The reaction: A + 3 B → D + F was studied and the following mechanism was determined. A + B C (fast) C + B → D + E (slow) E + B → F (very fast) The species, C, is properly described as

Answer:

pH= 12

Explanation:

Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:

KOH⇒ K⁺ + OH⁻

Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:

[OH⁻]= [KOH]

In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):

Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol

moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles

The molality of the solution is the number of moles of KOH per liter of solution:

V= 250.0 ml x 1 L/1000 ml= 0.250 L

M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M

Now, we calculate pOH:

pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2

Finally, we calculate pH from pOH:

pH + pOH = 14

⇒pH = 14 - pOH= 14 -2 = 12

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Answer:

Rubidium oxide

Explanation:

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M

Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M

GIven that :

Ka = [tex]1.9 x 10^{-5}[/tex]

Thus; it's pKa = 4.72

[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]

[tex]pH =4.72 + log(1.1376)[/tex]

[tex]pH =4.72 + 0.05598[/tex]

[tex]pH =4.77598[/tex]

pH ≅ 4.80

The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.

pH of the given solution will be used by using the following equation:

pH = pKa + log[conjugate base] / [weak acid]

Given chemical reaction will be represented as:

HN₃ + OH⁻ → N₃⁻ + H₂O

Moles will be calculated as:

n = M×V, where

M = molarity

V = volume

Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375  mol

Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol

From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:

0.00375 - 0.001995 = 0.001755 mol

And 0.001995 mol of N₃⁻ is preduced by the reaction.

Total volume of the solution = 0.025 +  0.0133 = 0.0383  L

To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:

[N₃⁻] = 0.001995 mol / 0.0383  L = 0.0521 M

[HN₃] = 0.001755 mol / 0.0383  L = 0.0458 M

Ka for HN₃ = 1.9 × 10⁻⁵

pKa = -log( 1.9 × 10⁻⁵ ) = 4.72

On putting all these values on the above equation, we get

pH = 4.72 + log (0.0521) / (0.0458)

pH = 4.80

Hence, pH of the solution is 4.80.

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Answer:

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The complete ionic equation includes all the ions and molecular species.

C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Answer:

Coals

Explanation:

The land feature that supports the theory of continental drift is ; ( C ) coal fields

Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.

Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.

Hence we can conclude that the land feature that supports the theory of continental drift is coal fields

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Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]

[tex]M=200g/mol[/tex]

Thus the molar mass of the unknown compound is 200 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × molality

The molality can be defined as the moles of solute per kg solvent

Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant ×  [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)

Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12

[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]

0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]

Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol

Molecular mass of solute = 199.626 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

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Answer:

The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.

Explanation:

The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.

A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.

Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.

SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.

Answer:

The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ

Explanation:

The reaction whose standard enthalpy is required, as obtained from the internet is

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

The formation reaction for some of the reactants and products are given as

C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ

To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction

ΔH (CO₂(g)) = -393.5 kJ

ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)

ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)

ΔH (CO(g)) = ? (This isn't given)

But it can be calculated from the second given reaction

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

Heat of reaction = ΔH(products) - ΔH(reactants)

Heat of reaction = -566.0 kJ

ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ

ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))

Hence, we have

-566 = -787 - [2 × ΔH (CO₂(g))]

2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ

ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Heat of reaction = ΔH(products) - ΔH(reactants)

ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]

= -393.5 + 0 = -393.5 kJ

ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ

Heat of reaction = -393.5 - (-352.3) = -41.2 kJ

Hope this Helps!!!

Answer:

Explanation:

(R)-5-octyne-4-ol

C8H140

Correct name for 5-octyne is 3-octane.

Structure is attached below.

Octane is a hydrocarbon and an alkane with the chemical formula C₈H₁₈, and the condensed structural formula CH₃(CH₂)₆CH₃. Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.

The structure for 5-octyne is given below.

Correct name for it is 3-octane.

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Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Answer:

[Ar] 4s²

Explanation:

Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.

The full electronic configuration for calcium is given as;

1s²2s²2p⁶3s²3p⁶4s²

The condensed electronic configuration is given as;

[Ar] 4s²

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

Answer:

Kb = 1.77x10⁻⁵

Explanation:

When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

And the dissociation constant, Kb, for this equilibrium is:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

To find Kb you need to find the concentration of each species. The equilibrium concentrations are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate.

You can know [OH⁻] and, therefore, X, with pH of the solution, thus:

pH = -log [H⁺] = 11.612

[H⁺] = 2.4434x10⁻¹²

As 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]

4.0926x10⁻³ = [OH⁻] = X

Replacing, concentrations of the species are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

[NH₃] = 0.9459M

[NH₄⁺] = 4.0926x10⁻³M

[OH⁻] = 4.0926x10⁻³M

Replacing in Kb expression:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]

The branch of science which deals with the chemical bond is called Chemistry.

The correct answer to the question is  [tex]Kb = 1.77*10^{-5[/tex]

Explanation:

When NH₃, is acts as a  weak base it forms an equilibrium with water the reaction occurs is:

[tex]NH_3(aq) + H_2O(l) ---><NH_4^+(aq) + OH^-(aq)[/tex]

The formula we gonna use is as follows:-

[tex]Kb = \frac{[NH_4^+] [OH^-]}{[NH_3]}[/tex]

The data is given in the question is as follows:-

Where X stands for reaction coordinate.

After solving the ph of the compound the value is as follows:-

Putting the value in the formula.

[tex]Kb = \frac{[4.0926*10^{-3}M] [4.0926*10^{-3}M]}{[0.9459M]}[/tex]

After solving the equation the value of Kb is [tex]1.77*10^{-5[/tex]

Hence, the correct answer is [tex]1.77*10^{-5[/tex]

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Answer:

Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹

Explanation:

Argon is a monoatomic gas that behaves as an ideal gas at 298K.

Using the first law of thermodinamics you can obtain:

Work, Q, for constant pressure molar heat capacity,CP:

CP = (5/2)R

For constant-volume molar heat capacity,CV:

CV = (3/2)R

That means:

2CP/5 = 2CV/3

3/5 = CV / CP

As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:

3/5 = CV / CP

3/5 = CV / 20.79 J K ⁻¹mol⁻¹

12.47 J K ⁻¹mol⁻¹ = CV

Answer:

To the left.

Explanation:

Step 1: Write the balanced reaction at equilibrium

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

Step 2: Calculate the reaction quotient (Qc)

Qc = [SO₃]² / [SO₂]² × [O₂]

Qc = 0.254² / 0.500² × 0.00855

Qc = 30.2

Step 3: Determine in which direction will proceed the system

Since Qc > Kc, the system will shift to the left to attain the equilibrium.

Answer:

Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred

Explanation:

Answer:

Explanation:

1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂

In this reaction , oxidation number of Cl decreases  from 0 to -1  so it is reduced  and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .

2 )

Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄

In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.

3 )

CaCO₃ = CaO + CO₂

In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.

So only first reaction is oxidation - reduction reaction.

2nd option is correct.

The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.

It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.

Hence, the correct option is b.

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Explanation:

Before proceeding we have to understand what electrolytes and non electrolytes are;

An electrolyte is a substance that produces an electrically conducting solution when dissolved. An electrolyte is a compound that can dissociate into ions.

Non electrolytes: A substance whose molecules in solution do not dissociate to ions and thus do not conduct an electric current

Going through the options;

1. LiOH

This is a compound of hat would dissociate into Li+ and OH-.  This is an electrolyte.

2. C4H2O4

This is an organic compound. Gnerally organic acids are non electrolytes, with the exception og the acids. This is a nonelectrolyte.

3. LiBr

This is an electrolyte because it would dissociate into Li+ and Br-  ions.

4. HNO3

HNO3 is a strong acid. Because it is a strong acid it will dissociate completely into its ions (H+ and NO3-). Therefore we consider HNO3 to be a strong electrolyte.

Answer:

because it is

Explanation:

In the reaction above, we clearly see that a chemical reaction took place but what kind of reaction you might ask?

Looks like it is Thermal decomposition of copper carbonate because decomposition reaction takes place due to the added heat. Decomposition in simple terms in chemical breakdown and here this result is obtained by adding heat.

1) What happened to the lime water?

Although not pictured above, but my assumption is that lime water turned milky or turbid because when CO2 comes in presence of limewater they react to form a percipitate of Calcium carbonate which is the milky color that you get.

2) What does this prove?

- My understanding would be that it proves that CO2 was formed, and that most metal carbonates undergo thermal decomposition into metal oxide and carbon dioxide, and also that a reaction took place since new products were made.

Answer:

RbOH

Explanation:

For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:

[tex]AOH~->~A^+~+~OH^-[/tex]

With this in mind we can write the reaction for each substance:

[tex]HCOOH~->~HCOO^-~+~H^+[/tex]

[tex]RbOH~->~Rb^+~+~OH^-[/tex]

[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]

[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]

The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.

I hope it helps!

RbOH or rubidium hydroxide is a base.  

• RbOH or rubidium hydroxide is the inorganic compound.  

• It comprises hydroxide anions and an equal number of rubidium cations.  

• Rubidium hydroxide, like other strong bases is highly corrosive.  

• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.  

• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.  

• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.  

Thus, RbOH is a base.

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Answer:

...

Explanation:

in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together

saturated- solid at room temperature

unsaturated= liquid at room temperature

two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats

Answer:

I think we will know a lot more about the universe and the things around us. We may also know a lot more about other planets, such as Mars and Saturn, and we might also know a lot more about other stars in the universe.

Explanation:

Answer:

CaF2 + CO3- ----> CaCO3 + 2 F-

Explanation:

The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.

Calcium carbonate is a quaternary salt

Answer:

please look at the picture below.

Explanation:

Answer:

See explanation

Explanation:

For this question, we have to remember the definition of a meso-compound. In a meso-compound, we will have chiral carbons but we don't optical activity. This is due to the symmetry, if we have symmetry in a substance with chiral carbons the optical activity is nullified. So, if we want to find the meso-compounds we have to find symmetry planes in the molecule.

A symmetry plane is an imaginary cut that can divide the molecule in two equal parts. We have to draw the molecule first (see figure 1) and then we can try to find the symmetry planes.

With this in mind, the only compounds with symmetry planes are:

b) cis-1,3-dimethylcyclohexane

d) cis-1,4-dimethylcyclohexane

See figure 2 to more explanations

I hope it helps!

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.