Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.

Answers

Answer 1

Answer:

1.94 × 10⁻³

Explanation:

Step 1: Calculate the concentration of H⁺ ions

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M

Step 2: Calculate the acid dissociation constant (Ka) of the acid

For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.

Ka = [H⁺]²/Ca

Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³


Related Questions

Determine the equilibrium constant, Keq, at 25°C for the reaction
2Br- (aq) + I2(s) <--> Br2(l) + 2I- (aq)



Eocell = (0.0257/n) lnKeq, Calculate Eocell from Use this equation to calculate K value.

Eo (I2/I-) = +0.53, Eo (Br2/Br-) = +1.07,

Answers

Explanation:

The given chemical reaction is:

[tex]2Br^- (aq) + I_2(s) <-> Br_2(l) + 2I^- (aq)[/tex]

[tex]E^ocell=oxidation potential of anode + reduction potential of cathode\\[/tex]

The relation between Eo cell and Keq is shown below:

[tex]deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell[/tex]

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

[tex]E^ocell= (-1.07+0.53)V\\=-0.54V[/tex]

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

[tex]ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3[/tex]

Answer:

Keq=6.13x10^33

Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.

What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase

Answers

Casease! Good luck!

how many of the electrons in a molecule of ethane are not involved in bondind​

Answers

Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons

#LETS STUDY

#BRAINLEST LOVE❣️

any two functions of crystals

Answers

Answer:

1. Participating in calcium homeostatis storage of calcium.

2. High capacity calcium (Ca) regulation and protection against herbivory

[tex]\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } [/tex]

_________________

[tex] \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .[/tex]

__________________

[tex]\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral  \: crystals \: bone  \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.[/tex]

The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.

Answers

Answer:

Explanation:

this guy on brainly already did it:

Alleei

Virtuoso

4.8K answers

37.1M people helped

Answer :  The work done for path A and path B is -685.3 J and -478.1 J  respectively.

Explanation :

To calculate the work done for path A :

First we have to calculate the moles of the gas.

where,

= initial pressure of gas  = 2.57 atm

= initial volume of gas  = 3.42 L

n = moles of gas  = ?

R = gas constant = 0.0821 atm.L/mol.K

T = temperature of gas  = 298 K

Now put all the given values in the above formula, we get:

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

where,

w = work done on the system = ?

n = number of moles of gas  = 0.359 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 298 K

= initial volume of gas  = 3.42 L

= final volume of gas  = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path A is, -685.3 J

To calculate the work done for path B :

The formula used for isothermally irreversible expansion is :

where,

w = work done

= external pressure = 1.19 atm

= initial volume of gas = 3.42 L

= final volume of gas = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path B is, -478.1 J

The compound IF5 contains Question 16 options: polar covalent bonds with partial negative charges on the F atoms. ionic bonds. polar covalent bonds with partial negative charges on the I atoms. nonpolar covalent bonds.

Answers

Answer:

See explanation

Explanation:

The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.

The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.

Consider the following reaction:

CO(g)+2H2(g)⇌CH3OH(g)

A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.

Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answers

Answer:

26.6

Explanation:

Step 1: Calculate the molar concentrations

We will use the following expression.

M = mass solute / molar mass solute × liters of solution

[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M

[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M

[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M

Step 2: Make an ICE chart

        CO(g) + 2 H₂(g) ⇄ CH₃OH(g)

I        0.184      0.227           0

C         -x           -2x             +x

E     0.184-x   0.227-2x        x

Since [CH₃OH]e = x, x = 0.0523

Step 3: Calculate all the concentrations at equilibrium

[CO]e = 0.184-x = 0.132 M

[H₂]e = 0.227-2x = 0.122 M

[CH₃OH]e = 0.0523 M

Step 4: Calculate the equilibrium constant (Kc)

Kc = [CH₃OH] / [CO] [H₂]²

Kc = 0.0523 / 0.132 × 0.122² = 26.6

Sometimes in lab we collect the gas formed by a chemical reaction over water . This makes it easy to isolate and measure the amount of gas produced.
Suppose the CO, gas evolved by a certain chemical reaction taking place at 50.0°C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 132. mL. Calculate the mass of CO, that is in the collection tube. Round your answer to 2 significant digits.

Answers

Answer:

0.17 g

Explanation:

Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.

So, number of moles, n = volume of gas, v/molar volume, V

n = v/V where v = 132 mL = 0.132 L and V = 22.4 L

So, substituting the values of the variables into the equation, we have

n = v/V

n = 0.132 L/22.4 L

n = 0.005893 mol

We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol

Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO

m = nM

m = 0.005893 mol × 28 g/mol

m = 0.165004 g

m ≅ 0.17 g to 2 significant digits

If a 0.320 mM solution of MnO41- has an absorbance of 0.480 at 525 nm in a 1.000 cm cell. What is the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength?

Answers

Answer:

Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.

Explanation:

Now A = el, el=const

Then,

[tex]A2 / A1 = C2/ C1\\\\0.49/ 0.48 = C2 / 0.32\\\\C2 = 0.3266[/tex]

A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.

Answers

Answer:

density of second liquid = 650 kg/m³

Explanation:

Given that:

The volume of the plastic block submerged inside the water  = 0.5 V

The force on the plastic block  = [tex]\rho_1V_1g[/tex]

[tex]= 0.5p_1 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

W [tex]= 0.5p_1 V_g[/tex]

[tex]\rho Vg = 0.5p_1 V_g[/tex]

[tex]\rho = 0.5 \rho _1[/tex]

where;

water density [tex]\rho _1[/tex] = 1000

[tex]\rho = 0.5 (1000)[/tex]

[tex]\rho = 500 kg/m^3[/tex]

In the second liquid, the volume of plastic block in the water = (100-23)%

= 77% = 0.7 V

The force on the plastic block is:

[tex]= 0.77p_2 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

[tex]W = 0.77p_2 V_g[/tex]

[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]

[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]

42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)​

Answers

Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.

Explanation:

Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:

--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.

--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens

--> The ability of carbon atoms to form single, DOUBLE or triple bonds.

The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.

Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.

What is "X" in the following reaction?

Answers

x is the answer that we do not know yet!!:)

You should set out support, like a cork ring or clamp, before removing the glassware from a glassware kit to place the glassware in and to stop it from _________. Thoroughly check that the glasswar is________ and that it does not have any _______before using it.

Answers

Answer:

(A) Slipping and breaking

(B) Clean and dry

(C) Cracks

Explanation:

This describes the process of unpacking a glassware for use.

You should set out support like a cork ring or clamp (these are simple machines that'll hold the glassware in place) before removing the glassware from a glassware kit; to place the glassware in and to stop it from slipping and breaking.

Thoroughly check that the glassware is clean and dry and that it does not have any cracks, before using it.

How much heat capacity, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.499J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?

Answers

Answer:

56511.75 J

13506.3 Calories

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = amount of heat, m = mass of the iron, c = specific heat capacity of the iron, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 75 g, c = 0.499 J/g.°C, t₂ = 1535°C, t₁ = 25°C

Substitute these values into equation 1

Q = 75(0.499)(1535-25)

Q = 75(0.499)(1510)

Q = 56511.75 J

Q in Calories is

Q = (56511.75×0.239)

Q = 13506.3 Calories

1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.

Answers

Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.

Explanation:

A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.  

A system receives 425 J of heat from and delivers 425 J of work to its surroundings. What is the change in internal energy of the system (in J)?

Answers

Answer:

0 J

Explanation:

Applying,

ΔE = q+w................ Equation 1

Where ΔE = change in internal energy of the system, q = Heat of the system, w = work of the system.

Note: q is positive, while w is negative

From the question,

Given: q = 425 J, w = -425 J

Substitute these values into equation 1

ΔE = 425-425

ΔE = 0 J

Hence the change in internal energy of the system is 0 J

A one electron species, X m, where m is the charge of the one electron species and X is the element symbol, loses its one electron from its ground state when it absorbs 3.49 x 10-17 J of energy. Using the prior information, the charge of the one electron species is:_____________

a. +8
b. +2
c. +3
d. +1
e. +4

Answers

Answer:

Option C

Explanation:

From the question we are told that:

Difference in energy [tex]\delta E =3.49 * 10^{-17} J[/tex]

The Ground state Difference in energy at n=1

[tex]\delta E_g = 2.18 * 10^{-18} × Z^2[/tex]

Generally the equation for Difference in energy is mathematically given by

[tex]\delta E=\delta E_g[/tex]

Therefore

[tex]3.49 * 10^{-17} = 2.18 * 10^{-18} * Z^2[/tex]

[tex]Z^2=16[/tex]

[tex]Z=4[/tex]

Therefore

Charge on element Z Q_Z

[tex]Q_Z= Atomic\ no\. of\ element - No.\ of\ electrons\ of\ element[/tex]

[tex]Q_Z =4-1[/tex]

[tex]Q_Z=+3[/tex]

Option C

Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction

Answers

Answer:

The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"

Explanation:

Calculating the rate of the equation:

[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]

Rate:

[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]

Classify the processes as endothermic or exothermic.

a. Ice melting
b. Water condensing on surface
c. Baking a cake
d. The chemical reaction inside an instant cold pack.
e. A car using gasoline

Answers

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

a. Ice melting - endothermic

b. Water condensing on the surface - exothermic

c. Baking a cake - endothermic

d. The chemical reaction inside an instant cold pack - endothermic

e. A car using gasoline - exothermic

What is an exothermic and endothermic reaction?

An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.

While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.

In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.

The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.

Learn more about the exothermic process, here:

brainly.com/question/12321421

#SPJ5

Use the Ka values for weak acids to identify the best components for preparing buffer solutions with the given pH values.

Name Formula Ka
Phosphoric acid H3PO4 7.5 x 10^-3
Acetic acid CH3COOH 1.8 x 10^-5
Formic acid HCOOH 1.8 x 10^-4

pH 1.9 =_________
pH 5.0 = ________
pH 3.9= ________

Answers

Answer:

pH= 1.9 then [tex]H_{3} PO_{4}[/tex]

pH = 5.0 , [tex]CH_{3} COOH[/tex]

pH = 3.9 , HCOOH

As we know range left [tex]pH= pKa+/- 1[/tex]

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of carbonic acid and sodium hydroxide are combined. Assume excess base.

Answers

Answer:

[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up this net ionic equation, by firstly setting up the complete molecular equation as follows:

[tex]H_2CO_3(aq)+2NaOH(aq)\rightarrow Na_2CO_3(aq)+2H_2O(l)[/tex]

Thus, since carbonic acid is weak it merely ionizes whereas sodium hydroxides ionizes for the 100 % as it is strong; thus, we can write the complete ionic equation:

[tex]H_2CO_3(aq)+2Na^+(aq)+2OH^-(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Whereas sodium ions act as the spectator ones to be cancelled out for us to obtain:

[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Regards!

Read the chemical equation.
N2 + 3H2 - 2NH3
Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2
if all measurements are taken at the same temperature and pressure?
A 2.8 liters
B 3.2 liters
C 5.4 liters
D 6.3 liters

Answers

Answer:

A 2.8 liters

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇄ 2 NH₃

Step 2: Establish the appropriate volume ratio

At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.

Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen

4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L

Inter-molecular forces determine the _______________ properties while intra-molecular forces determine the ________ properties of compounds.

Answers

Answer:

Physical

Chemical

Explanation:

Intermolecular forces are the forces that hold the molecules of a substance together in a particular state of matter. They decide the physical properties of a substance.

The intra molecular forces are the bond forces that hold atoms together in molecules. The nature of this bonding determines the chemical properties of substances.

Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution?(report answer in grams and only three Sigg figs do not put the unit)

Answers

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

Question In nickel-cadmium batteries: Select the correct answer below: the anodes are nickel-plated and the cathodes are cadmium-plated the anodes are cadmium-plated and the cathodes are nickel-plated both the anodes and cathodes are plated with a nickel-cadmium alloy none of the above

Answers

Answer:

the anodes are cadmium-plated and the cathodes are nickel-plated

Explanation:

Nickel cadmium battery works on the principle as by the other cell. There is anode and a cathode which is separated by a separator (spiral shaped inside the case). The anode is negative and is cadmium plated while the cathode is positive and is nickel plated. An electrolyte is also used.

So the correct answer is : "The anodes are cadmium-plated and the cathodes are nickel-plated."

For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.

ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Answers

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

What direction would equilibrium moves towards based on the following if we increased the volume of the container.

[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]

Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume

Please help!

Answers

To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium

on the left:

we have 2 moles of A, 5 moles of B and 12 moles of C

which gives us a grand total of 19 gaseous moles

on the right:

here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid

giving us 14 gaseous moles on the right

Where does the reaction shift?

more gaseous moles means more space taken, because gas likes to fill all the space it can

if we have more volume, more gas can move around without colliding (reacting) with each other

Hence more volume favors the side with more gaseous moles

here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side

Answer:

Explanation:

given reversible chemical reaction:

2A(g) + 5B(g) + 12C(g)  ↔  14AC(g) + 5B(s)

chemicals in solid form do not take up a lot of volume so change in container volume has no effect

look at chemicals in gas form only:

the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19

the total no. of moles of products in gas form = 14

so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles

the ans is the equilibrium will move towards a) reactants

a chemist combines 4.9 g of nitrogen gas with 9.4 grams of nitrogen gas to form 11.4 g of ammonia 2.9 g of nitrogen is remaining

Answers

The results agree with the law of conservation of mass Explanation: The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.

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A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume

Answers

Answer:

3.00 L

Explanation:

P₁V₁ = P₂V₂

V₁ = 1.00 L

P₁ = (x) atm

P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]

V₂ = unknown

(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)

divide both sides by ( [tex]\frac{x}{3}[/tex] atm)

( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂

x cancels out

(1.00)(3) = V₂

V₂ = 3.00 L

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