Answer:
The correct answer is -341.2 kJ per mole.
Explanation:
The reaction given is:
CO (g) + Cl₂ (g) ⇔ COCl₂ (g)
Kp = 5.62 × 10³⁵
T = 25 °C or 298 K
The formula for calculating ΔG is,
ΔG° = -RTlnKp
ΔG° = -8.314 × 298 ln (5.62 × 10^35)
ΔG° = -203.9 kJ/mol
ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)
ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)
ΔG°f (COCl₂ (g)) = -341.2 kJ/mol
The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
The given equation for the chemical reaction is
CO(g) + Cl2(g) ⇌ COCl2(g)
At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]
Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:
[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]
where;
gas constant (R) = 8.314 × 10⁻³ kJ/K.mol∴
[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]
[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]
[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]
Thus, the standard free energy for the reaction is 203.95 kJ/mol
For a given reaction, the standard Gibbs free energy can be calculated by using the formula:
[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]
[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]
replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:
[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]
Collecting like terms, we have:
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]
Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
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what js the percent yield of lithium hydroxide from a reaction of 7.40 g of lithium with 10.2 g of water? the actual yield was measured to be12.1 g
Answer:
89%.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2Li + 2H2O —> 2LiOH + H2
Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.
This is illustrated below:
Molar mass of Li = 7 g/mol
Mass of Li from the balanced equation = 2 x 7 = 14 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol
Mass of LiOH from the balanced equation = 2 x 24 = 48 g
Summary:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O.
Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.
From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.
Therefore, H2O is the limiting reactant and Li is the excess reactant.
Next, we shall determine the theoretical yield of LiOH.
In this case we shall use the limiting reactant.
The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:
From the balanced equation above,
36 g of H2O reacted to produce 48 g of LiOH.
Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.
Therefore, the theoretical yield of LiOH is 13.6 g
Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:
Actual yield = 12.1 g
Theoretical yield = 13.6 g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 12.1/ 13.6 x 100
Percentage yield = 89%
Therefore, the percentage yield LiOH is 89%.
Consider the following practical aspects of titration.
(a) how can you tell when nearing the end point in titration?
(b) What volume of NaOH is required to permanently change the indicator at the end point?
(c) If KHP sample #1 requires 19.90 mL of NaOH solution to reach an end point, what volume is required for samples #2 and #3?
(d) if vinegar sample #1 requires 29.05 mL of NaOH solution to reach an endpoint, what volume is required for samole #2 and #3?
Answer:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
Explanation:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb2O E) NaF and SrO
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
In which pair do both compounds exhibit predominantly ionic bonding?
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
What is Avogadro's number?
Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2
Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
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Please help, Which type of molecule is shown below?
Answer:
Carbohydrate
Explanation:
A.pex
11. The mass (in grams) of FeSO4.7H2O required for preparation of 125 mL of 0.90 M
solution is:
(a) 16 g
(b) 25 g
(c) 13 g
(d) 31 g
(e) 43 g
Answer:
what does little birdie say in the birth of their differences lak lak lak nu pasand aayi baby sleep are no longer children all strong industry all strong baby to show the meaning of rice is here to get up from sleep meaning of lips is Hasan let the mother is saying the baby to sleep in a new
Taking into account the definition of molarity and the molar mass of the compound, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
In first place, you have to know tha molarity is a measure of the concentration of that substance that indicates the number of moles of solute present in the solution.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.
[tex]molarity=\frac{number of moles of solute}{volume of solution}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case, you know:
molarity= 0.90 Mnumber of moles of solute= ?volume= 125 mL= 0.125 L (being 1000 mL=1 L)So, by definition of molarity, the number of moles is calculated as:
[tex]0.90 M=\frac{number of moles of solute}{0.125 L}[/tex]
Solving:
number of moles of solute= 0.90 M× 0.125 L
number of moles of solute= 0.1125 moles
On the other side, molar mass is the mass of one mole of a substance, which can be an element or a compound. In this case, the molar mass of FeSO₄.7H₂O is 277.85 [tex]\frac{g}{mole}[/tex].
Then you can apply the following rule of three: if by definition of molar mass, 1 mole of the compound contains 277.85 g, 0.1125 mole contains how much mass?
[tex]mass=\frac{0.1125 moles*277.85 g}{1 mole}[/tex]
Solving:
mass ≅ 31 g
Finally, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
What is buffers and mention its importance?
Answer:
Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.
Explanation:
Buffers are use in the system to maintain the value of pH, and the contain the pH value is not to change.Buffer maintain the body of pH for the optimal activity,and they are solution of pH constant.Buffer in used in the lab and that to maintain growth of the micro tissues and the culture media.Buffer are used in maintain necessary optimal reaction activity,determine the indicator of solution with pH.Buffer capacity is that concentration to the buffering agent, is the very small increase,buffer capacity to the pH is 32% , of the maximum value of pH.Buffers in a acid regions to the desired of that value to the particular buffer agent.Buffers can be made from that a mixture of the base and acid, buffer can be a wide range of the obtained.Buffers that the pH calculation and they required to performed in the critic acid that the overlap over the buffer range.Describe how you would prepare 500ml of 40% (w/v) aqueous iodine solution.
[Atomic mass of iodine =127g/mol].
Answer:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Explanation:
Hello,
In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:
[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]
In such a way, we need to find mass of iodine, which is computed as:
[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]
Thereby, the procedure will be:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Best regards.
How many moles of oxygen are in 49.2 grams of carbon dioxide?
Answer:
Number of mole in Oxygen = 2.24 mol
Explanation:
Given:
Amount of carbon dioxide (CO₂) = 49.2 gram
Find:
Number of moles in Oxygen
Computation:
Molecular weight of CO₂ = 44 gram (Approx)
Number of mole in CO₂ = 49.2 / 44
Number of mole in CO₂ = 1.11818182 mol
CO₂ has 2 mole of Oxygen,
So,
Number of mole in Oxygen = 2 × 1.11818182
Number of mole in Oxygen = 2.23636364
Number of mole in Oxygen = 2.24 mol
In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to
Answer:
Zinc- anode
Copper- cathode
Sodium sulphate- salt bridge
Explanation:
A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.
In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e
Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)
Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.
Answer:
zinc=anode
copper=cathode
Explanation:
If a sample of C-14 initially contains 1.6 mmol of C-14, how many millimoles will be left after 2250 years
Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
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03/08/2020
Question
1. (a) State the definition of a chemical formula.
(b) What does it tell about a compound?
(c) What information is conveyed by the formulation H2SO4?
Explanation:
According to your question,
no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.
b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.
c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).
Hope it helps...
The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
The front curve of a spectacle lens is called?
Answer:
Corrective lense or just lens.
Explanation:
One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.
Answer:
a. the Kb of NO₂⁻ to find the hydroxide concentration.
Explanation:
When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.
NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)
Na⁺ comes from NaOH (strong base) so it doesn't react with water.
NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to find the hydroxide concentration.
What type of chemist exclusively studies most carbon compounds?
-biochemist
-physical chemist
-organic chemist
-inorganic chemist
Answer:
Organic chemist? I do not know.
Explanation:
Thanks you.
The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.
What is an organic chemist ?The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.
Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.
Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.
According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.
Thus, option C is correct.
To learn more about an organic chemist, follow the link;
https://brainly.com/question/1674101
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Consider this synthesis of isoamyl acetate based on this week's experimental methods, after refluxing the reaction mixture for 25 minutes, what is likely present in solution
Answer:
acetic acid and phosphoric acid
Explanation:
After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,
there will be no reactant ( 3-methylbutanol )left in the reaction mixture
Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4
Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
Read the article. Use your understanding to answer the questions that follow. What type of source is this article? primary or secondary and how do you know
Answer: C
Explanation:
The article was sourced from the Oak National Laboratory
Which reasons did you include in your response? Check all of the boxes that apply.
1. The article does not present original research.
and
3. The article has references to primary sources.
Answer:
C
Explanation:
Which reasons did you include in your response? Check all of the boxes that apply.
The article does not present original research.
The article summarizes other research.
The article has references to primary sources.
