write the chemistry of Epsom salt

Answers

Answer 1
if you’re talking about the formula it’s, MgSo4

Related Questions

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."

once formed, how are coordinate covalent bonds different from other covalent bonds?

Answers

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

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

What is true of all matter?
A. It pushes or pulls on objects.
B. You can see it.
C. It gives off heat energy.
D. It has mass.

Answers

It would be D (I think)

What is the energy change when 78.0 g of Hg melt at −38.8°C

Answers

Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

How can this product be achieved using the starting material shown?

Answers

Answer:

this product can be achieved using the starting material shown is by use of NaOH as catalyst.

Answer:

By using NaOH as catalyst.

Explanation:

This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.

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

Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force

Answers

Hydrogen bonding
In the secondary structure of a protein, hydrogen bonds between amino acids determine the configuration of the molecules.
In DNA, hydrogen bonds connect the nitrogenous bases (2 hydrogen bonds between adenine and thymine, 3 hydrogen bonds between guanine and cytosine)

Answer:

hydrogen bonding

Explanation:

just took the test :D

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

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]

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

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 force is behind us when we ride a bike?

Answers

Answer:

gravity, ground, friction, rolling resistance, and air resistance.

gravity and force which helps us to not a
fall and keep going

In the given range,at what temperature does oxy gen have the highest solubility?​

Answers

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.
25oC
Solubility of oxygen and oxygen compounds

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.

What is an emission spectrum?

A. The total amount of energy emitted by an element
B. The products created when an element is burned
C. The energy absorbed when an electron gains energy
D. The colors of light given off when an element loses energy​

Answers

Answer:

D

Explanation:

The electromagnetic radiation is emitted due to a particle moves from a higher to a lower energy state

An emission spectrum is the colors of light given off when an element loses energy​. Therefore, option D is correct.

What is emission spectrum ?

The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.

An emission spectrum is the range of radiations that are released in different places when electrons jump back and forth between higher and lower energy levels to achieve stability.

Since what you are seeing is the direct radiation produced by the source, this form of spectrum is also known as an emission spectrum. You can see all the colors in the Sun's spectrum because light from the Sun is produced at practically all energies in the visible spectrum.

Thus, option D is correct.

To learn more about emission spectrum, follow the link;

https://brainly.com/question/27268130

#SPJ2

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

Answers

Casease! Good luck!

What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.

Answers

Answer:

[tex]\alpha=17.7[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=13g[/tex]

Volume [tex]V=10mL[/tex]

Angle [tex]\theta=23[/tex]

Sample Tube=10cm

Generally the equation for concentration is mathematically given by

 [tex]C=m/v[/tex]

 [tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]

Therefore the Specific Rotation

 [tex]\alpha=frac{\theta }{m*l}[/tex]

 [tex]\alpha=frac{23 }{1.3*1.0}[/tex]

 [tex]\alpha=17.7[/tex]

A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.

Answers

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

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

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!

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)

Answers

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma

Answers

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

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]

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.

What is "X" in the following reaction?

Answers

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

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.

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/4254573-determine-if-the-results-of-the-following-word-problem-adhere-to.html

Identify the phase of the copper product after each reaction in the copper cycle.

The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )

Answers

Answer:

addition of HNO3 HNOX3 to Cu - Aueous

addition of H2SO4 HX2SOX4 to CuO - Aqueous

addition of Z n Zn to C u S O 4 CuSOX4 - Solid

addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid

heating of C u ( O H ) - Solid

Explanation:

Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.

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

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

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

Other Questions
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