Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
The boiling of water is a:_______.
a. chemical change because a gas (steam) is given off.
b. chemical change because heat is needed for the process to occur.
c. physical change because the water merely disappears chemical and physical damage.
d. physical change because the gaseous water is chemically the same as the liquid.
Answer:
D
Explanation:
trust me its correct i think
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
What happens in a double replacement reaction
Answer: D
Explanation: The elements in two compunds switch places
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
Could someone please help me with this chemistry question I will mark the correct answer as brainliest
Calculate the pH of this solution 0.0043 M of H2SO4=
Answer:
pH = - log [concentration]
pH = - log (0.0043M)
pH = 2.37
The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 × 10–4 mol L–1. Calculate the concentration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.
Answer:
The correct answer is 5.30 * 10^-4 mol per L.
Explanation:
Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.
Therefore, it can be written as,
C2/C1 = P2/P1
Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,
C2/(6.8*10^-4) = 0.78/1
C2 = 0.78 * 6.8*10^-4
C2 = 5.30 * 10^-4 mol per L.
Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
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Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.850 A that flows for 60.0 min.
Answer:
Mass of Ga = 0.73694 gram
Explanation:
Given:
Current = 0.850 A
Time = 60 minutes
Find:
Amount of gas deposit.
Computation:
Total charge = Current × Time in second
Total charge = 0.850 × 60 × 60
Total charge = 3,060 C
Mole of electron = Total charge / Faraday constant [Faraday constant = 96,485.3329]
Mole of electron = 3,060 / 96,485.3329
Mole of electron = 0.0317146
Moles of Ga = 1/3 [Mole of electron]
Moles of Ga = 1/3 [0.0317146]
Moles of Ga = 0.01057
Mass of Ga = molar mass × Moles of Ga
Mass of Ga = 69.72 × 0.01057
Mass of Ga = 0.73694 gram
At 25 oC, the rate constant for the first-order decomposition of a pesticide solution is 6.40 x 10-3 min-1. If the starting concentration of pesticide is 0.0314 M, what concentration will remain after 62.0 minutes at 25 oC? 3.12 x 10-2 M 47.4 M 2.11 x 10-2 M 4.67 x 10-2 M 8.72 M
Answer:
[tex]2.11\ * 10^{-2}[/tex] is the correct answer to the given question.
Explanation:
Given k=6.40 x 10-3 min-1.
According to the first order reaction .
The concentration of time can be written as
[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]
Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.
So [tex][\ A\ ]_{0}= 0.0314 M[/tex]
Putting this value into the above equation.
[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]
=0.211 M
This can be written as
[tex]=\ 2.11 *\ 10^{-2}[/tex]
2. In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a polar solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography
Water was poured over a large oil fire to extinguish it. What would happen and why?
Answer:
I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water
Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.
What happens when you pour water on an oil fire?A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.
Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.
The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.
Thus, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.
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A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
Answer:
0.031 m
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
Chemical change
Element
Explanation:
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³. What is the length of one side of the cube in cm?
Step 1: Convert the mass to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]
Step 2: Calculate the volume (V) of the cube
[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]
Step 3: Calculate the length (l) of one side of the cube
We will use the following expression.
[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The balanced chemical equation is:
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
This is a chemical change because new substances are formed.
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.
Who proposed the plum pudding model and what does it say about the structure of the atom
Answer:
J. J. Thomson
Explanation:
First proposed by J. J. Thomson in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
Which process absorbs the greatest amount of heat?
a. the cooling of 10 g of liquid water from 100°C to 0°C.
b. the heating of 10 g of liquid water from 0°C to 100°C.
c. the freezing of 10 g of liquid water the melting of 10 g of ice.
d. the condensation of 10 g of gaseous water.
Answer:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Explanation:
Hello,
In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.
Regards.
The process having the greatest amount of heat is:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Looking at all the options:The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.
The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.
Find more information about Heat here:
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A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added
Answer:
pH = 9.08
Explanation:
Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:
Q + H₂O ⇄ QH⁺ + OH⁻
Where pKb is 5.10
Using H-H equation for weak bases:
pOH = pKb + log₁₀ [QH⁺] / [Q]
The reaction of quinine with HCl is:
Q + HCl → QH⁺ + Cl⁻
Initial moles of quinine are 0.125 moles and moles added of HCl are:
0.05000L × (1.00mol / L) = 0.05000moles.
That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:
Q = 0.125mol - 0.050mol = 0.075 moles
QH⁺ = 0.050 moles
Replacing in H-H equation:
pOH = 5.10 + log₁₀ [0.050] / [0.075]
pOH = 4.92
As pH = 14 - pOJ
pH = 9.08Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *
Answer:
if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!
Explanation:
Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.
Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.
Oxidation half equation;
Mg(s)-----> Mg^2+(aq) + 2e
Reduction half equation;
O2(g) + 2e ------> 2O^2-(aq)
From the balanced reaction equation, two moles of electrons is transferred.
Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.
The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 6.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.904 g of precipitate, what is the molarity of lead(II) ion in the original solution
Answer:
[tex]M=0.0637M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]
Finally, the resulting molarity in 30.8 mL (0.0308 L):
[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]
Regards.
A sample of helium has a volume of 325 mL and a pressure of 655 mmHg. What will be the pressure, in mmHg, if the sample of helium is compressed to 125 mL (T, n constant)? (Show calculations.)
Answer:
1703 mmHg
Explanation:
Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:
(13/5)(655 mmHg) = 1703 mmHg
In the compound Fe2O3, iron's oxidation number is +3, and oxygen's oxidation
number is
Answer here
Answer: The oxygen's oxidation number is -2.
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as cation and oxygen is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]
The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.
ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]
Answer:
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Explanation:
Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]
Photon momentum = h/wavelength
[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]
Mass of H2 molecule m = molar mass/Avogadros number
[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]
Since momentum is conserved:
Photon momentum = H2 molecule momentum = mass x velocity of H2
[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]
velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]
Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2
[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
CAN SOMEONE HELP ME!!
Solutions of each of the hypothetical acids in the following table are prepared with an initial concentration of 0.100 M. Which of the four solutions will have the lowest pH and be most acidic? Explain please.
Acid pKa
HA 4.00
HB 7.00
HC 10.00
HD 11.00
a. HA
b. HB
c. HC
d. HD
e. All will have the same pH because the concentrations are the same.
Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Explanation:
We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.
[tex]pK_a = -log K_a[/tex]
This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.
As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.
In the given options, HA has the smallest [tex]pK_a[/tex] value.
Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Round to 3 significant figures.
1.4593
Answer : The correct answer is 1.46
Explanation :
The following rules are used to round off a number to the required number of significant figures:
(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.
(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.
(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.
(4) The same procedure is follow for decimal values.
As we are given, 1.4593
In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.
According to the rules, round off the given measurement in three significant figures as 1.46
Therefore, the correct answer is 1.46
9. Predict the major products formed when: (a) Toluene is sulfonated. (c) Nitrobenzene is brominated. (b) Benzoic acid is nitrated. (d) Isopropylbenzene reacts with acetyl chloride and AlCl3. If the major products would be a mixture of ortho and para isomers, you should so state.
Answer:
a) ortho-para isomers predominates
b) 3-nitrobenzoic acid ( meta isomer predominates)
c) 3-bromo nitrobenzene ( meta isomer predominates)
d) the ortho- para isomers predominates
Explanation:
a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.
b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.
c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.
d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .
In TLC chromatography of plant pigments, why do different pigments travel up the plate at different rates
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!