Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
To determine the number of molecules of XeF₆,
First, we will determine the number of moles of F₂ present in the 12.9L of F₂
From the ideal gas equation
PV = nRT
Where P is the pressure
V is the volume
n is the amount of substance ( number of moles)
R is ideal gas constant
T is the temperature
From the question,
P = 2.6 atm
V = 12.9 L
R = 0.082057 L atm mol⁻¹ K⁻¹
T = 298 K
Putting these values into the equation
PV = nRT
2.6 × 12.9 = n × 0.082057 × 298
33.54 = n × 24.452986
∴ n = 33.54 ÷ 24.452986
n = 1.3716116 moles
The number of moles of F₂ present is 1.3716116 moles
From the given equation of reaction
Xe(g) +3F₂(g)→XeF₆(g)
1 mole of Xe reacts with 3 moles of F₂ to produce 1 mole of XeF₆
∴ 1.3716116 moles of F₂ will give (1.3716116/3) moles of XeF₆
Hence, the number of moles of XeF₆ produced = 0.457204 moles
To determine the number of molecules formed,
From the formula
Number of molecules = number of moles × Avogadro's number
∴ Number of molecules of XeF₆ formed = 0.457204 × 6.02214 ×10²³
Number of molecules of XeF₆ formed = 2.7533 × 10²³ molecules
Number of molecules of XeF₆ formed ≅ 2.75 × 10²³ molecules
Hence, the number of molecules of XeF₆ that are formed from 12.9 L of F₂ (at 298 K and 2.6 atm) is 2.75 × 10²³ molecules.
The correct option is E) 2.75 × 1023 molecules XeF6
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The rate constant k of the second-order reaction CH3CHO→CH4+CO is 6.73×10−3Lmol s. The concentration of CH3CHO at t=50.0 seconds is 0.151 mol/L. What was the initial concentration of CH3CHO?
Answer:
The initial concentration of ethanal was 0.1590 mol/L.
Explanation:
Integrated rate law for second order kinetic:
[tex]k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})[/tex]
k = Rate constant =[tex]6.73\times 10^{-3} L mol s[/tex]
t = Time elapsed = 50.0 s
[tex][A]_o[/tex] =initial concentration of ethanal
[A] = Concentration of ethanal left after time t = 0.151 mol/L
On substituting the value:
[tex]6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})[/tex]
[tex][A]_o=0.1590 mol/L[/tex]
The initial concentration of ethanal was 0.1590 mol/L.
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]
When titrating a strongmonoprotic acid and KOH at 25°C, theA) pH will be less than 7 at the equivalence point.B) pH will be greater than 7 at the equivalence point.C) titration will require more moles of base than acid to reach the equivalence point.D) pH will be equal to 7 at the equivalence point.E) titration will require more moles of acid than base to reach the equivalence
Answer:
D => pH will be equal to 7 at equivalence point
Explanation:
For Strong Acid + Strong Base titrations, pH = 7 as neither ion of the salt produced will undergo hydrolysis as would weak electrolyte titrations.
BOTH CASES ARE PRESENTED FOR CONTRAST ...
Stong monoprotic acid being titrated with NaOH ...
=> HX + NaOH => NaCl + H₂O
=> NaCl => Na⁺ + Cl⁻
=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)
=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)
This leaves only the Auto Ionization of Water as the reaction affecting the pH of the solution at the equivalence point of a strong acid + strong base titration. That is ...
HOH ⇄ H⁺ + OH⁻ & [H⁺] = [OH⁻] = 1 x 10⁻⁷M
pH = -log[H⁺] = -log(1 x 10⁻⁷) = -(-7) = 7
----------------------------------
Weak Acid + Strong Base titration => pH > 7 at equivalence point
Assume => HA = weak acid
=> HA + NaOH => NaA + H₂O
=> NaA => Na⁺ + A⁻ & A⁻ is the conjugate base of a weak acid HA
=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)
=> A⁻ + H₂O => HA + OH⁻ => Excess OH⁻ at equivalence pt => pH > 7
-------------------------------------
Weak Base + Strong Acid titration => pH < 7 at equivalence point
Weak Bases => ammonia (NH₃) or ammonia derivatives (RNH₂)* in water.
(ammonia in water) => :NH₃ + H₂O => NH₄OH ⇄ NH₄⁺ + OH⁻
(ammonia derivative in water) => RN:-H₂ + H₂O => RNH₃OH ⇄ RNH₃⁺ + OH⁻
Titration of weak base with strong acid ...
=> NH₄OH + HX => NH₄X + H₂O
=> NH₄X => NH₄⁺ + X⁻
=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)
=> NH₄⁺ + HOH ⇄ NH₄OH + H⁺ => Weak base is in molecular form with excess hydronium ions (H₃O⁺ = H⁺) at equivalence point => pH < 7.
----------------------
*RNH₂ is a primary amine used in the illustration, but the above process will also occur for secondary (R₂N:-H) and tertiary amines (R₃N:) in water also.
When 1.50g of Ba is added to 100g of water in a container open tothe atmosphere, the reaction shown below occurs and the temperatureof the resulting solution rises from 22 degrees to 33.10 degrees.If the specific heat of the solution is 4.18J/(g*C), calculatedelta H for the reaction, as written.Ba (s)+2H2O(l) yields Ba(OH)2(aq)+H2
Answer:
- 431.15 kJ/mol.
Explanation:
Firstly, we can calculate the amount of heat (Q) released by the solution using the relation:Q = m.c.ΔT,
where, Q is the amount of heat released from the solution (Q = ??? J).
m is the mass of solution (m = 1.5 g + 100 g = 101.5 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 33.1°C - 22°C = 11.1°C).
∴ Q = m.c.ΔT = (101.5 g)(4.18 J/g.°C)(11.1°C) = 4709.4 J.
To find ΔH:∵ ΔH = Q/n
no. of moles of Ba (n) = mass/atomic mass = (1.50 g)/(137.3270 g/mol) = 0.011 mol.
∴ ΔH = - Q/n = (4709.4 J)/(0.011 mol) = - 431.15 kJ/mol.
The negative sign is not from calculation, but it is an indication that the reaction is exothermic.
Calcium hydride (CaH2) reacts with water to form hydrogen gas: CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g) How many grams of CaH2 are needed to generate 48.0 L of H2 gas at a pressure of 0.995 atm and a temperature of 32 °C?
Answer:
40.g CaH2
Explanation:
1. ideal gas law(PV = nRT) → use ideal gas law first when volume is given
P = 0.995atm
V = 48.0L H2
n = ?
R = 0.0821L atm/molK
T = 32 + 273 = 305K
n = (0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) → do not simplify as small decimals might change the answer
2. Conversions
2 H2 and 1 CaH2 → 1/2
(mole of H2) x 1/2 x (molar mass of CaH2)
(0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) x 1/2 x (40.08 + 2.02) = 40.g CaH2
the longer answer will be 40.14887882 but as the minimum sigfig given in the question is 2, it is 40.g CaH2.
Hope it helped!
In which main energy level does the 'd' sublevel first appear? K (first main energy level) L (second main energy level) M (third main energy level) N (fourth main energy level)
Answer:
M (third main energy level)
Explanation:
The third main energy level bears the first appearance of the 'd' sublevel. The principal quantum number(n) depicts the main energy levels in which an orbital is located. It takes values of n=1,2,3,4,5..... and it can be represented by the shells k,l,m,n.......
The subshells in these main orbitals are represented by s,p,d and f. For the K shell, the principal quantum number is m and its sublevel notations are s,p and d. This is where the d-sublevel first appears.
The 'd' sublevel first appears in the third main energy level (M). This is because the electron configuration in atoms is organized into main energy levels and sublevels, which define an electron's distance from the nucleus and energy.
Explanation:The 'd' sublevel first appears in the third main energy level, also labeled as M. The electron configuration in atoms is organized into main energy levels and sublevels, which help define an electron's relative distance from the nucleus and its energy. The main energy levels are generally labeled K, L, M, N, etc., starting from the nucleus. Each energy level has one or more sublevels: s, p, d, f. The 's' sublevel appears in all main energy levels, the 'p' sublevel starts from the second (L), and the 'd' sublevel commences from the third (M) main energy level. Therefore, the 'd' sublevel does not exist in the first (K) or second (L) main energy levels.
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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:
A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your answer with 1 decimal place and no units (the understood unit in this problem is units).
Answer:
1,081.1 units of heparin should be given to 298 lb person.
Explanation:
We are given:
Weight of the person = 298 lb = 135.143 kg
Conversion factor used:
1 lb = 0.4535 kg
Number of unit of heparin to be given in an hour = 8.0 units/kg
Number of units given to the patient weighing 135.143 kg :
[tex]8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units[/tex]
1,081.1 units of heparin should be given to 298 lb person.
Classify each of these reactions. Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g)Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g) acid–base neutralization precipitation redox none of the above 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq)2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq) acid–base neutralization precipitation redox none of the above CaO(s)+CO2(g)⟶CaCO3(s)CaO(s)+CO2(g)⟶CaCO3(s) acid–base neutralization precipitation redox none of the above KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s)KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s) acid–base neutralization precipitation redox none of the above
The first reaction is a redox reaction, the second and third reactions are none of the above, and the fourth reaction is a precipitation reaction.
Explanation:The reactions asked in the question can be classified as follows:
Ba(ClO₃)₂(s)⟶BaCl₂(s)+3O₂(g): This is a redox reaction. It involves a transfer of electrons which is characterized by changes in oxidation states. Here, chlorine is reduced from +5 in ClO₃⁻ to -1 in Cl⁻, and oxygen is oxidized from -2 in ClO₃⁻ to 0 in O₂.2NaCl(aq)+K₂S(aq)⟶Na₂S(aq)+2KCl(aq): This is a type of double displacement reaction known as 'metathesis', but it can't be classified as acid-base neutralization, redox, or precipitation, so it would fall under none of the above.CaO(s)+CO₂(g)⟶CaCO₃(s): It's a combination reaction resulting in the formation of a single product, calcium carbonate. Given the options, this reaction would also be classified as none of the above.KOH(aq)+AgNO₃(aq)⟶KNO₃(aq)+AgOH(s): This is a precipitation reaction where soluble ions in solution react to form an insoluble product, AgOH(s), which precipitates out of solution.Learn more about Chemical reaction classification here:https://brainly.com/question/8117294
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The first reaction is an oxidation-reduction (combustion) reaction, the second reaction is a precipitation reaction, the third reaction is an acid-base neutralization reaction, and the fourth reaction is also an acid-base neutralization reaction.
Explanation:The first reaction, Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g), is a decomposition reaction also known as oxidation-reduction (combustion). The solid compound breaks down into a solid product and a gas.
The second reaction, 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq), is a precipitation reaction. The combination of two aqueous solutions forms an insoluble product.
The third reaction, CaO(s)+CO2(g)⟶CaCO3(s), is an acid-base neutralization reaction. The solid oxide reacts with a gas to form a solid carbonate.
The fourth reaction, KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s), is also an acid-base neutralization reaction. The aqueous solutions react to form a solid hydroxide and an aqueous salt.
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Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu Speed of light = 2.9979 × 10^8 m/s Calculate the nuclear binding energy (absolute value) of 3Li^6. which has an atomic mass of 6.015126 amu. J/mol.
Answer: The nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]
Explanation:
For the given element [tex]_3^6\textrm{Li}[/tex]
Number of protons = 3
Number of neutrons = (6 - 3) = 3
We are given:
[tex]m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu[/tex]
M = mass of nucleus = [tex](n_p\times m_p)+(n_n\times m_n)[/tex]
[tex]M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu[/tex]
Calculating mass defect of the nucleus:
[tex]\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol[/tex]
Converting this quantity into kg/mol, we use the conversion factor:
1 kg = 1000 g
So, [tex]0.032694g/mol=0.032694\times 10^{-3}kg/mol[/tex]
To calculate the nuclear binding energy, we use Einstein equation, which is:
[tex]E=\Delta mc^2[/tex]
where,
E = Nuclear binding energy = ? J/mol
[tex]\Delta m[/tex] = Mass defect = [tex]0.032694\times 10^{-3}kg/mol[/tex]
c = Speed of light = [tex]2.9979\times 10^8m/s[/tex]
Putting values in above equation, we get:
[tex]E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol[/tex]
Hence, the nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]
The binding energy of the lithium nucleus is 2.94 * 10^14 J/mol.
What is binding energy?The term binding energy refers to the energy that hold the nucleons in the atom together.
We know that the atomic mass of the Li is 6.015126 amu. Note that there are three protons and three neutrons. Hence;
Mass of protons= 3(1.00728 amu) = 3.02184
Mass of neutrons = 3(1.00866 amu) = 3.02598
Mass defect = (3.02184 + 3.02598) - 6.015126 amu = 0.032694 amu = 0.032694 g/mol
Now;
E = mc^2 = (0.032694 g/mol * (3 * 10^8)^2) = 2.94 * 10^14 J/mol
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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]).
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
Which of the following reactions will not occur as written? Which of the following reactions will not occur as written? Sn (s) + 2AgNO3 (aq) → 2Ag (s) + Sn(NO3)2 (aq) Zn (s) + Pb(NO3)2 (aq) → Pb (s) + Zn(NO3)2 (aq) Co (s) + 2HI (aq) → H2 (g) + CoI2 (aq) Mg (s) + Ca(OH)2 (aq) → Ca (s) + Mg(OH)2 (aq) Co (s) + 2AgCl (aq) → 2Ag (s) + CoCl2 (aq)
Answer: Option (4) is the correct answer.
Explanation:
According to the reactivity series, a more reactive metal has the ability to replace a less reactive metal in a chemical reaction.
It is known that calcium is more reactive than magnesium. So, in a chemical reaction magnesium can never replace calcium.
For example, [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex]
Therefore, this given reaction is not possible. But rest all given reactions are possible.
Thus, we can conclude that out of the given options [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex] reaction will not occur as written.
OPTION D.
The reaction Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written because in the activity series of metals, magnesium (Mg) is more reactive than calcium (Ca), and hence, Mg cannot displace Ca from its compound.
Explanation:In the given set of reactions, Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written. The explanation is based on activity series of metals. In this series, magnesium (Mg) is more reactive than calcium (Ca). So, Mg cannot displace Ca from its compound. In other words, a more reactive metal can displace a less reactive metal from its compound in a solution, but not the other way round. The other reactions would occur as represented since in each of them the free metal is more reactive than the one in the compound.
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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.
Directions: Using the definition of molarity, the given balanced equations, and stoichiometry, solve the following problems.
Sodium chloride solution and water react to produce sodium hydroxide and chlorine gas according to the following balanced equation: 2NaCl(aq) + 2H20(l) <-> 2Na0H(aq) + Cl2(g)
a. How many liters of 0.4 M sodium chloride do you need in order to have 3.0 moles of chlorine gas?
b. Find the number of moles of water needed to produce 3.0 L of chlorine gas at STP.
Answer:
15L of 0.40M NaCl(aq) solution
Explanation:
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)
2Na⁺(aq) + 2Clˉ(aq) + 2H₂O(l) → 2Na⁺(aq) + 2OHˉ(aq) + Cl₂(g)
Na⁺(aq) is a spectator ion in the given reaction and does not enter into the reaction process…
Net Ionic Equation is then 2Clˉ(aq) + 2H₂O(l) → 2OHˉ(aq) + Cl₂(g)
From Rxn, 2 moles Clˉ(aq) is needed to produce 1 mole of Cl₂(g)
Therefore, 6 moles Clˉ(aq) is needed to produce 3 moles of Cl₂(g)
That is, 6 moles NaCl(aq) → 6 moles Clˉ(aq) = 0.40M x V(NaCl)liters
V(NaCl) liters = 6 moles Clˉ(aq)/(0.40mole/liter) = 15 liters of 0.40M NaCl(aq)
7.5 liters of 0.4M sodium chloride solution are needed to produce 3.0 moles of chlorine gas, and 0.134 moles of water are needed to produce 3.0 liters of chlorine gas at STP.
Explanation:The question is related to molarity and stoichiometry. Let's solve part (a) and (b).
Part A:
Molarity is represented by moles of solute / liters of solution. The balanced equation gives a 1:1 ratio between NaCl and Cl2. The question gives us 3.0 moles of Cl2 gas, implying that we need the same moles of NaCl.
So, Moles = Molarity x Volume, Volume = Moles / Molarity, Volume = 3.0 moles / 0.4 M = 7.5 liters of NaCl solution are needed.
Part B:
For part (b), we see from the stoichiometry of the balanced reaction that the number of moles of water is equal to the number of moles of chlorine. According to the ideal gas law (PV=nRT), at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4L. So, 3.0L of Cl2 gas corresponds to 3.0 L / 22.4 L/mole = 0.134 moles. Therefore, you'll need 0.134 moles of water.
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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.
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
An amino acid A, isolated from the acid-catalyzed hydrolysis of a peptide antibiotic, gave a positive ninhydrin test and had a specific optical rotation (HCl solution) of 37.5 mL/(g·dm). Compound A was not identical to any of the 20 essential amino acids. The isoelectric point of compound A was found to be 9.4. Compound A can be prepared by the reaction of L-glutamine with Br2 in NaOH followed by acidification. Suggest a structure for A.
A possible structure for compound A is γ-Bromoglutamic acid (2-amino-4-bromobutanedioic acid).
What is this structure?
Ninhydrin test: A positive ninhydrin test indicates the presence of a primary or secondary amine. Glutamine has a primary amine, which would explain the positive test.
Specific optical rotation: The specific rotation (37.5 mL/(g·dm)) suggests chirality, which glutamine also possesses.
Not one of the 20 essential amino acids: This eliminates most common amino acids.
Isoelectric point (pI) of 9.4: This is consistent with γ-bromoglutamic acid, as the additional bromine group introduces a new acidic side chain (pKa ≈ 4.5) that lowers the pI compared to glutamine (pI ≈ 5.6).
Synthesis from L-glutamine and Br2/NaOH: This reaction is known to selectively brominate at the γ-carbon of glutamine, making it a plausible route for obtaining γ-bromoglutamic acid.
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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Which one of the following is an oxidation-reduction reaction?
NaOH + HNO3 --> H2O + KNO3
NaOH + HNO3 --> H2O + KNO3
SO3 + H2O --> H2SO4
CaCl2 + Na2CO3 --> CaCO3 + 2 NaCl
CH4 + 2 O2 --> CO2 + 2 H2O
Al2(SO4)3 + 6 KOH --> 2 Al(OH)3 + 3 K2SO4
Answer:
CH4 + 2 O2 --> CO2 + 2 H2O
Explanation:
CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...
O₂ → CO₂ + 4e⁻ ==> oxidation
O₂ + 4e⁻ → H₂O ==> reduction
2O₂ + 4e⁻ → CO₂ + H₂O + 4e⁻ ==> Net oxidation-reduction
=> 4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.
Then...
Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...
2CH₄ + 2O₂ + 8e⁻ → 2CO₂ + 2H₂O + 8e⁻
Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...
2CH₄ + 2O₂ → 2CO₂ + 2H₂O
Answer:
CH₄ + 2O₂ ⟶ CO₂ + 2H₂O
Explanation:
To identify an oxidation-reduction reaction, you must determine the oxidation number of every atom involved in the reaction and see if it changes.
The only reaction where two elements change oxidation number is the oxidation of methane.
Here's the oxidation number of every atom involved:.
[tex]\stackrel{\hbox{-4}}{\hbox{C}}\stackrel{\hbox{+1}}{\hbox{H}}_{4} +\stackrel{\hbox{0}}{\hbox{O}}_{2} \, \longrightarrow \, \stackrel{\hbox{+4}}{\hbox{C}}\stackrel{\hbox{-2}}{\hbox{O}}_{2} + \stackrel{\hbox{+1}}{\hbox{H}}_{2}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]
We see that some elements change oxidation numbers.
C: -4 ⟶ +4; increase in oxidation number = oxidation
O: 0 ⟶ -2; decrease in oxidation number = reduction
H: +1 ⟶ +1; no change.
The reaction is an oxidation-reduction reaction, because carbon is oxidized, and oxygen is reduced.
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.
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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When 10.g of CH3COOH is combusted in a sealed calorimeter, it releases enough energy to heat 2000. g of water from 23.5 °C to 34.3 °C. a. Calculate the energy released per 10 g of CH3COOH. b. Calculate the energy released per mole of CH3COOH.
Answer:
a. 90.288 kJ.
b. - 54.06 kJ/mol.
Explanation:
a. Calculate the energy released per 10 g of CH₃COOH.
We can calculate the amount of heat (Q) released to water using the relation:Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = ??? J).
m is the mass of water (m = 2000.0 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 34.3°C - 23.5°C = 10.8°C).
∴ Q = m.c.ΔT = (2000.0 g)(4.18 J/g.°C)(10.8°C) = 90288 J = 90.288 kJ.
b. Calculate the energy released per mole of CH₃COOH.
To find ΔH:∵ ΔH = Q/n
no. of moles of CH₃COOH (n) = mass/atomic mass = (10.0 g)/((60.052 g/mol) = 0.167 mol.
∴ ΔH = - Q/n = - (90.288 kJ)/(0.167 mol) = - 54.06 kJ/mol.
The negative sign is not from calculation, but it is an indication that the reaction is exothermic.
(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) Suppose 0.750 L of 0.480 M H2SO4 is mixed with 0.700 L of 0.290 M KOH. What concentration of sulfuric acid remains after neutralization?
These are two questions and two answers
Answer:
Question 1:
H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)Question 2:
0.201 MExplanation:
Question 1:
The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.
The products of an acid-base reaction are salt and water.
This is the sketch of such neutralization reaction:
1) Word equation:
sulfuric acid + potassium hydroxide → potassium sulfate + water↑ ↑ ↑ ↑
acid base salt water
2) Skeleton equation (unbalanced)
H₂SO₄ + KOH → K₂SO₄ + H₂O#) Balanced chemical equation (including phases)
H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answerQuestion 2:
1) Mol ratio:
Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:
1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)2) Moles of H₂SO₄:
V = 0.750 literM = 0.480 mol/literM = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol3) Moles of KOH:
V = 0.700 literM = 0.290 mol/literM = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol4) Determine the limiting reagent:
a) Stoichiometric ratio:
1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH
b) Actual ratio:
0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH
Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.
5) Amount of H₂SO₄ that reacts:
Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒
x = 0.203 / 2 = 0.0677 mol of H₂SO₄
6) Concentration of H₂SO₄ remaining:
Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 molTotal volume = 0.700 liter + 0.750 liter = 1.450 literConcetration = MM = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer
The balanced neutralization reaction between H2SO4 and KOH is H2SO4 + 2KOH → K2SO4 + 2H2O. To calculate the concentration of sulfuric acid remaining after neutralization, use the moles of KOH reacted and the mole ratio in the balanced equation.
Explanation:(a) The neutralization reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) in aqueous solution can be represented as:
H2SO4 + 2KOH → K2SO4 + 2H2O
(b) To determine the concentration of sulfuric acid remaining after neutralization, we need to calculate the moles of KOH reacted using the given volumes and concentrations. Since H2SO4 has a 1:2 mole ratio with KOH in the equation, half of the moles of KOH reacted will be the moles of sulfuric acid neutralized. Therefore, the concentration of sulfuric acid remaining can be calculated by subtracting the moles of sulfuric acid neutralized from the initial concentration.
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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]
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.
You want to remove as much CO2 gas as possible from a water solution. Which of the following treatments would?
a. cool the solution
b. filter the solution
c. boil the solution
d. aerate the solution
Are hydrocarbons approved for retrofit applications
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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