Based on the values of δH°rxn, reactions can be classified as thermodynamically favorable or unfavorable. Reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.
Explanation:Based on the values of δH°rxn, we can determine whether a reaction is thermodynamically favorable or unfavorable. A negative δH°rxn indicates a thermodynamically favorable reaction, while a positive δH°rxn indicates a thermodynamically unfavorable reaction. Let's analyze the given reactions:
ag+(aq) + li(s) → ag(s) + li+(aq): δH°rxn = -384.4 kJ/mol (negative, thermodynamically favorable)fe(s) + 2na+(aq) → fe2+(aq) + 2na(s): δH°rxn = 392.3 kJ/mol (positive, thermodynamically unfavorable)2k(s) + 2h2o(l) → 2koh(aq) + h2(g): δH°rxn = -393.1 kJ/mol (negative, thermodynamically favorable)Therefore, reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.
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Calculate the freezing point of a solution that contains 8.0 g of sucrose (c12h22o11) in 100 g of h2o. kf for h2o = 1.86c/m
The freezing point is the temperature at which the fluid freezes to a solid form. The freezing point of the solution is -0.435 degrees celsius.
What is the freezing point?The freezing point is the product of the molality, van 't Hoff factor, and the cryoscopic constant. It is given as,
[tex]\rm \Delta T_{F} = K_{F} \times b\times i[/tex]
Given,
Mass of water = 0.1 kg
Mass of sucrose = 8.0 gm
Moles of sucrose are calculated as:
[tex]\begin{aligned}\rm moles &= \dfrac{8.0}{342.3}\\\\&= 0.0233 \;\rm mol \end{aligned}[/tex]
The molality of sucrose is calculated as:
[tex]\begin{aligned}\rm b &= \dfrac{\text{moles of sucrose}}{\text{mass of water}}\\\\&= \dfrac{0.0233 \;\rm mol}{0.1}\\\\&= 0.233 \;\rm m\end{aligned}[/tex]
The freezing point depression is calculated as:
[tex]\begin{aligned}\rm \Delta T &= 0.233 \;\rm m \times 1.86\; ^{\circ} \;\rm C/m\\\\&= 0.435 ^{\circ}\;\rm C\end{aligned}[/tex]
Therefore, the freezing point of a solution is -0.435 degrees celsius.
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help again pleaseee!!
The three beakers shown below contain solutions of [cof6]3–, [co(nh3)6]3+, and [co(cn)6]3–. based on the colors of the three solutions, which compound is present in each of the beakers?
The colors of the compounds in the beakers could potentially be: [CoF6]3– (green), [Co(NH3)6]3+ (yellow-orange), and [Co(CN)6]3– (red), although colors can vary based on different conditions.
Explanation:The question asks us to identify the compounds present in three beakers based on the color of the solution. While this is generally impossible to answer with absolute certainty without more information or additional tests, we can make a fair guess based on some known color-characteristics of these compounds.
[CoF6]3– is generally green due to the color of the Cobalt(III) ion.
[Co(NH3)6]3+ is usually yellow to orange in color.
FInally, [Co(CN)6]3– commonly has a strong red color.
Note that in reality, actual colors can vary depending on concentration, temperature, and other factors.
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What is the limiting reactant if 0.5 g Al is reacted with 3.5 g CuCl2?
What is the mass of 2.30 moles of magnesium chloride, mgcl2?
Please help kind of confused here.
In the early 1900s, Henry Gwyn-Jeffreys Moseley added to the development of the modern periodic table. Which answer best explains how Moseley started his investigations?
1. by reviewing the work of those who contributed to the previous versions of the table
2. by making an entirely new organizational scheme for the previous versions of the table
3. by analyzing the work of a select group of contributors to the previous versions of the table
4. by rejecting the work of all those who had previously developed the earlier versions of the table
Answer:
The correct answer is 1. Moseley started his investigations by reviewing the work of those who contributed to the previous versions of the table.
Explanation:
Henry Gwyn Jeffreys Moseley was a British physicist. His most important scientific contribution was the proof of the correctness of the concept of atomic numbers in chemistry.
In 1913, using X-ray spectroscopy, he found a systematic relationship between the wavelength and the atomic number. Previously, it was assumed that the atomic number was an arbitrary number based on the order of the atomic masses, but had to be changed to bring an element into the right place in the periodic table. Moseley's discovery showed that atomic numbers had an experimentally measurable basis. In addition, he showed that there were gaps in numbers 43, 61 and 75 (known today as the radioactive elements technetium and promethium, as well as the stable but rare rhenium). His work was further proof of the then controversial atomic theory.
If I cut the height of an object in half, what will happen to its GPE?
A. Not enough info
B. Double
C. Disappear
D. Halve
If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.
Explanation:If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.
GPE is determined by the height of the object and its mass. When you cut the height in half, the GPE is reduced by half because the potential energy is directly proportional to the height.
For example, if a book has a certain GPE at a certain height, cutting the height in half will result in the book having half the GPE it had before.
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Which change in oxidation number represents reduction?
A) –1 to +1
B) –1 to –2
C) –1 to +2
D) –1 to 0
James is looking at a parallel circuit plan for lighting. There is a battery providing the power. There are switches labeled A,B,C,D that can be turned on to close the circuit. Which switch does not have to be on for light 3 to function?
For light 3 to function, switch D does not have to be on. This diagram shows a parallel circuit that provides more than one way for the current to return to the power source.
Glacier and erosion in a sentence
Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(aq, 0.0155 m)∥∥ag+(aq, 3.50 m)||ag(s) net cell equation: sn +2ag^{+}->sn^{2+} +2ag sn+2ag+⟶sn2++2ag special δσω λμπ reset ( ) [ ] xyxyyyx⟶↽−−⇀ • (s) (l) (aq) (g) calculate e∘cell, δg∘rxn, δgrxn, and ecell at 25.0 ∘c, using standard potentials as needed
Answer:
E°cell = 0.94 V
Ecell = 1.00 V
ΔG = -1.9 × 10⁵ J
ΔG° = -1.8 × 10⁵ J
Explanation:
Let's consider this electrochemical cell:
Sn(s)|Sn²⁺(aq,0.0155M)||Ag⁺(aq, 3.50M)|Ag(s)
The corresponding half-reactions are:
Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻ E°red = -0.14 V
Reduction (cathode): 2 Ag⁺(aq) + 2 e⁻ → 2 Ag(s) E°red = 0.80 V
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an = 0.80 V - (-0.14 V) = 0.94 V
We can find the cell potential using the Nernst equation.
Ecell = E°cell - (0.05916/n) . log Q
Ecell = 0.94 V - (0.05916/2) . log ([Sn²⁺]/[Ag⁺]²)
Ecell = 1.00 V
We can find ΔG and ΔG° using the following expressions.
ΔG = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 × 10⁵ J
ΔG° = -n.F.E°cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 × 10⁵ J
The net cell equation for the electrochemical cell is Sn(s) + 2Ag+ -> Sn₂+ + 2Ag. To find the standard cell potential, standard free energy change, free energy change, and cell potential at 25.0 °C, we use standard reduction potentials and the Nernst equation. Standard free energy change is calculated with ΔG°rxn = -nFE°cell, while the cell potential under non-standard conditions is found using the Nernst equation.
Explanation:The net cell equation for the given electrochemical cell is Sn(s) + 2Ag+ (aq) → Sn₂+ (aq) + 2Ag(s). To calculate the standard cell potential (E°cell), standard free energy change (ΔG°rxn), free energy change (ΔGrxn), and the cell potential (Ecell) at 25.0 °C, we can use the standard reduction potentials and the Nernst equation. The standard cell potential is calculated by subtracting the standard reduction potential of the anode from that of the cathode.
The standard free energy change can be calculated from the standard cell potential using the formula ΔG°rxn = -nFE°cell, where n is the number of moles of electrons transferred in the reaction, and F is the Faraday constant. The cell potential under non-standard conditions (Ecell) can be determined using the Nernst equation, which incorporates the concentration of the ionic species involved in the half-reactions.
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Which element has the electron configuration shown above
Click the "draw structure" button to launch the drawing utility. draw any one of the skeletal structures of a 2° alkyl bromide having the molecular formula of c6h13br and two stereogenic centers. indicate chirality by using wedge and hashed wedge notation. lone pairs do not need to be shown.
Identify the nuclide produced when uranium-238 decays by alpha emission: 238 92u→42he + ?
The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This process involves the emission of an alpha particle, which causes uranium-238 to lose 2 protons and 2 neutrons, resulting in thorium-234.
Explanation:The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This decay process involves the release of an alpha particle from uranium-238 nucleus. An alpha particle is equivalent to a helium nucleus - it contains 2 protons and 2 neutrons. Therefore, when uranium-238 (which has 92 protons and 146 neutrons) emits an alpha particle, it loses 2 protons and 2 neutrons, transforming into a new element with 90 protons and 144 neutrons, which is thorium-234 (234 90Th).
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Answer how many moles of sodium will react with 2.6l of cl2 gas at 1.15 atm
What is the oxidation state for the oxygen atom in na 2 o 2 ?
What is the molar concentration of a phosphoric acid (h3po4) solution if 55.5 ml of the solution requires 35.5 ml of .150 m naoh to completely react with the phosphoric acid?
Acetyl-coa and succinyl-coa are both high-energy thioesters, but their chemical energy is put to different uses. explain how the chemical energy is used in the two cases.
The half-life of cobalt-60 is 5.20 yr. how many milligrams of a 2.000-mg sample remain after 10.50 years?
Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenediaminecobalt(iii)
When the concentration of a is doubled,the rate for the reaction:2a+b → 2cquadruples.whentheconcentrationofbisdoubledtherateremainsthesame.whichmechanismbelowisconsistentwiththeexperimentalobservations?
Choose all the answers that apply. Muscles in the small intestine are _____. striated voluntary involuntary smooth cardiac
Answer:
The answer is smooth and Involuntary.
Explanation:
Smooth muscles are involuntary which means they automatically do a function without the brain telling it to.
Name the salt k4[pt(co3)2f2] given that the carbonate ion acts as a monodentate ligand in the complex. the oxidation number of platinum is +2. enter the name.
Answer:
Potassium dicarbonatedifluoroplatinate (II)
Explanation:
Hello,
Based on the IUPAC rules, the given compound is a complex called potassium dicarbonatodifluoroplatinate (II) as long as the carbonate ion is present twice as a monodentate ligand therefore it is preceded by a "di" prefix. In addition, two fluorines are also present with the "di" prefix as well as a platinum which complete the anionic section including +2 as the platinum oxidation state.
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Consider the following reversible reaction. mc030-1.jpg What is the equilibrium constant expression for the given system?
If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after the reaction has reached completion? 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(g)
The answer is:
the mass of the excess reactant leftover after the reaction has reached completion is 90.135 grams
The explanation:
According to the reaction equation:
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)
when m is the mass of C₂H₆ m(C₂H₆) = 152 g
So we need to get the number of moles of C₂H₆
n(C₂H₆) = mass C₂H₆ / molar mass of C₂H₆(M)
and when the molar mass of C₂H₆ = 30 g/mol
so, by substitution:
n(C₂H₆) = m(C₂H₆) / M(C₂H₆).
n(C₂H₆) = 152 g / 30 g/mol.
n(C₂H₆) = 5.067 mol.
Then
when the mass of O₂ m(O₂) = 231 g
so we need to get the number of moles of O₂
when nO₂ = mass O₂/ molar mass of O₂
when molar mass of O₂ = 32 g /mol
So, by substitution:
n(O₂) = 231 g / 32 g/mol.
n(O₂) = 7.218 mol
So O₂ is the limiting reactant
according to the chemical reaction we can get the molar ratio between the O₂and C₂H₆:
n O₂ : n C₂H₆ → 7.218 : n C₂H₆
7 : 2 7 : 2
∴ n(C₂H₆) = 2 * 7.218 mol / 7
∴ n(C₂H₆) = 2.0625 mol
The number of moles remaining n(C₂H₆) = 5.067 mol - 2.0625 mol
∴ n (C₂H₆) = 3.0045 mol
So the mass remains = moles remains * molar mass of C₂H₆
∴ m (C₂H₆) = 3.0045 mol * 30 g/mol = 90.135 g
Complete the following table.
Acid Molarity Moles of H⁺ released per liter
HCl 1 ____
H2SO4 1 ____
H3PO4 1 ____
H2SO4 0.5 ____
H3PO4 3 ____
HNO3 2 ____
Complete the following table.
Acid Molarity Moles of H⁺ released per liter
HCl 1
H2SO4 1
H3PO4 1
H2SO4 0.5
H3PO4 3
HNO3 2
[Answer]
1
2
3
1
9
2
What is the molarity of 500 ml of a solution containing 85.0g AgNO3?
Answer: 1 M
Explanation:
Molarity : It is defined as the number of moles of solute present per liter of the solution.
Formula used :
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{85.0g}{170g/mol}=0.5moles[/tex]
[tex]V_s[/tex] = Volume of solution in ml
Now put all the given values in the formula of molarity, we get
[tex]Molarity=\frac{0.5moles\times 1000}{500ml}=1mole/L[/tex]
Therefore, the molarity of solution will be 1 M.
Consider the electrolysis of molten barium chloride (bacl2). (a) write the half-reactions. include the states of each species..
Molten barium
chloride is separetes:
BaCl₂(l) →
Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted
because of heat:
BaCl₂(l) →
Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive and the cathode is negative.
The electrolysis of molten barium chloride involves reduction of barium ions and oxidation of chloride ions, requiring two electrons to move through the circuit for one unit of the reaction.
Explanation:During the electrolysis of molten barium chloride (BaCl2), barium ions (Ba2+) are reduced at the cathode, and chloride ions (Cl-) are oxidized at the anode. The half-reactions, showing the movement of electrons through the circuit, are as follows:
Cathode (reduction): Ba2+(l) + 2e- → Ba(l)Anode (oxidation): 2Cl-(l) → Cl2(g) + 2e-The overall cell reaction is obtained by combining the half-reactions and balancing the electrons:
Ba2+(l) + 2Cl-(l) → Ba(l) + Cl2(g)
For each mole of barium ions reduced, two electrons are involved in the transfer. Similarly, for each mole of chlorine gas produced, two electrons are given up by chloride ions. As such, there would be two electrons that moved through the circuit for one unit of the reaction to occur.
. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?
A student does not filter his/her saturated solution before titrating. will the calculated ksp probably be too high, too low, or unaffected? why?
Not filtering a saturated solution before titrating will likely cause the calculated Ksp to be a) too high
When a student does not filter their saturated solution before titrating, the calculated Ksp (solubility product constant) will likely be affected.
Specifically, the Ksp will probably be too high.
This is because the undissolved solid present in the unfiltered solution will falsely contribute to the ionic concentration being measured during the titration, leading to an overestimation of the dissolved ions and thus a higher calculated Ksp.
As a result, to ensure accurate determination of the Ksp, it is essential to filter the solution to remove any undissolved solute before conducting the titration.
Complete question is - A student does not filter his/her saturated solution before titrating. will the calculated ksp probably be (a) too high, (b) too low, or (c) unaffected? why?