Draw the products formed when both cis- and trans-but-2-ene are treated with OsO4, followed by hydrolysis with NaHSO3 H2O. Explain how these reactions illustrate that syn dihydroxylation is stereospecific.

Answer:

0.56 mL

Explanation:

Volume = mass ÷ density

Volume = 1.5 ÷ 2.7 g/mL

Volume = 0.5555555556 = 0.56 mL

The volume of the aluminum block is 0.56 mL.

Hope this helps. :)

The volume of aluminum block is 0.556 mL.

The density of a substance is its mass per unit volume.

It given by formula:

[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} [/tex]

Given:

Density = 2.7 g/mL

Mass= 1.5 g

To find:

Volume=?

On substituting the values in the above formula:

[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} \\\\\text{Volume}=\frac{\text{Mass}}{\text{Density}} \\\\\text{Volume}=\frac{1.5}{2.7} \\\\\text{Volume}=0.556mL[/tex]

Thus, the volume of the aluminum block is 0.556mL.

Learn more:

brainly.com/question/898827

Answer:

C2H4O3

Explanation:

We would have to do some preparations between before solving it the normal way. The main goal is to get the masses of the Individual elements. So here goes;

We can get the mass of C from CO2 using the following steps:

1 mole of CO2 has a mass of 44g (Molar mass) and contains 12g of C.

How did we know the molar mass of CO2 is 44g?

Easy. 1 mole of C = 12, 1 mole of O = 16

But we have two O’s so the total mass of O = (2 * 6) = 32

Total mass of CO2 = mass of C + Mass of O = 12 + 32 = 44

So if 44g of CO2 contains 12g of C, how much of C would be present in 35.21g CO2.

12 = 44

X = 35.21

X = (35.21 * 12) / 44 = 9.603g

We can also get the mass of H from H2O. 1 mole of H2O has a mass of 18g and contains 2g of H.

How did we know the molar mass of H2O is 18g?

Easy. 1 mole of H = 1, 1 mole of O = 16

But we have two H’s so the total mass of H = (2 * 1) = 2

Total mass of H2O = mass of H + Mass of O = 2 + 16 = 18

So how much of H would be present in 14.42g of H2O?

2 = 18

X =14.42

X = (14.42 * 2 ) / 18 = 1.602g

Now we have the masses of C and H. But the question says the compound contains the C, H and O.

So we still have to calculate the mass of Oxygen. We obtain this from;

Mass of Compound = Mass of Carbon + Mass of Oxygen + Mass of Hydrogen

Mass of Oxygen = Mass of compound – (Mass of Carbon + Mass of Hydrogen)

Mass of Oxygen = 30.42 – (9.603 + 1.602)

Mass of Oxygen = 30.42 - 11.205  = 19.215

Now we have all the masses so we are good too go. Let’s have our table.

Elements Carbon (C) Hydrogen (H) Oxygen (O)

Mass        9.603             1.602           19.215

                0.800            1.602           1.2001 (Divide by molar mass)

                1                    2                   1.5       (Divide by lowest number)

                2                    4                   3        (Convert to simple integers by * 2)

The Empirical formula of the compound is C2H4O3

Explanation:

a)element=argon

number of electrons=18

it's an inert gas,thus it is fully filled

b)element=cobalt

number of electrons=27

valence electron=1

c).element=nitrogen

number of electrons=7

valence electrons=3

d)element=stanium

number of electrons=50

valence electrons=2

Answer: The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol

Explanation:

The balanced reaction for combustion of isooctane is:

[tex]C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]

We are given:

[tex]\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_{C_8H_{18}(l)}=-258.07kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(8\times (-393.51))+(9\times (-285.8))]-[(1\times (-258.07))+(\frac{25}{2}\times (0))]\\\Delta H^o_{rxn}=-5462.2kJ/mol[/tex]

The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol

Answer:

A

Explanation:

The group that has one valence electron is the first group, also known as the alkali metals.

Answer:

The  percentage is    % [tex]Fe^{2+[/tex]   [tex]= 57.3[/tex]%

Explanation:

From the question we are told that  

       The volume of blood in the human body  is  [tex]V = 5.0 0 \ L[/tex]

       The  concentration of  [tex]Fe^{2+[/tex] is  [tex]C_{F} = 1.10 *10^{-5} \ M[/tex]

        The volume  of  NaCN  ingested is  [tex]V_N = 9.00 \ mL = 9.00 *10^{-3} \ L[/tex]

       The concentration of  NaCN ingested is  [tex]C_N = 21.0 \ mM = 21.0 *10^{-3} \ M[/tex]

The  number of moles of   [tex]Fe^{2+[/tex] in the blood is  

                    [tex]N_F = C_F * V[/tex]

substituting values  

                    [tex]N_F = 1.10 *10^{-5} * 5[/tex]

                    [tex]N_F = 5.5*10^{-5} \ mols[/tex]

The  number of moles of  [tex]CN^{-}[/tex] ingested is  mathematically evaluated as

           [tex]N_C = C_N * V_N[/tex]

substituting values    

          [tex]N_C = 21*10^{-3} * 9 *10^{-3}[/tex]

          [tex]N_C = 1.89 *10^{-4} \ mols[/tex]

The balanced chemical equation for the reaction  between   [tex]Fe^{2+[/tex] and   [tex]CN^{-}[/tex]  is  represented as

          [tex]Fe^{2+} + 6 CN^{-} \to [Fe(CN)_6]^{2-}[/tex]

From this  reaction we see that  

         1 mole  of    [tex]Fe^{2+[/tex]  will react with 6  moles of  [tex]CN^{-}[/tex]

=>         x  moles of  [tex]Fe^{2+[/tex] will react with   [tex]1.89 *10^{-4} \ moles[/tex] of  [tex]CN^{-}[/tex]

Thus  

         [tex]x = \frac{1.89 *10^{-4} * 1}{6}[/tex]

        [tex]x = 3.15 *10^{-5}[/tex]

Hence the percentage  of  [tex]Fe^{2+[/tex]  that reacted is  mathematically evaluated as

       %  [tex]Fe^{2+[/tex]   [tex]= \frac{3.15 *10^{-5}}{5.5*10^{-5}} * 100[/tex]

        %  [tex]Fe^{2+[/tex]   [tex]= 57.3[/tex]%

Answer:

Three atoms are attached to the central atom in NF3.

Explanation:

The central atom is always regarded as the atom having the least electronegativity in a molecule or ion. We can decide on what atom should be the central atom by comparing the relative electro negativities of the atoms in the molecule or ion.

If we consider NF3, we can easily see that nitrogen is less electronegative than fluorine, hence nitrogen is the central atom in the molecule. We can also observe from the molecular model that three atoms of fluorine were attached to the central atom. Hence there are three atoms attached to the central atom in the molecule NF3.

Answer:

E. HCl

Explanation:

Cl atom does not have enough electronegativity to make enough positive charge on H.

HCl is the compound which doesn't have hydrogen bonds. This is because of

the higher size of the chlorine atom.

There is no hydrogen bond because of the high size of the chlorine.

Chlorine have electrons with a very low density. It is also very

electronegative which explains why the formation of hydrogen bonds in the

compound HCl is not possible.

Instead, HCl has covalent bonds in which electron is shared between the

hydrogen and  chlorine to achieve a stable configuration.

Read more on https://brainly.com/question/10462894

Answer:

The correct answer is (d) 24.8 Torr

Explanation:

When a solute is added to a solvent, the water pressure of the solution is lower than the vapor pressure of the pure solvent. This is called vapor pressure lowering and it is given by the following expression:

Psolution= Xsolvent x Pºsolvent

We have to calculate Xsolvent (mole fraction of solvent) which is given by the number of moles of solute divided into the total number of moles.

First, we calculate the number of moles of solute and solvent. The solute is glucose (C₆H₁₂O₆), and its number of moles is calculated from the mass and the molecular weight (MM):

MM (C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

moles of glucose= mass/MM= (72.9 g)/)(180 g/mol)= 0.405 moles

The solvent is water (H₂O) and again we calculate the number of moles as follows:

MM(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol

moles of water= mass/MM= (115 g)/(18 g/mol)= 6.389 moles

Now, we calculate the total number of moles (nt):

nt= moles of glucose + moles of water= 0.405 moles + 6.389 moles= 6.794 moles

The mole fraction of water (Xsolvent) is given by:

Xsolvent= moles of water/nt= 6.389 moles/6.794 moles= 0.940

Finally, the vapor pressure of water over the solution will be the following:

Psolvent= Xsolvent x Pºsolvent= 0.940 x 26.4 Torr= 24.8 Torr

Answer:

a. Acid B

b. Acid B

c. lower

Hope this helps you

Answer:

4.50 × 10⁻⁴ mol L⁻¹ s⁻¹

Explanation:

Step 1: Write the balanced equation

2 NO(g) + Cl₂(g) → 2 NOCl(g)

Step 2: Establish the appropriate molar ratio

The molar ratio of NO(g) to NOCl(g) is 2:2, that is, when 2 moles of NO(g) are consumed, 2 moles of NOCl(g) are formed.

Step 3: Calculate the rate of consumption of NO(g)

The rate of formation of NOCl(g) is 4.50 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of consumption of NO(g) is:

[tex]\frac{4.50 \times 10^{-4}molNOCl}{L.s} \times \frac{2molNO}{2molNOCl} = \frac{4.50 \times 10^{-4}molNO}{L.s}[/tex]

Answer:

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Explanation:

When Na₂A is in equilibrium with water, the reactions that occurs are:

2Na⁺ + A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + 2Na⁺(aq) + OH⁻(aq)

As sodium ion doesn't react:

A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)

Kb1 = KwₓKa2 = 1x10⁻¹⁴/ 6.19x10⁻⁹ = 1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

And HA⁻ will be in equilibrium:

HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)

Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ = 1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

In the reaction, you have 2 equilibriums, for the first reaction, concentrations in equilibrium are:

[HA⁻] = X

[OH⁻] = X

[A²⁻] = 0.236M - X

Replacing in Kb1:

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

1.6155x10⁻⁶ = [X] [X] / [0.236-X]

3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²

3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0

Solving for X

X = -0.00195 → False solution. There is no negative concentrations

X = 0.001952.

Replacing, concentrations for the first equilibrium are:

[HA⁻] = 0.001952

[OH⁻] = 0.001952

[A²⁻] = 0.234

Now, in the second equilibrium:

[HA⁻] = 0.001952 - X

[OH⁻] = X

[H₂A] = X

Replacing in Kb1:

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0

Solving for X

X = -5.04x10⁻⁷ → False solution. There is no negative concentrations

X = 5.0409x10⁻⁷

Replacing, concentrations for the second equilibrium are:

[HA⁻] = 0.001951M

[OH⁻] = 5.0409x10⁻⁷M

[H₂A] =  5.0409x10⁻⁷M

Thus, you have concentrations of H2A, HA−, and A2−

Now, for pH, the sum of both productions of [OH⁻] is:

[OH⁻] = 0.0019525

pOH = -log[OH⁻] = 2.709

As 14 = pH+ pOH

11.29 = pH

As per the question the pH and the cons of the H2A, the HA−, and the A2−, at equilibrium for a 0.236 M.

Learn more about the A2−, at equilibrium.

brainly.com/question/17086012.

Answer:

a) Conjugate base F– b) Conjugate acid K+ c) Conjugate acid NH4+ d) Conjugate base NO2- e) Conjugate base HCOO– f) Conjugate acid CH4+

Explanation:

Acid will produce Conjugate base

Base will produce Conjugate acid.

Answer:

a. The acid HF: F-

b. The base KOH: H2O

c. The base NH3: NH4+

d. The acid HNO3: NO3-

e. The acid HCOOH: COOH-

f. The base CH3NH2: CH3NH3+

Explanation:

Answer:

B - The terrestrial planets (except Earth) and the gas giants are not habitable.

~ .

Based on the nature of the planets in the solar system, the statement which accurately describes the habitability of the planets is; the terrestrial planets (except Earth) and the gas giants are not habitable.

The planets refers to large massive bodies which revolve in orbits around a star.

The solar system consists of the sun as the star and the planets revolving around it.

Of the planets revolving around the sun, only the earth is habitable.

Habitability refers to the ability of life to found in a particular place.

Therefore, the statement which accurately describes the habitability of the planets is; the terrestrial planets (except Earth) and the gas giants are not habitable.

Learn more about the planets at: https://brainly.com/question/2513966

Answer:crossing over

Explanation:

Answer:

Crossing Over and Independent Assortment.

Explanation:

Crossing over: In Prophase I of Meiosis I, homologous chromosomes line up their chromatids and "cross-over", or exchange corresponding segments of DNA with each other. This produces genetic variation by allowing more combinations of genes to be produced.

Independent Assortment: In Anaphase I of Meiosis I, homologues separate and move to opposite sides of the cell. Resulting cells have one chromosome from each pair of homologous chromosomes. However, WHICH chromosome that each cell gets is completely random.

Answer:

114 ppm

Explanation:

Data obtained from the question include:

Conc. of compound in mol/L = 1.5×10¯⁴ mol/L

Molar mass of compound = 760 g/mol

Conc. in ppm =..?

Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:

Conc. in mol/L = conc. in g/L / Molar mass

1.5×10¯⁴ = conc. In g/L / 760

Cross multiply

Conc. in g/L = 1.5×10¯⁴ x 760

Conc. in g/L = 0.114 g/L

Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:

1 g/L = 1000 mg/L

Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L

Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:

1 mg/L = 1 ppm

Therefore, 114 mg/L = 114 ppm

From the calculations made above,

1.5×10¯⁴ mol/L Is equivalent to 114 ppm.

Answer:

D. exothermic reaction

Explanation:

In an exothermic reaction, the reactants are at a higher energy level than the products.

Answer:

Initial concentration = 2.30 M

Explanation:

**********

Full Question;

For this question, consider the reaction A-->B with a rate constant k=7.17x10^-4 m/s.

**********

From the units of the rate constant, we can tell that this is a Zero order reaction.

k = 7.17 x 10^-4 m/s

t = 15 minutes = 15 * 60 = 600s (Converting to seconds)

concentration of A to decrease by 28%.

[A] = Final Concentration

[A]o = Initial Concentration

[A] = [A]o - ( [A]o * 28 ) / 100

[A] = [A]o - 0.28[A]o

[A] = 0.72 [A]o

For a zero order reaction, we have;

[A] = [A]o -kt

Substituting the new value of [A], we have;

0.72 [A]o = [A]o - (7.17 x 10^-4 * 900)

0.28 [A]o = (7.17 x 10^-4 * 900)

[A]o = 2.30 M

Answer:

Ksp = 7.4x10⁻⁷

Explanation:

Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.

Ksp of Zn(OH)₂ is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

Ksp = [Zn²⁺] [OH⁻]²

And the molar solubility, X, is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

                 ⇄ X + 2X

Because X are moles of substance dissolved.

Ksp = [X] [2X]²

Ksp = 4X³

As molar solubility, X, is 5.7x10⁻³mol/L:

Ksp = 4X³

Ksp = 4 (5.7x10⁻³mol/L)³

Explanation:

Half life = 12.3years

Time =  30 years

Basically half life is the amount of time taken for the intial concentration to be reduced to half.

First half life = 12.3 years = 10/2 = 5 Kg left

Second half life = 24.6  years= 5/2 = 2.5 Kg left

Third half life = 36.9 years = 2.5 / 2 = 1.25 Kg left

This means that after 30 years, the amount of tritium left would e betweem 1.25kg to 2.5 kg.

The number of tritium that will be left after 30 years is  1.844.

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance Also the number of years should be 30 years

So,

= 10*2^(-30/12.3)

= 1.844

Therefore, we can conclude that The number of tritium that will be left after 30 years is  1.844.

Learn more about substance here: https://brainly.com/question/24840674

Answer:

81.0°C

Explanation:

Kb benzene = 2.5°C/m

The addition of a solute to a pure solvent produce an elevation in boiling point regarding to boiling point of pure solvent. This phenomenon follows the equation:

ΔT = Kb×m×i

where ΔT represents the increasing in boiling point, Kb is the elevation boiling point constant of the solvent (2.5°C/m for benzene), m is molality of solution (Moles solute / kg solvent) and i is Van't Hoff factor (1 for a non-electrolyte solute as naphthalene).

Moles of 11.5g of naphthalene (Molar mass: 128.17g/mol) are:

11.5g × (1mol / 128.17g) = 0.0897 moles of naphthalene in 250.0g = 0.250kg of benzene.

Molality is:

0.0897 moles of naphthalene / 0.250kg of benzene = 0.359m

Replacing in the equation:

ΔT = Kb×m×i

ΔT = 2.5°C/m×0.359m×1

ΔT = 0.90°C

That means the solution prepared has an elevation in boiling point of 0.90°C. As boiling point of pure benzene is 80.10°C, boiling point of the solution is:

80.10°C + 0.90°C =

Answer:

9.308

Explanation:

The computation of the pH of the given solution is shown below:

But before we need to determine the HI molarity which is

[tex]Molarity\ of\ HI = \frac{moles}{volume \ in\ L}[/tex]

[tex]= \frac{4.63\times10^{-2}}{0.250}[/tex]

= 0.1852  M

Now

As we know that

[tex]NH_3 + HI = NH_4I[/tex]

So,

[tex]NH_4I = HI = 0.1852 M[/tex]

Now the molarity of [tex]NH_3[/tex] left is

= 0.397 - 0.1852

= 0.2118

[tex]pOH = pKb + log (\frac{NH_4I}{NH_3})[/tex]

[tex]= 4.75 + log(\frac{0.1852}{0.2118})[/tex]

= 4.692

Now as we know that

pH = 14 - pOH

= 14 - pOH

= 14 - 4.692

= 9.308

We simply applied the above equations

Answer: 8368 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed or released =?

c = specific heat capacity of water = [tex]4.184J/g^0C[/tex]

Initial temperature of water = [tex]T_i[/tex] = [tex]20^0C[/tex]

Final temperature of water = [tex]T_f[/tex]  = [tex]60^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(60-20)^0C=40^0C[/tex]

Putting in the values, we get:

[tex]Q=50.0g\times 4.184J/g^0C\times 40^0C=8368J[/tex]

Thus energy in Joules required is 8368.

Answer:

1.563 moles of SO3.

Explanation:

We begin by calculating the number of mole present in 100g of sulphur dioxide, SO2. This can be obtained as follow:

Molar mass of SO2 = 32 + (16x2) = 64g/mol

Mass of SO2 = 100g

Mole of SO2 =..?

Mole = mass/Molar mass

Mole of SO2 = 100/64

Mole of SO2 = 1.563 mole

Now, we can obtain the number of mole of sulphur trioxide, SO3 produce from the reaction as follow:

2SO2 + O2 —> 2SO3

From the balanced equation above,

2 moles of SO2 reacted to produce 2 moles of SO3.

Therefore, 1.563 moles of SO2 will also react to produce 1.563 moles of SO3.

Therefore, 1.563 moles of SO3 is obtained from the reaction.

Answer: The concentration of [tex]CaSO_4[/tex]  in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]

The equation for the ionization of [tex]CaSO_4[/tex]  is given as:

[tex]K_{sp}[/tex] of [tex]CaSO_4[/tex]  = [tex]9.0\times 10^{-6}[/tex]

By stoichiometry of the reaction:

1 mole of  [tex]CaSO_4[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 1 mole of [tex]SO_4^{2-}[/tex]

When the solubility of [tex]CaSO_4[/tex] is S moles/liter, then the solubility of [tex]Ca^{2+}[/tex] will be S moles\liter and solubility of [tex]SO_4^{2-}[/tex] will be S moles/liter.

[tex]K_{sp}=[Ca^{2+}][SO_4^{2-}][/tex]

[tex]9.0\times 10^{-6}=[s][s][/tex]

[tex]9.0\times 10^{-6}=s^2[/tex]

[tex]s=3.0\times 10^{-3}M[/tex]

Thus concentration of [tex]CaSO_4[/tex]  in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]

Answer:

A one-step mechanism involving a transition state that has a carbon partially bonded to both chlorine and oxygen

Explanation:

The compound CH3Cl is methyl chloride. This is a nucleophilic substitution reaction that proceeds by an SN2 mechanism. The SN2 mechanism is a concerted reaction mechanism. This means that the departure of the leaving group is assisted by the incoming nucleophile. The both species are partially bonded to opposite sides of the carbon atom in the transition state.

Recall that an SN2 reaction is driven by the attraction between the negative charge of the nucleophile (OH^-) and the positive charge of the electrophile (the partial positive charge on the carbon atom bearing the chlorine leaving group).

Answer:

[tex]r_{N_2}=-0.1725M/s[/tex]

Explanation:

Hello,

In this case, by means of the law of mass action, we firstly write the described chemical reaction:

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

Thus, as ammonia is being formed at 0.345 M/s, nitrogen will be disappearing at (consider law of mass action):

[tex]r_{NH_3}=0.345M/s\\\\\frac{1}{-1} r_{N_2}=\frac{1}{-3}r_{H_2}=\frac{1}{2} r_{NH_3}\\\\r_{N_2}=-\frac{1}{2} r_{NH_3}=-\frac{1}{2} *0.345M/s\\\\r_{N_2}=-0.1725M/s[/tex]

Best regards.

Answer:

0.901 M

Explanation:

The concentration of the CO(g) in the equilibrated mixture is shown below:

[tex]K_C = \frac{H_2\times CO_2}{CO\times H_2O}[/tex]

where,

[tex]K_C[/tex] = equilibrium constant

And, we placing these above values to the formula shown as above

So, the concentration of CO(g) is

[tex]= \frac{3.33 \times 0.134}{0.66\times 0.750}[/tex]

= [tex]\frac{0.44622}{0.495}[/tex]

= 0.901 M

We simply applied the above formula in order to determine it and so that the  correct answer could arrive

Answer:

[tex]V_{CO_2}=16.0mL[/tex]

Explanation:

Hello,

In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:

[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]

We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:

[tex]n_{CO}V_{CO}=n_{CO_2}V_{CO_2}[/tex]

Thus, solving for the yielded volume of carbon dioxide we obtain:

[tex]V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL[/tex]

Best regards.