Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above

Answers

Answer 1

Answer:

Option D. 17.5

Explanation:

Equiibrium is: CO + 2H₂  ⇄  CH₃OH

1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.

Initially we have 0.42 moles of CO and 0.42 moles of H₂

If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.

So in the equilibrium we may have:

0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂

Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen

Finally 0.13 moles of methanol, are found after the equilibrium reach the end.

Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²

0.13 / (0.29 . 0.16²)

Kc = 17.5


Related Questions

Suppose that a certain atom possesses only four distinct energy levels. Assuming that all transitions between levels are possible, how many spectral lines will this atom exhibit

Answers

Answer:

Following are the response to the given question:

Explanation:

The number of shells

n = 4

Calculating the spectral line:

[tex]= \frac{n(n-1)}{2}\\\\ = \frac{4(4-1)}{2} \\\\= \frac{4\times 3}{2}\\\\ = \frac{12}{2}\\\\ = 6[/tex]

why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen​

Answers

Answer:

Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.

Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.  

The nucleophile in these reactions are new and called enols and enolates.

Explanation:

The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.  

Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.  

Aldehyde hydrogens not given Greek leters.  

α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.  

Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.  

The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.  

Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.  

The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.  

The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.

The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.

Aromaticity can also stabilize the enol tautomer over the keto tautomer.

Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.  

Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.

half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________

Answers

Answer:

Cu+(aq)--->Cu2+(aq) + e- : oxidation

reason: there is loss of electrons.

I2(s) + 2e--->2I-(aq) : reduction

reason: There is reduction of electrons.

What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?

Answers

Answer:

The correct approach is "12.25°C".

Explanation:

Given:

Mass of lead,

mc = 245 g

Initial temperature,

tc = 300°C

Mass of Aluminum,

ma = 150 g

Initial temperature,

ta = 12.0°C

Mass of water,

mw = 820 g

Initial temperature,

tw = 12.0°C

Now,

The heat received in equivalent to heat given by copper.

The quantity of heat = [tex]m\times s\times t \ J[/tex]

then,

⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]

⇒             [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]

⇒             [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]

⇒                         [tex]43903.5 = 3582.185 T[/tex]

⇒                                  [tex]T = 12.25^{\circ} C[/tex]

g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr

Answers

Answer:

1.27 × 10⁵ L

Explanation:

Step 1: Given data

Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °C

Step 2: Convert the temperatures to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 21 °C + 273.15 = 294 K

K = -48 °C + 273.15 = 225 K

Step 3: Calculate the final volume of the balloon

We will use the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂/ T₁ × P₂

V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr

V₂ = 1.27 × 10⁵ L

CAN HF USED TO CLEAVE ETHERS EXPLAIN

Answers

Answer:

no

Explanation:

Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.

Write the functional isomers of C2H6O?

Answers

Answer:

See explanation

Explanation:

Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.

The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).

Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.

How to solve this problem step by step

Answers

Answer:

[tex]V_2= 736mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:

[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]

Thus, we solve for the final volume by solving for V2 as follows:

[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]

Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:

[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]

Regards!

How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?

Answers

Answer:

The right answer is "3 g".

Explanation:

Given:

Initial mass substance,

[tex]M_0=24 \ g[/tex]

By using the relation between half lives and amount of substances will be:

⇒ [tex]M=\frac{M_0}{2^n}[/tex]

        [tex]=\frac{24}{2^3}[/tex]

        [tex]=3 \ g[/tex]

Thus, the above is the correct answer.

Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.

Answers

Answer:

Explanation:

The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.

Consider the following equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover.
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g) ΔHrxn = −1790kJ
If a bottle of nail polish remover contains 143 g of acetone, how much heat would be released by its complete combustion? Express your answer to three significant figures.

Answers

Molar mass of Acetone

C3H6O3(12)+6+1658g/mol

Now

1 mol releases -1790KJ heat .

Moles of Acetone:-

143/58=2.5mol

Amount of heat:-

2.5(-1790)=-4475kJ

Chemical reactions can exhibit different rate constants at differing: Select the correct answer below: initial concentrations volumes of container temperatures none of the above

Answers

Explanation:

Chemical reactions can exhibit different rate constants at differing:

i)initial concentrations

ii)volumes of container

iii) temperatures

iv)none of the above.

The rate constant of a reaction is constant at a particular temperature.

It is not depending on the initial concentration of the reactants. It varies with temperature.

Thus, among the given options the correct answer is Temperatures.

Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.

What is a rate constant?

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.

Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.

For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.

The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.

On the other hand, the volume of the container, initial concentration does not affect the rate constant.

Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.

To learn more about rate constant, click here:

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9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain​

Answers

Answer:

See explanation

Explanation:

Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.

When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.

When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.

How many atoms are in .45 moles of P4010

Answers

Answer:

5×6.02×1023

Explanation:

there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010

C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).

Answers

Answer:

Al^3+

Explanation:

Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.

Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.

If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;

Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)

How many grams of sodium nitrate (NaNO3) are needed to
prepare 100 grams of a 15.0 % by mass sodium nitrate
solution?

Answers

Answer:

15.0 g

Explanation:

15.0% =0.150

100.0 g × 0.150= 15.0g

Sodium nitrate is "an inorganic compound with the formula of NaNO₃.

What is an inorganic compound?

Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".

15% = 0.15

100.0 g × 0.15= 15g

Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.

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Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr

Answers

Answer:

I do not speak Spanish.

Explanation:

Which of the following are examples of physical properties of ethanol? Select all that apply.

The boiling point is 78.37°C

It is a clear, colorless liquid

It is flammable

It is a liquid at room temperature

Answers

Ethanol is: flammable, liquid at room temperature, the boiling point is 78.37 ° C.

2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25

Answers

Answer:

5.25 moles of protons. Option e

Explanation:

Reaction between phosphoric acid and sodium hydroxide is neutralization.

We can also say, we have an acid base equilibrium right here:

H₃PO₄  +  3NaOH →  Na₃PO₄  +  3H₂O

Initially we have 5.25 moles of base.

We have data from the acid, to state its moles:

M = mol/L, so mol = M . L

mol = 1.75 moles of acid

If we think in the acid we know:

H₃PO₄  →  3H⁺  +  PO₄⁻³

We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)

If we have 1.75 moles of acid, we may have

(1.75 . 3) /1 = 5.25 moles of protons

These moles will be neutralized by the 5.25 moles of base

H₃O⁺  +  OH⁻  ⇄  2H₂O     Kw

In a titration of a weak acid and a strong base, we have a basic pH

PLEASE HELP ASAP MOLES TO MOLECULES

Answers

Answer:

4.77mol is the correct answer

4.77 mol, is the answer:)

During the postabsorptive state, metabolism adjusts to a catabolic state.

a. True
b. False

Answers

Answer:

The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).

The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.

Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.

So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.

Balance the following chemical equation.

CCl4 -> ___ C+ ___ Cl2

Answers

Answer:

Explanation:

CCl4 => C + 2Cl2

Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).

Answers

Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

Explanation:

Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M

[tex]K_{a} = 1.0 \times 10^{-8}[/tex]

Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of  [tex]Cu(H_{2}O)^{2+}_{6}[/tex]  is as follows.

                               [tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for [tex]K_{a}[/tex] value is as follows.

[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]

Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]

Formula to calculate pH is as follows.

[tex]pH = -log [H^{+}][/tex]

Substitute the values into above formula as follows.

[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]

Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

What is the correct order for the reactions that produce the following transformation. a. (1) H2/Lindlar (2) CH3CO2OH b. (1) H2/Lindlar (2) O3, Zn, HCl c. (1) H2/Pd (2) CH3CO2OH d. (1) Na, NH3 (2) CH3CO2OH

Answers

Answer:

Explanation:

Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)

When perchloric acid (HClO4) reacts with tetraphosphorus decaoxide, phosphoric acid and dichlorine heptaoxide are produced.

a. Trei
b. False

Answers

Answer:

я не знаю ответа :(

Explanation:

Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value.

a. True
b. False

Answers

Answer:

True

Explanation:

When two atoms are at infinite distance from each other, the both atoms posses high energy.

However, as they begin to approach each other, the distance between them gradually decreases and so does their energy.

A point is eventually reached when the potential energy curve reaches its minimum value. The internuclear distance between the two atoms at this point is called the bond length of the system.

A gas mixture, with a total pressure of 300. torr, consists of equal masses of Ne (atomic weight 20.) and Ar (atomic weight 40.). What is the partial pressure of Ar, in torr

Answers

Answer:

The partial pressure will be "100 torr".

Explanation:

Given:

[tex]P_{Ar} = 300 \ torr[/tex]

By assuming Ar and Ne having 50 gm each, we get

mol of Ne = [tex]\frac{50}{20}[/tex]

                = [tex]2.5 \ mol[/tex]

mol of Ar = [tex]\frac{50}{40}[/tex]

               = [tex]1.25 \ mol[/tex]

now,

[tex]n_T= mol.A_r+mol.N_e[/tex]

     [tex]=1.25+2.5[/tex]

     [tex]=3.75[/tex]

then,

[tex]X_{Ar}=\frac{n_{Ar}}{n_T}[/tex]

       [tex]=\frac{1.25}{3.75}[/tex]

       [tex]=0.33[/tex]

hence,

The partial pressure of Ar will be:

⇒ [tex]P_{Ar} = P_T\times X_{AT}[/tex]

By substituting the values, we get

           [tex]=300\times 0.33[/tex]

           [tex]=100 \ torr[/tex]

The partial pressure of Ar in the mixture is 99.9 torr

Let the mass of both gas be 10 g

Next, we shall determine mole of each gas.

For Ne:

Mass = 10 g

Molar mass of Ne = 20 g/mol

Mole of Ne =?

Mole = mass / molar mass

Mole of Ne = 10 / 20

Mole of Ne = 0.5 mole

For Ar:

Mass = 10 g

Molar mass of Ar = 40 g/mol

Mole of Ar =?

Mole = mass / molar mass

Mole of Ar = 10 / 40

Mole of Ar = 0.25 mole

Next, we shall determine the mole fraction of Ar

Mole of Ne = 0.5 mole

Mole of Ar = 0.25 mole

Total mole = 0.5 + 0.25 = 0.75 mole

Mole fraction of Ar =?

[tex]mole \: fraction \: = \frac{mole}{total \: mole} \\ \\ mole \: fraction \: of \:Ar = \frac{0.25}{0.75} \\ \\ mole \: fraction \: of \:Ar = 0.333 \\ \\ [/tex]

Finally, we shall determine the partial pressure of Ar

Mole fraction of Ar = 0.333

Total pressure = 300 torr

Partial pressure of Ar =?

Partial pressure = mole fraction × total pressure

Partial pressure of Ar = 0.333 × 300

Partial pressure of Ar = 99.9 torr

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Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;

1. Capa (0,5 ponto)
2. Índice ou Sumário (0,5 ponto)
3. Texto do trabalho
a) Introdução (1,0 ponto)
b) Objetivos (0,5 ponto)
c) Método (0,5 ponto)
d) Desenvolvimento: Fundamentação Teórica (5,0 pontos)
e) Conclusão (1,0 ponto)
4. Bibliografia (1,0 ponto)

Answers

Answer:

fjskeowkcnekvo Dee five votes come vote for dog even r

DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL 9.10 mL 3. Final volume of the AgNO3 solution 9.10 mL 17.25 mL 4. Volume of AgNO3 solution used 9.10 mL 8.15 mL Line 3 – Line 2 5. Volume of AgNO3 solution in liters _____ L _____ L 6. Molarity of AgNO3 solution 0.01 M 0.01 M (given) 7. Number of moles of AgNO3 ______ mol _____ mol (Line 5 × Line 6) 8. Number of mol of NaCl present in pretzel ______ mol _____ mol (Line 7) number of mol NaCl = number of mol AgNO3 9. Mass of NaCl present in the titrated sample ______ gram _____ gram (Line 8) × 58.5 g/mol

Answers

Answer:

1. 1.00 gm

2. 50 ml

3. 38.93 ml

4. 11.07 ml

5. 0.01107 L

6. 0.010 moles / L

7. 0.0001107 moles

8. 0.0001107 moles

9. 0.00647042 grams

Explanation:

Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.

An endothermic reaction will start when the required
energy is received from the environment or solution.
AH
activation
thermal
kinetic

Answers

Answer:

A: ΔH

Explanation:

Endothermic reactions are this that occur as a result of absorption of heat energy from the surroundings by the reactants to form new products.

Thus, we can say it is one with an increase in enthalpy (ΔH) of the system.

Thus, option A is correct.

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