answer:

B.

Explanation:

A chemical reaction that has the general formula of nA → (A)n is best classified as a polymerization reaction.

Answer:

B. Polymerization

Explanation:

I'm just smart

Answer:

1.8x10¹⁷ molecules of CO are in each breath we take

Explanation:

Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.

A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.

In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:

2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =

[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take

A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.

Now CO molecules in each breath is

[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]

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Answer:

H2SO4 with concentrated HCl

Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Answer:

[tex]Speed = 3.30 \frac{km}{hr}[/tex]

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

[tex]Speed = \frac{Distance}{Time}[/tex]

Substitute 1 mile for distance and 8.92 minutes for time

[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]

Convert Miles to Kilometres

If

[tex]1\ mile = 1609\ m[/tex];

Then

[tex]1\ mile = \frac{1609\ km}{1000}[/tex]

[tex]1\ mile = 1.609\ km[/tex] --- (1)

Convert minutes to hour

[tex]1\ minutes = 0.0167\ hour[/tex]

Multiply both sides by 8.92

[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]

[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)

By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have

[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]

[tex]Speed = 3.304309 \frac{km}{hr}[/tex]

[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated

Answer:

3.0x10⁻²M

Explanation:

Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:

Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻

When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:

ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:

1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

1.4x10⁻⁵ = [2X]² [X]

1.4x10⁻⁵ = 4X³

3.5x10⁻⁶ = X³

0.015 = X

As [Ag⁺] is 2X:

[Ag⁺] = 0.030 = 3.0x10⁻²M

The answer is:

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

Answer:

-A nucleus with a neutron:proton ratio of 1.49

-The nucleus of Sb-123

-A nucleus with a mass of 187 and an atomic number of 75

Answer:

The rate would have doubled

Explanation:

Answer:

6.5 mg/L.

Explanation:

Step one: write out and Balance the chemical reaction in the Question above:

NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.

Step two: Calculate or determine the number of moles of AgCl.

So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:

Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.

Step three: Calculate or determine the number of moles of NiCl2.

Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.

Step four: detemine the mass of NiCl2.

Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.

Step five: finally, determine the concentration of NiCl2.

1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.

Answer:

The correct answer is 206 ml.

Explanation:

Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,  

M₁V₁ = M₂V₂

Now putting the values in the formula we get,  

0.0887 × 69.6 = 0.0224 M × V₂

V₂ = 0.0887 × 69.6 / 0.0224

V₂ = 275.6 ml

Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,  

= 275.6 ml - 69.6 ml = 206 ml

Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]

Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:

[tex]\Delta H=\Delta E+\Delta(PV)[/tex]

So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.

Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.

Analysing each reaction above:

Reaction A has the same number of moles in reagents and products, so  enthalpy change and internal energy change will be equal;

Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.

Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.

Answer:

14.297 g

Explanation:

From the question;

1 mo of the compound requires 1320.0 kJ

From the molar mass;

1 ml of the compound weighs 30.55g

How many grams requires 617.30kJ?

1 ml = 1320

x mol = 617.30

x = 617.30 / 1320

x = 0.468 mol

But 1 mol = 30.55

0.468 mol = x

x = 14.297 g

Answer:

the chemical equation of ethanol as fuel is

C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)

the preparation process of ethanol from cassava is

cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol

Answer:

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

[tex]\frac{2.9g}{3.1g} *100[/tex]

(a) The product in solid state would be the precipitate. Hence, the precipitate would be Ca3(PO4)2

(b) From the balanced equation of the reaction: 3 moles of Ca(NO3)2 is required for 1 mole of Ca3(PO4)2

If there are just 0.030 moles of Ca(NO3)2, then"

3 moles = 1

0.030 moles =    1 x 0.030/3

                         = 0.01 moles of Ca3(PO4)2

In other words, 0.01 moles of the precipitate would be expected to be produced from 0.030 moles of calcium nitrate.

(c) 0.01 moles solid (Ca3(PO4)2) is expected. Mass of Ca3(PO4)2 expected:

      mass   = mole x molar mass

molar mass of Ca3(PO4)2 = 310.18 g/mol

mass of Ca3(PO4)2 expected to be produced = 0.01 x 310.18

                                                                       = 3.1018 g

Hence, 3.1018g of solid is expected to be produced.

(d) Percentage yield = actual yield/theoretical yield x 100

                          = 2.9/3.1018 x 100

                               = 93.5%

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Answer:

6.93

Explanation:

Step 1: Given data

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

Answer:

Mass = 11.688g

Explanation:

Volume = 2.00L

Molar concentration = 0.100M

Mass = ?

These quantities are relatted by the following equation;

Conc = Number of moles / volume

Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 0.2mol * 58.44g/mol

Mass = 11.688g

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.