Answer:
275. °CExplanation:
1) Data:
a) Constant volume
b) Initial pressure: P₀
c) Intitial temperature: T₀ = 0.987°C = 0.987 + 273.15 K = 274.137 K
d) Final pressure: P₁ = 2 P₀
e) Final temperature: unknown, T₁
2) Applied principles:
Gay - Lussac's law: at constant volume, the pressure and temperature of the gases are in direct proportion:⇒ P / T = constante ⇒ P₁ / T₁ = P₀ / T₀
3) Solution:
a) Solve for T₁
T₁ = P₁ T₀ / P₀b) Substitute the data:
T₁ = 2P₀ × 274.137 K / P₀ = 548.274 Kc) Converto to °C:
T₁ = 548.274 - 273.15 = 275.124 °C ≈ 275. °C (three significant figures)To keep a balloon inflated at the same volume when the pressure exerted on it is doubled, the temperature must be increased according to Gay-Lussac's Law. After performing the necessary calculations, we found that the new required temperature should be approximately 275.124°C.
Explanation:This question relates to Gas Laws, specifically Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature, assuming the volume stays constant. In this scenario, we need to work out the new temperature when the initial pressure is doubled.
First, let's convert the initial temperature from Celsius to Kelvin. Kelvin = Celsius + 273.15, so, 0.987°C = 274.137K.
According to Gay-Lussac's Law (P₁/T₁ = P₂/T₂ where P is pressure and T is temperature), if P2 = 2*P1 then the new temperature T₂ = 2× T₁, as the volume remains constant.
So, T₂ (in Kelvin) = 2 × T₁ = 2 × 274.137K = 548.274K
To convert T2 back to Celsius, we subtract 273.15 from the Kelvin temperature: T₂ (in Celsius) = 548.274K - 273.15 = 275.124°C.
So, if the pressure exerted on the balloon is doubled, the temperature needs to be approximately 275.124°C to keep the balloon inflated at the same volume.
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What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Assume that the contribution of protons from H2SO4 is near 100 %.Ba(OH)2(aq)+H2SO4(aq)→
Final answer:
The net ionic equation for the reaction between barium hydroxide and sulfuric acid is Ba2+(aq) + SO42-(aq) → BaSO4(s), which shows the formation of an insoluble precipitate of barium sulfate.
Explanation:
The correct net ionic equation for the reaction between barium hydroxide and sulfuric acid is as follows:
Ba2+(aq) + SO42−(aq) → BaSO4(s)
In this equation, Ba2+(aq) represents the barium ion in aqueous solution, SO42−(aq) denotes the sulfate ion in aqueous solution, and BaSO4(s) is the precipitate of barium sulfate. It is important to remember that net ionic equations must be balanced by both mass and charge, and in this case, the equation follows these rules with a product that is insoluble in water.
Aluminum chloride can be formed from its elements:
(i) 2Al(s) + 3Cl2 (g) ⟶ 2AlCl3 (s) ΔH° = ? Use the reactions here to determine the ΔH° for reaction (i):
(ii) HCl(g) ⟶ HCl(aq) ΔH(ii) ° = −74.8 kJ
(iii) H2 (g) + Cl2 (g) ⟶ 2HCl(g) ΔH(iii) ° = −185 kJ
(iv) AlCl3 (aq) ⟶ AlCl3 (s) ΔH(iv) ° = +323 kJ/mol
(v) 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g) ΔH(v) ° = −1049 kJ
Answer: The [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of [tex]AlCl_3[/tex] is:
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)[/tex] [tex]\Delta H^o_{formation}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]HCl(g)\rightarrow HCl(aq.)[/tex] [tex]\Delta H_1=-74.8kJ[/tex] ( × 6)
(2) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex] [tex]\Delta H_2=-185kJ[/tex] ( × 3)
(3) [tex]AlCl_3(aq.)\rightarrow AlCl_3(s)[/tex] [tex]\Delta H_3=+323kJ[/tex] ( × 2)
(4) [tex]2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)[/tex] [tex]\Delta H_4=-1049kJ[/tex]
The expression for enthalpy of formation of [tex]AlCl_3[/tex] is,
[tex]\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ[/tex]
Hence, the [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.
The change in enthalpy for the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) is -832.2 kJ, calculated by manipulating and summing the enthalpy changes of the given reactions using Hess's Law.
Explanation:The change in enthalpy (∆H°) of the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we want to find the enthalpy change of reaction (i) using reactions (ii) - (v).
To achieve this, let's manipulate the reactions as follows:
1. Reverse reaction (ii) and multiply by 6. This will give it the same number of HCl (moles) as in reaction (v): 6HCl(aq) ⟶ 6HCl(g), ∆H′ = +74.8 kJ * 6 = +448.8 kJ.
2. Multiply reaction (iii) by 3 to match the number of H2 (moles) and HCl in reaction (v): 3H2 (g) + 3Cl2 (g) ⟶ 6HCl(g), ∆H′′ = -185 kJ * 3 = -555 kJ.
Reaction (iv) remains unchanged: AlCl3 (aq) ⟶ AlCl3 (s), ∆H′′′ = +323 kJ.
3. Using reaction (v) in the forward direction: 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g), ∆H′′′′ = -1049 kJ.
Summing these manipulated reactions will give you reaction (i): ∆H° = ∆H′ + ∆H′′ + ∆H′′′ + ∆H′′′′ = 448.8 kJ - 555 kJ + 323 kJ - 1049 kJ = -832.2 kJ
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The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA. How many potassium ions pass through if the ion channel opens for 1.0 ms? What is the current density in the ion channel?
Answer:
Explanation:
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and
calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane
known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective
for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA.
Part A. How many potassium ions pass through if the ion channel opens for 1.0 ms?
Part B. What is the current density in the ion channel?
Solution: In 1.0 ms, the charge that passes through is
Q = I ∆t =
( 1.8 × 10−12 A ) (1.0 × 10−3 s )
= 1.8 × 10−15 C
Since each ion has a +1 charge (measured in electron charges), this represents
NK+ = Q e = (1.8 × 10−15 C ) (1.60 × 10−19 C) = 11250
The current density is calculated from the current and the size of the channel.
J = I A = ( 1.8 × 10−12 A ) ( π (0.30 × 10−9 m/2)2 ) = 2.55 × 107 A/m2
A) The number of potassium ions that will pass through the ion channel; 11250
B) The current density in the ion channel = 2.55 × 10⁷ A/m²
Given data;
Measurement with microelectrodes = 0.30-nm- diameter
K⁺ = 1.8 pA
a) calculate the number of potassium ions that passes through the ion channel
given that the channel opens for 1.0 ms
first step ; determine the value of charge ( Q )
Q = I*Δt = ( 1.8 × 10⁻¹² A )*(1.0 × 10⁻³ s ) = 1.8 × 10⁻¹⁵ C
where ; I = 1.8 × 10⁻¹² A
Δt = 1.0 × 10⁻³ s
next step : determine number of K⁺ ions passing through
NK⁺ = Q*e = ( 1.8 × 10⁻¹⁵ C )*( 1.60 × 10⁻¹⁹ C) = 11250
∴ number of K⁺ ions passing through = 11250 ions.
B) determine the current density in the ion channel
current density = current * size of channel
= ( 1.8 * 10⁻¹² ) * ( π * ( 0.30 * 10⁻⁹ )²
= 2.55 × 10⁻⁷ A/m²
Hence we can conclude that the number of potassium ions that will pass through the ion channel; 11250 and The current density in the ion channel = 2.55 × 10⁷ A/m²
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A 47.1 g sample of a metal is heated to 99.0°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at 21.4°C. The final temperature of the water is 24.5°C. Which metal was used?
Answer : The metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]C_1[/tex] = specific heat of metal = ?
[tex]C_1[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of metal = 47.1 g
[tex]m_2[/tex] = mass of water = 120 g
[tex]T_f[/tex] = final temperature of water = [tex]24.5^oC[/tex]
[tex]T_1[/tex] = initial temperature of metal = [tex]99^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]21.4^oC[/tex]
Now put all the given values in the above formula, we get
[tex]47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC[/tex]
[tex]c_1=0.44J/g^oC[/tex]
Form the value of specific heat of metal, we conclude that the metal used in this was iron.
Therefore, the metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).
An open flask sitting in a lab fridge looks empty, but it is actually filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?
Answer: [tex]0.67\times 10^{23}[/tex] molecules are contained in the flask.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P= Pressure of the gas = 1 atm
V= Volume of the gas = 2.50 L
T= Temperature of the gas = 273 K
R= Value of gas constant = 0.0821 Latm/K mol
[tex]n=\frac{PV}{RT}=\frac{1\times 2.50L}{0.0821 \times 273}=0.11moles[/tex]
1 mole of gas contains=[tex]6.022\times 10^{23}[/tex] molecules
0.11 moles of gas contains=[tex]\frac{6.022\times 10^{23}}{1}\times 0.11=0.67\times 10^{23}[/tex] molecules
[tex]0.67\times 10^{23}[/tex] molecules are contained in the flask.
Given the two reactions H2S⇌HS−+H+, K1 = 9.58×10−8, and HS−⇌S2−+H+, K2 = 1.58×10−19, what is the equilibrium constant Kfinal for the following reaction? S2−+2H+⇌H2S Enter your answer numerically.
Answer:
K(net) = K₁ x K₂ = 1.51 x 10⁻²⁶
Explanation:
Calculate the enthalpy of the following reaction: 4 B (s) + 3 O2 (g) → 2 B2O3 (s) given the following pertinent information: (A) B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ (B) 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ (C) H2 (g) + LaTeX: \frac{1}{2} 1 2 O2 (g) → H2O (l), ΔHoC = −285 kJ (D) H2O (l) → H2O (g), ΔHoD = +44 kJ
Answer:
4B + 3O₂ => 2B₂O₃; ΔH° = -3673Kj
Explanation:
Work these type problems in pairs of rxns… That is, add Rxn-1 & Rxn-2 => Rxn-1,2; then add Rxn-3 to Rxn-1,2 => Rxn- 1,2,3. Rxn-4 is not needed to obtain target rxn.
Target Rxn => 4B + 3O₂ => 2B₂O₃
Given …
(1) B₂O₃ + 2H₂O => 3O₂ + B₂H₆
=> reverse and double
=> 2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O
(2) 2B + 3H₂ => B₂H₆ => double and add to Rxn-1 => 4B + 6H₂ => 2B₂H₆
2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O
4B + 6H₂ => 2B₂H₆
________________________
∑(1,2) 4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O; ΔH°₁₂ = -2035Kj + (+72Kj) = -1963Kj
(3) => H₂ + ½O₂ => H₂O
=> reverse and multiply by 6, then add to (1,2) => 6H₂O => 6H₂ + 3O₂
6H₂O => 6H₂ + 3O₂
4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O
________________________
∑[(1,2,3) 4B + 3O₂ => 2B₂O₃; ΔH°₁₂₃ = -1963Kj + 6(-285Kj) = -3673Kj
Hydroxyapatite, Ca5(PO4)3(OH), is the main mineral component of dental enamel, dentin, and bone, and thus has many medical uses. Coating it on metallic implants (such as titanium alloys and stainless steels) helps the body accept the implant. In the form of powder and beads, it is used to fill bone voids, which encourages natural bone to grow into the void. Hydroxyapatite is prepared by adding aqueous phosphoric acid to a dilute slurry of calcium hydroxide. (a) Write a balanced equation for this preparation. (b) What mass of hydroxyapatite could form from 100. g of 85% phosphoric acid and 100. g of calcium hydroxide?
Answer:
(a) 5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O.
(b) 135.62 g.
Explanation:
(a) Write a balanced equation for this preparation.
The balanced equation for5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O,
that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) "hydroxyapatite" and 9 mol of H₂O.
(b) What mass of hydroxyapatite could form from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide?
Firstly, we need to calculate the no. of moles of 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide:The no. of moles of 100.0 g of 85% phosphoric acid:
∵ The percent of phosphoric acid = 85%.
∴ The actual mass of phosphoric acid in solution = 85.0 g.
∴ no. of moles of phosphoric acid = mass/molar mass = (85.0 g)/(97.994 g/mol) = 0.87 mol.
The no. of moles of 100.0 g of calcium hydroxide:
∴ no. of moles of calcium hydroxide = mass/molar mass = (100.0 g)/(74.093 g/mol) = 1.35 mol.
From the stichiometry, Ca(OH)₂ react with H₃PO₄ with (5: 3) molar ratio.
∴ 1.35 mol of calcium hydroxide "limiting reactant" react completely with 0.813 mol of phosphoric acid "excess reactant" with (5: 3) molar ratio.
Now, we can find the no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite":
Using cross multiplication:
5 mol of Ca(OH)₂ produce → 1 mol Ca₅(PO₄)₃(OH), from stichiometry.
1.35 mol of Ca(OH)₂ produce → ??? mol Ca₅(PO₄)₃(OH).
∴ The no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (1.35 mol)(1 mol)/(5 mol) = 0.27 mol.
Finally, we can get the mass of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (no. of moles)(molar mass) = (0.27 mol)(502.3 g/mol) = 135.62 g.
The equation of the reaction is:
5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O.Based on the data provided, the mass of hydroxyapatite formed is 135.62 g.
What are minerals?Minerals are substances which contain metallic elements combined with other elements found in the earth's crust.
Example of a mineral is Hydroxyapatite, Ca5(PO4)3(OH).
The balanced equation for the preparation of hydroxyapatite is given below:
5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O,that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) and 9 mol of H₂O.
The mass of hydroxyapatite prepared from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide? is calculated as follows:
100.0 g of 85% phosphoric acid contains 100 * 85/100 g of phosphoric acid = 85.0 g.
moles of phosphoric acid in 85 g = mass/molar mass
molar mass of phosphoric acid = 97.994 g/mol
moles of phosphoric acid in 85 g = 85.0 g/97.994 g/mol
moles of phosphoric acid in 85 g = 0.87 moles
moles of calcium hydroxide in 100.0 g = mass/molar mass
molar mass of calcium hydroxide = 74.093 g/mol
moles of calcium hydroxide in 100.0 g = 100.0 g/74.093 g/mol
moles of calcium hydroxide in 100.0 g = 1.35 moles
From the equation of the reaction, Ca(OH)₂ react with H₃PO₄ in a 5: 3 molar ratio.
5 mol of Ca(OH)₂ produce 1 mole Ca₅(PO₄)₃(OH)
1.35 mol of Ca(OH)₂ will produce 1.35 * 1/5 moles of Ca₅(PO₄)₃(OH) = 0.27 moles
mass of 0.27 moles of hydroxyapatite = no. of moles * molar mass
molar mass of hydroxyapatite = 502.3 g/mol
mass of hydroxyapatite = 0.27 mol * 502.3 g/mol
mass of hydroxyapatite = 135.62 g
Therefore, the mass of hydroxyapatite produced is 135.62 g
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Many drugs are sold as their hydrochloric salts (R2NH2 Cl−), formed by reaction of an amine (R2NH) with HCl. a. Draw the major organic product formed from the formation of acebutolol with HCl. Acebutolol is a β blocker used to treat high blood pressure. Omit any inorganic counterions.b. Discuss the solubility of acebutlol and its hydrochloride salt in water.c. Offer a reason as to why the drug is marketed as a hydrochloride salt rather than a neutral amine.
Answer:
Here's what I find.
Explanation:
a. Structure
Acebutolol is a secondary amine (basic). It forms a substituted ammonium salt when treated with hydrochloric acid.
The structure of the salt is shown below, with a red arrow pointing toward the positive charge on the N atom.
b. Solubility
The formula of acebutolol is C₁₈H₂₈N₂O₄.
The amide, acetyl, and ether groups confer little solubility to the molecule.
The alcohol and secondary amine do confer some solubility, because they can donate and accept hydrogen bonds.
However, they can each overcome the hydrophobic properties of only three to five carbons, and acebutolol has 18 of them.
The free amine would be preferentially soluble in lipid material (fats)
The protonated amine is ionic and therefore much more soluble in aqueous media (e.g., blood).
c. Marketing
The drug must be delivered to the tissues of the heart, where it blocks the effects of adrenalin. The best way to do this is through the blood, so acebutolol is marketed as the hydrochloride salt.
Dumbledore decides to gives a surprise demonstration. He starts with a hydrate of Na2CO3 which has a mass of 4.31 g before heating. After he heats it he finds the mass of the anhydrous compound is found to be 3.22 g. He asks everyone in class to determine the integer x in the hydrate: Na2CO3·xH2O; you should do this also. Round your answer to the nearest integ
Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles = [tex]\frac{mass }{molar mass }[/tex]
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = [tex]\frac{3.22}{106}[/tex] = 0.03mole
Number of moles of H₂O = [tex]\frac{1.09}{18}[/tex] = 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2
After heating, the mass loss corresponds to 1.09 g of water, which, when compared to the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex], gives a mole ratio of approximately 2:1, making [tex]\( x = 2 \)[/tex].
To determine the integer x in the hydrate [tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex], we need to follow these steps:
1. Calculate the mass of the water lost during heating:
[tex]\[\text{Mass of water} = \text{Mass of hydrate} - \text{Mass of anhydrous compound}\][/tex]
Given:
[tex]\[\text{Mass of hydrate} = 4.31 \, \text{g}\][/tex]
[tex]\[\text{Mass of anhydrous compound (Na}_2\text{CO}_3\text{)} = 3.22 \, \text{g}\][/tex]
Thus:
[tex]\[\text{Mass of water} = 4.31 \, \text{g} - 3.22 \, \text{g} = 1.09 \, \text{g}\][/tex]
2. Calculate the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
The molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is calculated as follows:
[tex]\[\text{Molar mass of Na}_2\text{CO}_3 = 2 \times 23.0 + 12.0 + 3 \times 16.0 = 46.0 + 12.0 + 48.0 = 106.0 \, \text{g/mol}\][/tex]
[tex]\[\text{Moles of Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Molar mass of Na}_2\text{CO}_3} = \frac{3.22 \, \text{g}}{106.0 \, \text{g/mol}} = 0.0304 \, \text{mol}\][/tex]
3. Calculate the moles of water lost:
The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is:
[tex]\[\text{Molar mass of H}_2\text{O} = 2 \times 1.0 + 16.0 = 18.0 \, \text{g/mol}\][/tex]
[tex]\[\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of H}_2\text{O}} = \frac{1.09 \, \text{g}}{18.0 \, \text{g/mol}} = 0.0606 \, \text{mol}\][/tex]
4. Determine the mole ratio of water to [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
[tex]\[x = \frac{\text{Moles of water}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.0606 \, \text{mol}}{0.0304 \, \text{mol}} = 1.993 \approx 2\][/tex]
Thus, the integer x in the hydrate[tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex]is 2.
Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of solution in which the Na concentration is 0.10 M?
Answer: The number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Molarity of solution = 0.10 mol/L
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol[/tex]
The chemical reaction for the ionization of sodium sulfate follows the equation:
[tex]Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)[/tex]
By Stoichiometry of the reaction:
2 moles of sodium ions are produced by 1 mole of sodium sulfate
So, 0.10 moles of sodium ions will be produced by = [tex]\frac{1}{2}\times 0.1=0.05moles[/tex] of sodium sulfate.
Hence, the number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.
Show how to convert the temperature 84.7° C to Kelvin. Please include all steps and label the final
answer
Answer:
184.46
Explanation:
(84.7°c x 9/5)+32=184.46 °F
What are the major products produced in the combustion of C10H22 under the following conditions? Write balanced chemical equations for each. a. An excess of oxygen b. A slightly limited oxygen supply c. A very limited supply of oxygen d. The compound is burned in air
Answer: The chemical reactions are given below.
Explanation:
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
[tex]\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O[/tex]
If supply of oxygen gas is limited, it is known as incomplete combustion and carbon monoxide gas is also produced as a product.
For a: An excess of oxygenHere, complete combustion reaction takes place. The chemical equation follows:
[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO_2+22H_2O[/tex]
For b: A slightly limited oxygen supplyHere, incomplete combustion takes place and carbon monoxide is also formed.
[tex]C_{10}H_{22}+13O_2\rightarrow 5CO+5CO_2+11H_2O[/tex]
For c: A very limited supply of oxygenHere, incomplete combustion takes place and only carbon monoxide with water are formed as the products.
[tex]2C_{10}H_{22}+21O_2\rightarrow 20CO+22H_2O[/tex]
For d: The compound is burned in airWhen a compound is burned in air, it means that unlimited supply of oxygen is there. So, complete combustion reaction takes place and carbon dioxide gas is formed as a product.
[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO+22H_2O[/tex]
Hence, the chemical reactions are given below.
Answer:
WAIT SORRY I CANT DECLICK ill try to answer doe a.
Explanation:
What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2? Express you answer as a chemical equation including phases. View Available Hint(s)
Answer : The net ionic equation will be,
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The same number of ions present on reactant and product side which do not participate in a reactions.
The given balanced ionic equation will be,
[tex]MgSO_4(aq)+Pb(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+PbSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Mg^{2+}(aq)+SO_4^{2-}(aq)+Pb^{2+}(aq)+2NO^{3-}(aq)\rightarrow PbSO_4(s)+Mg^{2+}(aq)+2NO^{3-}(aq)[/tex]
In this equation, [tex]Mg^{2+}\text{ and }NO_3^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g)+Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 480 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.820 Cl2 1.27 COCl2 0.230 What is the equilibrium constant, Kp, of this reaction?
Answer:
0.2208 is the equilibrium constant,[tex]K_p[/tex].
Explanation:
Equilibrium constant is defined as ratio of concentration of products to the concentration of reactants raised to the power equal to their stoichiometric coefficients in balanced chemical equation. It is expressed as [tex]K_c[/tex]
If the equilibrium is in gaseous phase then instead of concentration take partial pressure of each compound.The It is expressed as [tex]K_p[/tex].
[tex]CO(g)+Cl_2(g)\rightarrow COCl_2(g)[/tex]
Partial pressure of the[tex] p_{[CO]} = 0.820 atm[/tex]
Partial pressure of the[tex] p_{[Cl_2]} = 1.27 atm[/tex]
Partial pressure of the [tex]p_{[COCl_2]} = 0.230 atm[/tex]
The expression of an equilibrium constant is given as:
[tex]K_p=\frac{p_{[COCl_2]}}{p_{[CO]}p_{[Cl_2]}}[/tex]
[tex]K_p=\frac{0.230 atm}{0.820 atm \times 1.27 atm}=0.2208 [/tex]
Iodine is prepared both in the laboratory and commercially by adding Cl2(g)Cl2(g) to an aqueous solution containing sodium iodide. 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI,NaI, must be used to produce 89.1 g89.1 g of iodine, I2?I2? mass: g NaI
Answer : The mass of sodium iodide used to produced must be, 105.22 grams.
Explanation : Given,
Mass of [tex]I_2[/tex] = 89.1 g
Molar mass of [tex]I_2[/tex] = 253.8 g/mole
Molar mass of [tex]NaI[/tex] = 149.89 g/mole
First we have to calculate the moles of [tex]I_2[/tex].
[tex]\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles[/tex]
Now we have to calculate the moles of [tex]NaI[/tex].
The balanced chemical reaction is,
[tex]2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)[/tex]
From the balanced chemical reaction, we conclude that
As, 1 mole of [tex]I_2[/tex] obtained from 2 moles of [tex]NaI[/tex]
So, 0.351 moles of [tex]I_2[/tex] obtained from [tex]2\times 0.351=0.702[/tex] moles of [tex]NaI[/tex]
Now we have to calculate the mass of [tex]NaI[/tex].
[tex]\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI[/tex]
[tex]\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g[/tex]
Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.
To produce 89.1 grams of iodine (I2), 178.2 grams of sodium iodide (NaI) must be used.
Explanation:To find the number of grams of sodium iodide (NaI) needed to produce 89.1 g of iodine (I2), we need to use the stoichiometric coefficients from the balanced chemical equation:
2NaI(aq) + Cl2(g) ⟶ I2(s) + 2NaCl(aq)
From the equation, we can see that 2 moles of NaI react to produce 1 mole of I2. The molar mass of NaI is 149.89 g/mol. We can set up a proportion:
(2 mol NaI / 1 mol I2) = (x g NaI / 89.1 g I2)
Solving for x:
x = (2 mol NaI / 1 mol I2) * (89.1 g I2 / 1 mol I2) = 178.2 g NaI
Therefore, 178.2 grams of sodium iodide (NaI) must be used to produce 89.1 grams of iodine (I2).
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Why are acid gases removed from the natural gases?
The removal of acid gases from natural gas is critical for ensuring the safety and longevity of the gas transportation and storage infrastructure, protecting human health and the environment, complying with regulations, and improving the quality and economic value of the natural gas product.
The removal of acid gases from natural gases is essential for several reasons:
1. Pipeline Corrosion Prevention: Acid gases, such as hydrogen sulfide (H2S) and carbon dioxide (CO2), can react with water to form acids. These acids, particularly sulfuric acid from H2S, can cause severe corrosion in pipelines and other transportation and storage facilities. Removing these gases prevents this corrosion, extending the lifespan of the infrastructure and reducing maintenance costs.
2. Safety and Health Concerns: Hydrogen sulfide is highly toxic and can be lethal at relatively low concentrations. It also has a strong, unpleasant odor. Removing H2S from natural gas is crucial to ensure the safety of workers and the public, as well as to comply with environmental and health regulations.
3. Product Quality: Acid gases can interfere with the burning properties of natural gas, leading to inefficient combustion and the potential for equipment malfunction. Removing these impurities improves the quality of the natural gas, ensuring it meets the specifications required for various applications, including its use as a fuel for power generation, industrial processes, and residential heating.
4. Environmental Protection: CO2 is a greenhouse gas, and its release into the atmosphere contributes to global warming. While natural gas combustion produces less CO2 than other fossil fuels like coal and oil, reducing the CO2 content in natural gas can further lessen its environmental impact. Additionally, the removal of sulfur compounds helps prevent the release of sulfur dioxide (SO2), a precursor to acid rain, during combustion.
5. Regulatory Compliance: Many countries have regulations that limit the amount of acid gases in natural gas. These regulations are in place to protect the environment and public health. Gas processing facilities must remove acid gases to comply with these standards.
6. Economic Value: Sulfur, which can be extracted from acid gases, has economic value and can be sold as a byproduct. The Claus process is a common method used to convert hydrogen sulfide into elemental sulfur. By recovering sulfur, the economic viability of natural gas processing is improved.
In summary, the removal of acid gases from natural gas is critical for ensuring the safety and longevity of the gas transportation and storage infrastructure, protecting human health and the environment, complying with regulations, and improving the quality and economic value of the natural gas product.
Studies indicate that the methyl radical is trigonal planar. Based on this, which of the following best describes the methyl radical? The carbon is sp2 hybridized and the unpaired electron occupies an sp2 orbital. The carbon is sp2 hybridized and the unpaired electron occupies a 2p orbital. The carbon is sp3 hybridized and the unpaired electron occupies an sp3 orbital. The carbon is sp3 hybridized and the unpaired electron occupies a 2p orbital.
Answer:
The carbon is sp² hybridized and the unpaired electron occupies a 2p orbital.
Explanation:
The central C atom has a trigonal planar geometry, so it is sp² hybridized.
The sp² bonds are formed by the hybridization of the 2s orbital with two 2p orbitals.
The unpaired electron goes into the unhybridized 2p orbital.
Complete the following reaction by drawing the missing reactant ("Structure A"). (Draw only one neutral organic compound in the empty box. "Ph" is an abbreviation for a benzene ring. Draw the benzene ring out in your answer.)
Answer:
what I dont get this bro
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0172 M solution. If the pH of the resulting solution is 2.39 at 20 °C, determine the pKa for the acid.
Answer:
9.7 x 10⁻⁴
Explanation:
HA ⇄ H⁺ + A⁻
C(eq) 0.0174 10⁻²·³⁹ 10⁻²·³⁹
=0.0041M =0.0041M
Ka = [H⁺][A⁻]/[HA] = (0.0014)²/(0.0174) = 9.7 x 10⁻⁴
Are hydrocarbons approved for retrofit applications
A single electron can have the following set of quantum numbers: n=3, l=2, ml=0, ms= -1/2.
True or False
Answer:
True => (3, 2, 0, -1/2) => 3d₀² electron
Explanation:
What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, C2H6? In this reaction, ethane is burned in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O). ____C2H6(g)+____O2(g)→____CO2(g)+____H2O(g)
Answer: The coefficients for balancing the given chemical equation are 2, 7, 4 and 6
Explanation:
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another form. This also means that total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.
The given balanced chemical equation follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
On reactant side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
On product side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
Hence, the coefficients for balancing the given chemical equation are 2, 7, 4 and 6
Complete and balance the molecular equation for the reaction of aqueous copper(II) chloride, CuCl2, and aqueous potassium phosphate, K3PO4. Include physical states. molecular equation: CuCl2(aq)+K3PO4(aq)⟶ Enter the balanced net ionic equation for this reaction. Include physical states. net ionic equation.
In the balanced molecular equation between CuCl₂(aq) and K₃PO₄(aq), the cations and anions exchanged, resulting in 6KCl(aq) and Cu₃(PO₄)₂(s). The net ionic equation, focusing only on participating substances is: 3Cu²⁺(aq) + 2PO₄³⁻(aq) → Cu₃(PO₄)₂(s).
Explanation:
The reaction between aqueous copper(II) chloride (CuCl2) and aqueous potassium phosphate (K3PO4) is a double displacement reaction, where the cation of one compound combines with the anion of the other compound. The balanced molecular equation for this reaction is: 2CuCl2(aq) + 3K3PO4(aq) → 6KCl(aq) + Cu3(PO4)2(s).
Now, for the net ionic equation, you would only include the species that participate in the reaction and exclude the spectator ions (ions that do not participate in the reaction). After breaking down the molecular equation to ions (excluding spectator ions) the net ionic equation of the reaction would be: 3Cu^2+(aq) + 2PO4^3-(aq) → Cu3(PO4)2(s).
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The balanced molecular equation is 3CuCl₂(aq) + 2K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq). The net ionic equation is 3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s). This represents the formation of copper(II) phosphate precipitate.
Complete and Balance the Molecular Equation
Let's complete and balance the molecular equation for the reaction of aqueous copper(II) chloride (CuCl₂) and aqueous potassium phosphate (K₃PO₄).
Molecular Equation:
CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)
First, we need to balance this equation:
Balance the copper (Cu) atoms: 3 CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)
Balance the potassium (K) atoms: 3 CuCl₂(aq) + 2 K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)
Write the Net Ionic Equation
To write the balanced net ionic equation, first write the complete ionic equation:
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 K⁺(aq) + 2 PO₄³⁻(aq) ⟶ 6 K⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
Next, identify and remove the spectator ions (K⁺ and Cl⁻), which are ions that do not participate in the reaction:
Net Ionic Equation:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)
This net ionic equation represents the formation of copper(II) phosphate precipitate from copper(II) ions and phosphate ions in solution.
The pH of a solution is measured eight times by one operator using the same instrument. She obtains the following data: 7.15, 7.20, 7.18, 7.19, 7.21, 7.20, 7.16, and 7.20. (a) Calculate the sample mean. Round your answer to 3 decimal places.
Answer:
7.186
Explanation:
The mean is the average of some given data from a sample point.
To calculate the mean, we use the formula below:
Mean = ∑fx/∑f
Where f = frequency
x = sample data
From the given pH of the solutions, we can form a table:
x f fx
7.15 1 7.15
7.16 1 7.16
7.18 1 7.18
7.19 1 7.19
7.20 3 21.6
7.21 1 7.21
Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49
∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8
The mean = [tex]\frac{57.49}{8}[/tex] = 7.18625 = 7.186
When 23.6 mL of 0.500 M H2SO4 is added to 23.6 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)
Answer:
-112 Kj/mole (diprotic acid)
Explanation:
The enthalpy of reaction is -55 KJ/mol.
We must first write down the equation of the reaction;
2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)
Then we compute the number of moles of H2SO4 = 23.6/1000 × 0.500 M = 0.012 moles
And the number of moles of KOH = 23.6/1000 × 1.00 M = 0.024 moles
From what we know in the reaction equation;
2 moles of KOH produces 2 moles of water
Therefore, 0.0026 moles of KOH produces 0.024 moles of water.
The total volume of solution is obtained by adding = 23.6 mL + 23.6 mL = 47.2 mL
Mass of water = density × volume = 1.00 g/mL × 47.2 mL =47.2 g
Using the formula;
ΔH = mcθ
Mass of solution (m) = 47.2 g
Specific heat capacity of solution (c) = 4.184 J/g·°C
Temperature difference(θ) = 30.17°C - 23.50°C = 6.67°C
Substituting values;
ΔH = -( 47.2 g × 4.184 J/g·°C × 6.67°C)/ 0.024 moles
ΔH = -(1.32 KJ/0.024 moles)
ΔH = -55 KJ/mol
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given a condensed structural formula, write a line-angle formula for (CH3)2CHCH(CH3)2
Answer : The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.
Explanation :
Condensed structural formula : It is a formula in which the lines are used between the bonded atoms and the atoms are also shown in the structural formula.
Line-angle formula : In the line-angle formula, the carbon atoms are represented at the corner and ends of the lines with some angle and the hydrogen atoms are hidden.
The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.
The line-angle formula for (CH3)2CHCH(CH3)2 is CH3 CH3 CH2 CH3 CH3.
Explanation:The line-angle formula for (CH3)2CHCH(CH3)2 can be written as:
CH3 CH3
|
CH2
|
CH3 CH3
This formula represents the structural arrangement of atoms in the molecule, with each line representing a bond between carbon atoms. The ends of the lines represent carbon atoms, and any additional atoms or groups attached to those carbon atoms are shown as branches off the main line.
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Given that the density of the saturated solution is found to be 1.16 g/mL. The molar mass of copper sulfate pentahydrate is 249.68 g/mol, calculate grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0oC (show calculations for full credit)
Explanation:
Since, it is given that density is 1.16 grams per milliliter and molar mass of copper sulfate pentahydrate is 249.68 g/mol.
Now, as we known that density is the amount of mass present in a unit volume.
Mathematically, Density = [tex]\frac{\text{molar mass}}{volume}[/tex]
Hence, calculate the volume as follws.
Density = [tex]\frac{\text{molar mass}}{volume}[/tex]
1.16 g/mL = [tex]\frac{249.68 g/mol}{volume}[/tex]
volume = 215.24 mL
As, 215.24 mL can dissolve in 249.68 g/mol of complex. So, in 100 mL volume amount dissolved will be calculated as follows.
[tex]\frac{249.68 g/mol}{215.24 mL} \times 100[/tex]
= 116 grams
Thus, we can conclude that 116 grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0[tex]^{o}C[/tex].
To calculate the grams of copper sulfate pentahydrate that dissolve in 100 g of water at 0°C, solubility data at that temperature is needed, which is not provided in the question.
Explanation:The student is asking to calculate the grams of copper sulfate pentahydrate that will dissolve in 100 grams of water at 0°C, given that the density of the saturated solution is 1.16 g/mL and the molar mass of copper sulfate pentahydrate is 249.68 g/mol. Since the volume of solution can be derived from the mass of the water and the density of the solution, we can calculate the amount of copper sulfate pentahydrate dissolved in this volume by using dimensional analysis. However, to perform this calculation, we need the solubility data for copper sulfate pentahydrate at 0°C, which is not provided in the question. Without this information, we cannot proceed with the calculation.
The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water?
In this chemistry problem, we first calculate the heat produced by the combustion of C6H6, and then we use the heat equation to find the final temperature of the water which comes out to be 53.14 °C.
Explanation:The subject of this question pertains to the field of chemistry, specifically physical chemistry dealing with energy changes in chemical reactions, and this seems to be a high school-level problem based on the complexities involved. The principle used here is q=mc∆T, which allows us to calculate how much heat is transferred when the water temperature changes.
First, the amount of heat produced by the combustion of 8.6 g of C6H6 should be calculated based on its enthalpy change (heat produced per mole). Given the heat of combustion, -6542 kJ for 2 moles of C6H6, the heat of combustion for 8.6g of C6H6 can be calculated by (8.6 g / (78.11 g/mol) mol ) * (-6542 kJ / 2 mol) = -227.11 kJ.
The heat absorbed by the water can then be calculated using the heat equation: q=mc∆T. Here, m = mass of water = 5691 g, c = specific heat capacity of water = 4.18 J/g°C. The heat is converted to kilojoules: q = -227.11 kJ * 1000 = -227110 J. Hence, the equation becomes -227110 = 5691*4.18*∆T. After rearranging and solving, the final temperature (T) will be 53.14 °C.
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The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.323 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K.
Answer : The equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.
Explanation : Given,
Moles of [tex]COCl_2[/tex] = 0.323 mole
Volume of solution = 1 L
Initial concentration of [tex]COCl_2[/tex] = 0.323 M
Let the moles of [tex]CO\text{ and }Cl_2[/tex] be, 'x'. So,
Concentration of [tex]CO[/tex] = x M
Concentration of [tex]Cl_2[/tex] = x M
The given balanced equilibrium reaction is,
[tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]
Initial conc. 0.323 M 0 0
At eqm. conc. (0.323-x) M (x M) (x M)
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]1.29\times 10^{-2}=\frac{(x)\times (x)}{(0.323-x)}[/tex]
By solving the term 'x', we get :
x = 0.0584
Thus, the concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] at equilibrium are :
Concentration of [tex]COCl_2[/tex] = (0.323-x) M = (0.323-0.0584) M = 0.2646 M
Concentration of [tex]CO[/tex] = x M = 0.0584 M
Concentration of [tex]Cl_2[/tex] = x M = 0.0584 M
Therefore, the equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.