Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.

Answers

Answer 1

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

Answer 2

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

     

Calculating the pH a) 0 mL        

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

   

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

c) 20 mL  

We can find the pH of the solution from the reaction of equilibrium (3).

 Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

Calculating the pH  

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

  Calculating the concentrations of NH₃ and NH₄⁺

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

Calculating the pH  

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

e) 36 mL  Finding the number of moles of NH₃ and NH₄⁺

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

                                   

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

 

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

Calculating the pH  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl  

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

Calculating the pH  

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

   

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Answer:

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Explanation:

A homolytic fission is said to have occurred when the breakage of a bond between two atoms leaves each of the bonding atoms with equal number of electrons. Homolytic fission often results in the creation of radicals.

Since homolytic fission yields two species with equal number of electrons(usually odd number of electrons), the products of such process can not be charged. They can not be nucleophiles because nucleophiles need to possess two electrons which can be shared with another chemical specie.

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A chemist measures the energy change
ΔH during the following reaction:
2NH3(g)→N2(g)+3H2(g)
ΔH=160kJUse the information to answer the following questions.This reaction is:__________.
a. endothermic
b. exothermic
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a. Yes, absorbed
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Answers

Answer:

For (1): The correct option is (a)

For (2): The correct option is (a) and 333.6 kJ of heat will be absorbed when 70.9 g of ammonia reacts.

Explanation:

There are 2 types of reactions that are classified based on enthalpy change:

Endothermic reactions: These are the reactions where heat is absorbed by the reaction. The change in enthalpy of the reaction, [tex]\Delta H_{rxn}[/tex] is positive for these reactions.

Exothermic reactions: These are the reactions where heat is released by the reaction. The change in enthalpy of the reaction, [tex]\Delta H_{rxn}[/tex] is negative for these reactions.

For (1):

For the given chemical reaction:

[tex]2NH_3(g)\rightarrow N_2(g)+3H_2(g);Delta H=160kJ[/tex]

As the change in enthalpy or heat of the reaction is positive. Thus, the reaction is an endothermic reaction because heat is absorbed by the reaction.

For (2):

When ammonia reacts, some amount of heat will be absorbed by the reaction. Thus, we can say the heat will be absorbed.

The number of moles is calculated by using the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ammonia = 70.9 g

Molar mass of ammonia = 17 g/mol

Using equation 1:

[tex]\text{Moles of ammonia}=\frac{70.9g}{17g/mol}=4.17mol[/tex]

By stoichiometry of the reaction:

If 2 mole of ammonia reacts, the heat absorbed is 160 kJ

So, if 4.17 moles of ammonia reacts, the heat absorbed will be = [tex]\frac{160kJ}{2mol}\times 4.17mol=333.6kJ[/tex]

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Answer:

Explanation:

The density of pure water is 1 gram per 1 milliliter or one cubic cm. By knowing the density of water we can use it in dilution equations or to calculate the specific gravity of other solutions.

It can also help us determine what other substances are made of using the water displacement experiment. This is done by observing how much water is displaced when an object is submerged in the water. As long as you know the density of the water, the mass of the object being submerged and the volume of increase you can calculate the density of the object.

This was done by the great Archimedes in discovering what composed the kings crown.

Senario: 2 years ago, a fruit was smuggled into California on a plane from an exotic, far away land. The homeowner saw that the fruit had maggots and tossed it into the backyard, hoping the seed would grow. The larvae hatched out and moved throughout the area. This fictitious insect will destroy fruit and has the possibility of spreading disease killing the trees. The insect consumes plants in the Prunus species of stone fruits? Look up the plant genus Prunus.

Discussion: The insect has spread over a large area of Southern California, discovered at UC Riverside. What steps would you do to control or eradicate the destructive exotic insect?

PLZ HELP THX WITH COLLEGE LEVEL EXPERICENCE

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Answer:

The best existing methods of control—artemisinin-based drug treatment and insect control with chemical sprays and treated bednets—can reduce the burden of disease substantially, and can even eliminate the disease in some regions,

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Answers

Answer:

1.17 L of H₂

Explanation:

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Answers

Answer:

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Explanation:

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Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...    

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Which of the following lists describes characteristics of an acid? (3 points)
Bitter taste, high pH, and caustic
Sour taste, low pH, and dissolves metals
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Slippery, low pH, and caustic

Answers

Acids have sour taste, low pH and dissolves metals.

The list that describes the characteristics of an acid is that it has sour taste, low pH, and dissolves metals.

Characteristics of an acid

An acid is a chemical substance that has the ability to donate hydrogen ions when involved in a chemical reaction.

The following are the characteristics or features of an acid:

They have a sour taste when tasted.

When measured using a pH scale it is less than 7(low pH).

They react with active metals to yield hydrogen gas.

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Answer:

Scientific method.

Explanation:

Scientific method is the way taken to acquire scientific knowledge. It includes experiments, statistical analysis of existing data, and all kinds of observations of the world around us, while theoretical research is based on deriving certain theories about the world from basic principles, in a mathematical or logical way. The scientific method applies to both types of research, and emphasizes that scientific research is objective, that it can be verified by other scientists, and that knowledge is not acquired without context, but in a way that leads to a greater understanding of previous research and the world. we live in. To contribute to this, researchers are expected to clearly record both their findings and the methods they use to arrive at the results.

A chemist combines 33 g of methane with 289 g of oxygen to from 189 g of carbon dioxide and 30 g of water. Determine if the results of the following word problem adheres to the Law of Conservation of Mass.

Answers

Answer:

The correct answer is - no not adhere to the law of mass conservation.

Explanation:

According to the law of mass conservation in an isolated system, the mass can not be created or destroyed and in a chemical or physical change, the mass of products should be always equal to the mass of reactants.

On the basis of the law the mass of the chemical reaction-

Mass of products = mass of reactants

33 g of methane + 289g of oxygen = 189g of carbon dioxide + 30g of water

322g ≠ 219 g

which means this reaction does not adhere to the law of conservation.

What is the formula of the compound Pentasilicon trioxide ?

Answers

Answer: the molecular formula of  trioxide is ClOClO3 or Cl2O4

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Identify a process that is NOT reversible. Identify a process that is NOT reversible. melting of snow baking of bread deposition of carbon dioxide freezing water melting of aluminum

Answers

Answer:

Identify a process that is NOT reversible.

Melting of snow

baking of bread

deposition of carbon dioxide

freezing water

melting of aluminum

Explanation:

A physical change is the one in which there is a change only in its physical state, color, the appearance of the substance. But the chemical composition of the substance remains unchanged.

It is a temporary change and can be reversed easily.

For example:

melting, freezing, deposition etc.

Baking is a permanent change and the chemical composition of the substance changes.

Hence, among the given options, baking of bread is not a reversible change.

Out of the following all are physical changes except baking of bread and physical changes are reversible so the process  which is not reversible is baking of bread.

What are physical changes?

Physical changes are defined as changes which affect only the form of a substance but not it's chemical composition. They are used to separate mixtures in to chemical components but cannot be used to separate compounds to simpler compounds.

Physical changes are always reversible using physical means and involve a change in the physical properties.Examples of physical changes include melting,boiling , change in texture, size,color,volume and density.Magnetism, crystallization, formation of alloys are all reversible and hence physical changes.

They involve only rearrangement of atoms and are often characterized to be changes which are reversible.

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Which of the following statements correctly explains why bromination reactions are more selective than chlorination reactions.
a. bromine radical is less stable than chlorine radical, so it is more reactive and less choosy
b. bromine radical is more stable than chlorine radical, so it is more reactive and less choosy
c. bromine radical is more stable than chlorine radical, so it is less reactive and more choosy
d. bromine radical is less stable than chlorine radical, so it is less reactive and more choosy
e. relative radical stability is 3' radicals > 2" radicals> 1 radicals when bromine radicals snatch hydrogens from alkanes, but when chlorine radicals snatch hydrogens the resulting alkyl radical stability is 3 radicals < 2 radicals< 1' radicals

Answers

Answer: A bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.

Explanation:

A chlorine atom being more electronegative in nature is able to attract a hydrogen atom more readily towards itself as compared to a bromine atom.

Since bromine is less electronegative in nature so bromine will be more selective as a hydrogen abstracting agent. As a result, bromine radical is more stable in nature than chlorine radical.

Thus, we can conclude that bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.

What is the volume of a flask containing 0.199mol of Cl2at a temperature of 313K and a pressure of 1.19atm

Answers

Answer:

43.0 L

Explanation:

Step 1: Given and required data

Moles of chlorine gas (n): 0.199 molTemperature (T): 313 KPressure (P): 1.19 atmIdeal gas constant (R): 0.0821 atm.L/mol.K

Step 2: Calculate the volume of the flask (V)

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 1.99 mol × (0.0821 atm.L/mol.K) × 313 K / 1.19 atm = 43.0 L

The answer is 4.30 L

determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)

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Notch gfi.hu,ivjmgfcmhtxumgfhkvHbyrdyu,

Which diagram correctly depicts the trend in electronegativity?

a.
b.
c.
d. ​

Answers

Answer: B

Explanation: Reference this electronegativity graph!

The electronegativity increases across the period and decreases down the group. Thus, option B is correct.

Electronegativity can be defined as the tendency of an atom to gain or attract an electron. The electronegativity has been dependent on the size of the atom, as well as the atomic number and valence electrons.

The atom with the requirement of a less number of atoms to complete its octet can easily gain the electron and thereby have high electronegativity. The atomic size also plays a role in the electronegativity of the atom.

The atom with a bigger size has the lesser force of attraction from the nucleus and thus has difficulty attracting the electron, however, the smaller size atom can easily attract the electron with the attraction force from the nucleus.

Thus, the elements with smaller sizes and a high number of valence electrons are more electronegative. In the periodic table, on moving from left to right the valence electrons increase, thus the electronegativity increases.

On moving down the group, the element size increase, thus the electronegativity decreases down the group.

The electronegativity increases across the period and decreases down the group. Thus, option B is correct.

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Suppose a 0.042M aqueous solution of phosphoric acid (H3PO4) is prepared. Calculate the equilibrium molarity of HPO4^−2.

Answers

Answer:

2.89x10⁻⁵M = [HPO₄²⁻]

Explanation:

The equilibrium of H3PO4 in water occurs H2PO4-:

H3PO4(aq) + H2O(l) ⇄ H3O⁺(aq) + H2PO4⁻(aq)

pKa = 2.16. And as pKa = -log Ka; Ka = 10^-2.16

Ka = 6.9183x10⁻³ = [H3O⁺] [H2PO4⁻] / [H3PO4]

As both [H3O⁺] and [H2PO4⁻] comes from the same equilibrium,

[H3O⁺]=[H2PO4⁻] :

[H3O⁺] = X

[H2PO4⁻] = X

[H3PO4] = 0.042 - X

Where X is reaction coordinate

Replacing:

6.9183x10⁻³ = [X] [X] / [0.042 - X]

6.9183x10⁻³ = X² / 0.042 - X

2.905686x10⁻⁴ - 6.9183x10⁻³X - X² = 0

Solving for X:

X = -0.02M. False solution. There is no negative concentration.

X = 0.014M. Right solution

[H2PO4⁻] = 0.014M

In the second equilibrium:

H2PO4⁻(aq) + H2O(l) ⇄ HPO4-(aq) + H3O+(aq)

Based on the same principles of the last equilibrium:

pKa2 = 7.21

Ka2 = 6.166x10⁻⁸ = [HPO4-] [H3O+] / [H2PO4⁻]

[HPO4-] = X

[H3O+] = X

[H2PO4⁻] = 0.014M - X

6.166x10⁻⁸ = X² / [0.014M - X]

8.3623x10⁻¹⁰ - 6.166x10⁻⁸X - X² = 0

Solving for X:

X = -0.0000289485. False solution.

X =

2.89x10⁻⁵M = [HPO₄²⁻]

All light waves can be described in terms of their speed, frequency, and___

Answers

Answer:

all light waves can be described in terms of their speed, frequency and wavelength

Explanation:

Hope it helps u.....

the pressure of a sample of helium in a 0.150 L container is 1520 torr. if the helium is compressed to the volume of 0.012 L without changing the temperature what would be the pressure of the gas

Answers

Answer:

19000 torr

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 0.150 L

Initial pressure (P₁) = 1520 torr

Final volume (V₂) = 0.012 L

Temperature = constant

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

1520 × 0.150 = P₂ × 0.012

228 = P₂ × 0.012

Divide both side by 0.012

P₂ = 228 / 0.012

P₂ = 19000 torr

Thus, the final pressure of the gas is 19000 torr

What is oxygen's half-equation?

Answers

answer; 1/ 20_2[2-] +2e - ->0.

g 250. mL of a solution is created and has a molarity of 1.50 M. What is the molality of this solution if the volume and density of the pure solvent is about the same as the volume and density of the final solution and the density of the pure solvent is 1.15 g/mL

Answers

Answer:

1.30 m

Explanation:

First we calculate the number of moles of solute in the solution, using the definition of molarity:

Molarity = moles / litersConverting 250 mL ⇒ 250 / 1000 = 0.250 Lmoles = 1.50 M * 0.250 L = 0.375 moles

Then we calculate the mass of solvent, using the given volume and density:

1.15 g/mL * 250 mL = 287.5 gConverting 287.5 g to kg ⇒ 287.5 / 1000 = 0.2875 kg

Now we calculate the molality of the solution:

molality = moles of solute / kg of solvent = 0.375 mol / 0.2875 kgmolality = 1.30 m

Draw 2,3-dichloro octane

Answers

Answer:

Hi friend

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The length of a covalent bond depends upon the size of the atoms and the bond order.

a. True
b. False

Answers

Answer:

True

Explanation:

The length of covalent bond depends upon the size of atoms and the bond order.

g in the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your order

Answers

Answer:

3 > 2> 1

Explanation:

Aromatic compounds undergo electrophilic substitution reaction with several electrophiles.

Some substituted benzenes are more reactive towards electrophilic aromatic substitution than unsubstituted benzene.

Certain groups of substituents increase the ease with which an aromatic compound undergoes aromatic substitution.

If we look at the compounds closely, we will notice that only toluene leads to easy reaction with CH3Cl / AlCl3. Thus is due to the +I inductive effect of -CH3 which stabilizes the negatively charged intermediate produced in the reaction.

How many moles of carbon dioxide at STP will fit in a 50 liter container?

Answers

Answer:

n = 2.23 moles

Explanation:

Given the following data;

Standard temperature = 273 K

Standard pressure = 101.325 kPa

Volume = 50 liter

R = 8.314 J/mol·K  

To find the number of moles, we would use the ideal gas law formula;

PV = nRT

Where;

P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.

Making n the subject of formula, we have;

[tex] n = \frac {PV}{RT} [/tex]

Substituting into the formula, we have;

[tex] n = \frac {101.325*50}{8.314*273} [/tex]

[tex] n = \frac {5066.25}{2269.722} [/tex]

n = 2.23 moles

Therefore, 2.23 moles of carbon dioxide at STP will fit in a 50 liter container.

what class of organic compound is formed when cyclopentanone reacts with ethylamine in the presence of trace acid

Answers

The question is incomplete, the complete question is;

What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime

Answer:

imine

Explanation:

An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.

Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.

Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine

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