Final answer:
False statements include: a false notion that ∆Ssys must always be positive for a spontaneous reaction, and a misunderstanding about ∆Euniv increasing in spontaneous reactions. True comprehension of spontaneity involves ∆H and ∆S, affecting Gibbs free energy (∆G), where a negative ∆G indicates spontaneity.
Explanation:
Among the given statements regarding spontaneous chemical reactions, we need to identify those that are false. Here’s the analysis:
a) It is false that if a chemical reaction is spontaneous then the ∆Ssys must always be positive. Spontaneity can also depend on the temperature and ∆Hsys (enthalpy change) of the system, not just ∆Ssys (entropy change).
b) It is true that if a chemical reaction with a negative ∆Ssys is spontaneous, then ∆Ssurr must be positive. This balances the entropy change leading to a positive ∆Suniv (entropy change of the universe), which is a requirement for spontaneity.
c) The statement that if a chemical reaction is spontaneous then the ∆Euniv must be increasing is false. The energy of the universe remains constant; what matters for spontaneity is the entropy change of the universe (∆Suniv), not energy change.
d) The idea that we don't need to worry about spontaneous reactions because they are rare is false. Many important chemical and physical processes are spontaneous, including combustion, corrosion, and the melting of ice at room temperature.
The correct understanding of spontaneity in reactions requires knowledge of both enthalpy (∆H) and entropy (∆S) changes and their effects on the Gibbs free energy (∆G). Generally, for a process to be spontaneous, ∆G must be negative.
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2SO3(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: SO42−(aq)+Sn2+(aq)+ –––→H2SO3(aq)+Sn4+(aq)+ ––– Part B What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.
Answer:
The final balanced equation is :
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
Explanation:
[tex]SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq) [/tex]
Balancing in acidic medium:
First we will determine the oxidation and reduction reaction from the givne reaction :
Oxidation:
[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)[/tex]
Balance the charge by adding 2 electrons on product side:
[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-[/tex]....[1]
Reduction :
[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq) [/tex]
Balance O by adding water on required side:
[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]
Now, balance H by adding [tex]H^+[/tex] on the required side:
[tex]SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]
At last balance the charge by adding electrons on the side where positive charge is more:
[tex]SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]..[2]
Adding [1] and [2]:
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
The final balanced equation is :
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
The coefficients of the reactants and products in the balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution are 1, 1, 10, 1, 1, 5.
Explanation:The balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution can be written as:
SO₄²⁻(aq) + Sn₂⁺(aq) + 10H⁺(aq) → H₂SO₃(aq) + Sn₄+(aq) + 5H₂O(l)
The coefficients of the reactants and products in the balanced equation are 1, 1, 10, 1, 1, 5 respectively.
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Consider the two reduction half-reactions: Na+(aq) + e− LaTeX: \longrightarrow⟶ Na(s) Eo = −2.71 V Cl2(g) + 2 e− LaTeX: \longrightarrow⟶ 2 Cl−(aq) Eo = 1.36 VUse the electrode potentials above to calculate Eocell and ∆Gorxn for the reaction below, and determine if it is the reaction for a voltaic cell or an electrolytic cell. 2 Na+(aq) + 2 Cl−(aq) LaTeX: \longrightarrow⟶2 Na(s) + Cl2(g)
The cell potential (Eocell) of the given reaction is 4.07V and the Gibbs free energy(∆Gorxn) is -785.42 KJ/mol suggesting the reaction occurs spontaneously and the cell in consideration is a voltaic cell.
Explanation:To calculate the cell potential (Eocell), we use the formula Eocell = Eocathode - Eoanode. Here, for the reaction 2 Na+(aq) + 2 Cl−(aq)⟶2 Na(s) + Cl2(g), Na+ is being reduced to Na and Cl- is being oxidized to Cl2, therefore, Cl2 is the anode and Na+ is the cathode. So, the Eocell should be 1.36 V - (-2.71 V) = 4.07 V.
To calculate the Gibbs free energy (∆Gorxn), we use the formula ∆Gorxn = -n F Eocell, where n stands for the number of moles of electrons transferred in the redox reaction (which is 2) and F is the Faraday's constant (96500 C/mol). Therefore, ∆Gorxn = - (2 mol * 96500 C/mol * 4.07 V) = -785420 J/mol or -785.42 KJ/mol.
Since the E0cell value is positive, the reaction will occur spontaneously and represents a voltaic (or galvanic) cell which is a type of electrochemical cell where spontaneous redox reactions are used to convert chemical energy into electrical energy.
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completethe mass in grams of sample of carbon containing 68 atoms?
Answer:
1.356x10⁻²¹ grams.
Explanation:
mass of carbon containing 68 atoms = 12g Carbon/6.02x10²³atoms = 1.993x10⁻²³ g/atom x 68 atoms = 1.356x10⁻²¹ grams.
Draw the arrow-pushing mechanism of a generic esterification reaction: b) From the spectral data (NMR, IR, MS) you were given, identify the structure of your ester product. c) Based on your answer to part b), what is the structure of your starting alcohol? 2) Critical analysis (6 points): a) Fully assign the 1H NMR spectrum of your product (i.e. determine which peaks in the 1H NMR correspond to which hydrogens in the product). You will not receive full marks for determination of the unknown unless you assign the 1H NMR spectrum completely. b) You will notice that we gave you the m/z value of the M+ ion of your product on the EI (electron impact) mass spectrum, but the actual peak on the spectrum is very small (or even non-existent). Read Solomons chapter 9 and explain why the M+ peak is so small.
Answer:
See explaination for detailed answer
Explanation:
In the IR spectrum, the broad peak at 3322 cm-1 corresponds to OH stretching while the peaks at 2929-2961 cm-1 correspond to C-H stretching. Thus the presence of alcohol is evident in the IR spectrum.
The 13C NMR suggests the presence of seven C-atoms in this ester. The peak corresponding to carbonyl carbon appear most downfield at ~172 ppm. The other six peaks are in the aliphatic region suggesting an aliphatic ester.
In the 1H NMR, we see a singlet at 2.0 ppm with the integral value of 3. This singlet is characteristic to the protons of the acetate (CH3CO) group as seen in ethyl acetate. This suggests that the acetic acid was employed in this esterification reaction. Using this information along with what we know from 13C NMR we can be certain that the given alcohol contains 5 carbons (total 7 carbon – 2 carbon from acetate group). Therefore the starting material must be pentyl alcohol.
The 1H NMR peaks for the pentyl group are
The most downfield triplet at 4.1 ppm corresponding to OCH2. This is due to deshielding of the CH2 by the electronegative O-atom
The most upfield triplet at 0.9 ppm corresponding to CH3.
A multiplet at 1.3 ppm corresponding to CH2 CH2 which is attached toCH3 moiety
A pentate at 1.6 ppm corresponding to CH2 which is attached toOCH3 moiety
Therefore, the given alcohol is n-pentyl alcohol and the ester is pentyl acetate (Molar mass 130.0994)
See attachment for diagram
Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALYSIS AND CONCLUSIONS FOR DISCUSSION1. Calculate the [OH-] from the results of your titrations. Explain your calculations2. Calculate the [Ca2+]. Explain your calculations3. Calculate the Ksp for calcium hydroxide. Explain your calculations4. Temperature affects equilibrium in one direction or the other (depending on whether the system is Exo or Endothermic). Discuss5. What does the value of Ksp tell you in terms of equilibrium?6. The theoretical value for Ksp of Ca(OH)2 at 25°C is 9.0 X 10-6, compare and suggest reasons for the difference.
Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
[tex]K_{sp}[/tex] = [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
[tex]K_{sp}[/tex] = [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
[tex]K_{sp}[/tex] value describes the solubility of a particular ionic compound. The higher the [tex]K_{sp}[/tex] value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility
1. A scientist investigating the cell membrane concludes that a particular substance is transported
by active transport. Which observation would support this conclusion?
The substance moves along the concentration gradient toward a region of lower
concentration
O B. The transport of the substance into the cell is reduced or reversed in the absence of ATP.
The direction of transport changes when the external solution is changed from a
hypertonic solution to a hypotonic solution.
Transport proteins embedded in the plasma membrane assist in the transport of the
substance into the cell.
O
What’s the answer
Answer
The transport of the substance into the cell is reduced or reversed in the absence of ATP. I think........
Explanation:
In the molecule ClF5, chlorine makes five covalent bonds. Therefore, five of its seven valence electrons need to be unpaired. The orbitals with the same energy are known as degenerate orbitals. For example, the p subshell has three degenerate orbital, namely, px, py, and pz. How many degenerate orbitals are needed to contain seven electrons with five of them unpaired
Answer:
6 orbitals
Explanation:
Each orbital holds 2 electrons so for 5 of 7 orbitals to have unpaired electrons, five orbitals will have one electron and one orbital will have two electrons making 7 total electrons and 6 total orbitals.
The molecule ClF5 needs five degenerate orbitals to house seven electrons, with five of them unpaired. This set of orbitals becomes the five sp³d hybrid orbitals, similar to phosphorus pentachloride, created from the hybridization of atomic orbitals. Similar hybridization happens in boron where the fifth electron occupies one of three 2p degenerate orbitals.
Explanation:To accommodate seven electrons with five of them unpaired in the molecule ClF5, we need five degenerate orbitals. These are the 3s orbital, three 3p orbitals, and one 3d orbital to form the set of five sp³d hybrid orbitals, modeled after phosphorus pentachloride. They come from the hybridization of atomic orbitals, where valence electrons occupy the newly formed hybrid orbitals. In this case, with ClF5, the hybrid orbital overlaps with a fluorine orbital during the formation of the Cl-F bonds. The process is similar to how boron's fifth electron occupies one of the three degenerate 2p orbitals.
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What was the main use of constellations in ancient times?
o locate specific stars
O help navigate at night
O determine magnitude of stars
O predict time of the year
Answer:
the ans is option b(help navigate at night)
what do you think could be happening to the sand eels
Answer:
increasing fishing for them is thought to be causing problems for some of their natural predators, especially the auks which take them in deeper water. They are also tied as flies to catch fish.
Explanation:
Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What is the value of the solubility product constant for AB2?
Answer:
2.06 × 10⁻¹⁰
Explanation:
Let's consider the solution of a generic compound AB₂.
AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)
We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.
AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰
How many atoms does 32 grams of sulfur contain?
One hundred cubic meters of carbon dioxide initially at 150 oC and 50 bar is to be isothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar. Calculate i. The volume of the compressed gas ii. The work done to compress gas iii. The heat flow on compression assuming carbon dioxide (a) is an ideal gas (b) Obeys principle of corresponding states of section 6.6 (hint: See Illustration 6.6-2) (c) Obeys the Peng-Robinson equation of state (use the visual basic Peng
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
Image of page 3
Explanation:As revealed above, the stimuli connections are clearly stated
The decomposition of nitramide, O2NNH2, in water has the chemical equation and rate law O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H+] A proposed mechanism for this reaction is O2NNH2(aq)⥫⥬=k−1k1O2NNH−(aq)+H+(aq)(fast equilibrium) O2NNH−(aq)−→k2N2O(g)+OH−(aq)(slow) H+(aq)+OH−(aq)−→k3H2O(l)(fast) What is the relationship between the observed value of k and the rate constants for the individual steps of the mechanism?
Answer:
It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law
Explanation:
Please, observe the solution in the attached Word document.
The rate law is defined as the molar concentration of reactants raised to the power of their stoichiometric coefficients. The law states the dependency of chemical reactions on reactants.
In the given decomposition of nitramide in water, it can be stated that the third step of the reaction is fast, and therefore, does not contribute to the rate law.
The rate-determining step of a reaction is a slow step.
The decomposition reaction can be written as:
O₂NNH₂ [tex]\rightarrow[/tex] O₂NNH⁻ + H⁺ (fast equilibrium)
The value of K is:
[tex]\text{K}_{\text {eq}}&=\dfrac{K_1}{K_{-1}}&=\dfrac{\text[{O}_2\text{NNH}^-][\text{H}^+]}{\text[{O}_2\text{NNH}_{-2}]}[/tex]
[tex]\begin{aligned}\text{Rate}=\text{K}_2\frac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}_2]}\end{aligned}[/tex]
The second step of the reaction:
O₂NNH₂ [tex]\rightarrow[/tex] O₂N + OH⁻
The rate can be given as:
Rate = K₂ [O₂NNH⁻]
Substituting equation 1, we get:
[tex]\begin{aligned} \text[{O}_2\text{NNH}^-] &= \dfrac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}^+]}\end{aligned}[/tex]
Now, writing the equation in the rate constant, we get:
[tex]\text {K}&=\dfrac{K_2K_1}{K_{-1}}[/tex]
Thus, it can be concluded that the third step is very fast and does not contribute to the rate law.
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A metal bromide, MBr_2, is converted to a molten form at high temperature. Electrolysis of this sample with a current of 6.13A for 73.1 seconds results in deposition of 0.148 g of metal M at the cathode. The other product is bromine gas, Br_2(g), which is released at the anode. Determine the identity of metal M. Enter symbol of the element into clicker.
The molar mass of the metal M was calculated using Faraday's laws of electrolysis and found to be 63.65 g/mol, which closely corresponds to copper (Cu). Therefore, the identity of metal M is likely to be copper.
To determine the identity of the metal M, we will need to use Faraday's laws of electrolysis to calculate the molar mass of the metal M.
The charge (Q) that passed through the electrode can be calculated by the formula Q = It, where I is the current in amperes and t is the time in seconds. In this case, I = 6.13 A and t = 73.1 s, so Q = 6.13 × 73.1 = 448.203 C (coulombs).
Next, we will use the equivalent weight of metal M, which is the molar mass of M divided by the number of electrons transferred per ion of M during the electrolysis. For MBr₂, there are 2 moles of electrons transferred per mole of M (since the valency is 2).
Using Faraday's constant (approximately 96500 C/mol), we can determine the number of moles of M deposited by dividing the charge by Faraday's constant multiplied by the valency, which is moles of M = Q / (2 ×96500).
From this, we can calculate the molar mass: Molar mass of M = mass of M deposited / moles of M.
Using our given data, the moles of M deposited would be 448.203 C / (2 × 96500 C/mol) = 0.002325 moles. The molar mass would then be 0.148 g / 0.002325 mol = 63.65 g/mol. This molar mass is very close to that of copper (Cu), which is 63.55 g/mol. Thus, the correct answer is Cu.
The identity of the metal M is copper, Cu.
To determine the identity of metal M, we need to follow these steps:
1. Calculate the total charge (in coulombs) that passed through the molten salt during electrolysis using the equation [tex]\( Q = I \times t \)[/tex], where [tex]\( I \)[/tex] is the current in amperes and [tex]\( t \)[/tex] is the time in seconds.
2. Use Faraday's constant to find the moles of electrons that passed through the solution. Faraday's constant is [tex]\( 96485 \text{ C/mol} \).[/tex]
3. Relate the moles of electrons to the moles of metal M deposited, since the electrolysis of [tex]\( \text{MBr}_2 \)[/tex] involves the transfer of two moles of electrons for each mole of metal M produced.
4. Calculate the molar mass of metal M using the mass of metal M deposited and the moles of metal M calculated in the previous step.
5. Identify the metal M based on its molar mass.
Let's perform the calculations:
1. Total charge [tex]\( Q \)[/tex] is calculated as:
[tex]\[ Q = I \times t = 6.13 \text{ A} \times 73.1 \text{ s} = 448.563 \text{ C} \][/tex]
2. Moles of electrons [tex]\( n(e^-) \)[/tex] is given by:
[tex]\[ n(e^-) = \frac{Q}{F} = \frac{448.563 \text{ C}}{96485 \text{ C/mol}} \approx 4.65 \times 10^{-3} \text{ mol} \][/tex]
3. Since two moles of electrons are required to deposit one mole of metal M, the moles of metal M [tex]\( n(\text{M}) \)[/tex] is half the moles of electrons:
[tex]\[ n(\text{M}) = \frac{n(e^-)}{2} = \frac{4.65 \times 10^{-3} \text{ mol}}{2} \approx 2.325 \times 10^{-3} \text{ mol} \][/tex]
4. Molar mass [tex]\( M \)[/tex]of metal M is calculated as:
[tex]\[ M = \frac{m}{n} = \frac{0.148 \text{ g}}{2.325 \times 10^{-3} \text{ mol}} \approx 63.66 \text{ g/mol} \][/tex]
5. Based on the calculated molar mass, we can identify the metal M. The metal with a molar mass closest to [tex]\( 63.66 \text{ g/mol} \)[/tex] is copper, Cu, which has a molar mass of approximately[tex]\( 63.55 \text{ g/mol} \)[/tex].
Therefore, the identity of metal M is copper, Cu.
An adhesive dealer receives an order for 3000 lb of an adhesive solution containing 13% polymer by weight. On hand are 500 lb of 10% solution and very large quantities of 20% solution and pure solvent. Calculate the weight of each that must be blended together to fill this order. Use all of the 10% solution.
Answer:
500 lb of 10% solution1700 lb of 20% solution800 lb of pure solventExplanation:
The mass of polymer in the order recieved is:
3000 lb * 13/100 = 390 lbThe mass of polymer in the 10% solution is:
500 lb * 10/100 = 50 lbSo to solve this problem, we need to add (390-50) 340 lb of polymer that comes from the 20% solution.
We calculate the mass of the 20% solution that would contain 340 lb of polymer:
340 lb * 100/20 = 1700 lbSo far, in total we have a solution that weighs (1700+500) 2200 lb and contains 390 lb of polymer.
Thus, in order to reach the required mass of solution (and percentage of polymer by weight), we add (3000-2200) 800 lb of pure solvent.
If you have 40 grams of potassium nitrate in 100 grams of water at 20 C:
a.How many grams of potassium nitrate will dissolve?
b. How many grams of potassium nitrate will fall to the bottom of the container?
c. At this temperature, 20 C, is this solution saturated, unsaturated or
supersaturated? EXPLAIN.
d. At what temperature will the amount of potassium nitrate that fell to the bottom will also dissolve?
e. If this solution was heated to 90 C, would the solution be saturated, unsaturated or supersaturated? EXPLAIN.
Answer:
Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:
[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.
So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.
To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.
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In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the object is 40 g, the specific heat capacity of the chemical would be
Group of answer choices
25 J /g* °C
75 J /g* °C
100 J /g* °C
50 J /g* °C
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
g The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures in excess of 1100 K. We can conclude that a. ΔG° is negative for all temperatures. b. ΔH° is negative and ΔS° is negative. c. ΔH° is positive and ΔS° is positive. d. ΔH° is negative and ΔS° is positive. e. ΔH° is positive and ΔS° is negative.
Answer:
c. ΔH° is positive and ΔS° is positive.
Explanation:
Hello,
In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.
Best regards.
Final answer:
The reaction C(s) + CO2(g) → 2CO(g) being only spontaneous at high temperatures above 1100K indicates that the reaction has a positive ΔH° and a positive ΔS°, corresponding to option c.
Explanation:
The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures above 1100 K. This information allows us to infer the signs of ΔH° (enthalpy change) and ΔS° (entropy change). According to Gibbs free energy equation, ΔG = ΔH - TΔS, a reaction is spontaneous when ΔG is negative. Given that the reaction is only spontaneous at high temperatures, this suggests that ΔH° is positive (endothermic reaction) and ΔS° is positive (increase in disorder). At lower temperatures, the TΔS term is not sufficient to overcome the positive ΔH°, making ΔG positive and the reaction non-spontaneous. Hence, the correct conclusion is that the reaction is spontaneous at high temperatures because of a positive ΔH° and a positive ΔS°, indicating option c.
Calculate the cell potential for the galvanic cell in which the reaction Fe ( s ) + Au 3 + ( aq ) − ⇀ ↽ − Fe 3 + ( aq ) + Au ( s ) occurs at 25 ∘ C , given that [ Fe 3 + ] = 0.00150 M and [ Au 3 + ] = 0.797 M .
Final answer:
To calculate the cell potential for the given galvanic cell reaction, you can use the Nernst equation and the standard cell potential. Plug in the concentrations of Fe3+ and Au3+ and calculate the cell potential using the Nernst equation at 25°C.
Explanation:
The cell potential for the galvanic cell can be calculated using the Nernst equation, which is given by: Ecell = E°cell - (0.0592/n) * log(Q)
where E°cell is the standard cell potential, n is the number of electrons transferred in the reaction, and Q is the reaction quotient.
In this case, the reaction is: Fe(s) + Au3+(aq) -> Fe3+(aq) + Au(s)
The standard cell potential can be found in Appendix L. Plugging in the concentrations of Fe3+ and Au3+, as well as the standard cell potential, into the Nernst equation will give the cell potential at 25°C.
The cell potential for the given galvanic cell at 25°C is approximately 1.589 V.
This calculation involves determining standard reduction potentials, using the Nernst equation, and considering the reaction quotient.
To calculate the cell potential for the galvanic cell with the reaction:
Fe(s) + Au³⁺ (aq) → Fe³⁺ (aq) + Au (s) at 25°C with [Fe³⁺] = 0.00150 M and [Au³⁺] = 0.797 M, we can follow these steps:
Determine the standard reduction potentials:
Fe³⁺ (aq) + 3e⁻ → Fe (s), E° = -0.037 VAu³⁺ (aq) + 3e⁻ → Au (s), E° = 1.498 VCalculate the standard cell potential (E°cell) using the formula:
E°cell = E°cathode - E°anode
Here, E°cathode (Au) = 1.498 VE°anode (Fe) = -0.037 VE°cell = 1.498 V - (-0.037 V) = 1.535 VUse the Nernst equation to find the cell potential under non-standard conditions:
Ecell = E°cell - (0.0591 / n) * log(Q)Where n = 3, the number of electrons transferred, and Q is the reaction quotient.Q = [Fe³⁺] / [Au³⁺] = 0.00150 / 0.797Q ≈ 0.00188Plugging in the values:
Ecell = 1.535 V - (0.0591 / 3) * log(0.00188)log(0.00188) ≈ -2.727Ecell = 1.535 V - (0.0591 / 3) * (-2.727)Ecell = 1.535 V + (0.0591 * 2.727) / 3Ecell ≈ 1.589 VTherefore, the cell potential for the given galvanic cell at 25°C is approximately 1.589 V.
) Consider the starting materials and reagents. What do you expect to happen at the beginning of this reaction (Boxes 1-3 on the following page)? Explain your answer. Be sure to identify the species that act as electrophiles, nucleophiles, acids, or bases in Boxes 1-3 as part of your answer. (100 words max)
Answer:
Check the explanation
Explanation:
functional group found in the major organic product = alpha -beta unsaturated ketone
Reaction used to form this functional group = Michael condensation reaction
Also other reactions are - Aldol condensation , Robinson annulation reaction.
Kindly check the attached image below to see the step by step solution to the question above.
When H2(g) reacts with O2(g) to form H2O(g) , 242 kJ of energy are evolved for each mole of H2(g) that reacts. Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive. Use the SMALLEST INTEGER coefficients possible and put the energy term (including the units) in the last box on the appropriate side of the equation. If a box is not needed, leave it blank. 2H2 + O2 + 2H2O + An error has been detected in your answer. Check
The balanced thermochemical equation for the reaction where Hydrogen gas reacts with Oxygen gas to form Water with the release of 242 kJ energy is 2H2(g) + O2(g) → 2H2O(g) ΔH = -242 kJ. This indicates an exothermic reaction.
Explanation:The balanced thermochemical equation of the reaction where Hydrogen gas (H2) reacts with Oxygen gas (O2) to form Water (H2O) and 242 kJ (kilojoules) energy is released per mole of H2, can be formulated as follows:
2H2(g) + O2(g) → 2H2O(g) ΔH = -242 kJ
The negative sign before the energy value indicates that it's an exothermic reaction, meaning that energy is released during the reaction.
Learn more about Thermochemical Equation here:https://brainly.com/question/10023358
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The burning of a sweet potato sample generates 102.5 kJ of heat. This heat was used to raise the temperature of 300 grams of water from its initial temperature of 15.5 degrees Celsius. Considering the specific heat of water is 4.184 J/g-degrees-C, what is the final temperature of the water
Answer: 97.2°c
Explanation:Given
Heat added Q=102.5KJ
Mass m=300g
Iniatial temperature t1=15.5°c
Final temperature t2=?
Specific heat capacity c=4.184
Recall Q=mc(t2-t1)
102500=300*4.184(t2-15.5)
t2-15.5=81.66
T2=97.2°c
Answer:
97.16°C
Explanation:
Step 1:
Data obtained from the question. This includes:
Heat (Q) = 102.5 kJ = 102.5 x 1000 = 102500J
Mass (M) = 300 g
Initial temperature (T1) = 15.5°C
Specific heat capacity (C) = 4.184 J/g°C
Final temperature (T2) =?
Change in temperatures (ΔT) = T2 - T1 = T2 - 15.5
Step 2:
Determination of the final temperature.
Applying the following equation:
Q = MCΔT
We can easily calculate the value of the final temperature as follow:
Q = MCΔT
Q = MC(T2 - T1)
102500 = 300 x 4.184 (T2 - 15.5)
102500 = 1255.2 x (T2 - 15.5)
Divide both side by 1255.2
(T2 - 15.5) = 102500/1255.2
T2 - 15.5 = 81.66
Collect like terms
T2 = 81.66 + 15.5
T2 = 97.16°C
Therefore, the final temperature of the water is 97.16°C
The reactant concentration in a zero-order reaction was 5.00×10−2 M after 195 s and 2.50×10−2 M after 385 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.
Answer : The rate constant for this reaction is, [tex]3.65\times 10^{-3}Ms^{-1}[/tex]
Explanation :
To calculate the rate constant for zero order reaction, the expression used is:
[tex]\ln [A]=-kt+\ln [A_o][/tex]
where,
[tex][A_o][/tex] = initial concentration
[tex][A][/tex] = final concentration = [tex]5.00\times 10^{-2}M[/tex] at 195 s
[tex][A][/tex] = final concentration = [tex]2.50\times 10^{-2}M[/tex] at 385 s
k = rate constant = ?
Now put all the given values in the above expression, we get:
[tex]\ln (5.00\times 10^{-2})=-k\times 195+\ln [A_o][/tex] ............(1)
and,
[tex]\ln (2.50\times 10^{-2})=-k\times 385+\ln [A_o][/tex] ............(2)
Subtracting 1 from 2, we get:
[tex]\ln (2.50\times 10^{-2})-\ln (5.00\times 10^{-2})=-k\times 385+\ln [A_o]+k\times 195-\ln [A_o][/tex]
[tex]\ln (2.50\times 10^{-2})-\ln (5.00\times 10^{-2})=-k\times 385+k\times 195[/tex]
[tex]-0.693=-190k[/tex]
[tex]k=3.65\times 10^{-3}Ms^{-1}[/tex]
Therefore, the rate constant for this reaction is, [tex]3.65\times 10^{-3}Ms^{-1}[/tex]
A chemist fills a reaction vessel with 0.623g mercurous chloride(Hg2Cl2) solid, 0.645M mercury (I) (Hg2^2+)aqueous solution, and 0.905M chloride (Cl-) aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Hg2Cl2(s) ⇌ Hg2^2+ (aq) + 2Cl^- (aq)
Answer:
1.34 KJ
Explanation:
To calculate the reaction free energy for the chemical reaction, find the solution below.
Aqueous aluminum sulfate is mixed with aqueous ammonium hydroxide
Answer:
Al2(SO4)3 + 6(NH3•H2O) → 2Al(OH)3 + 3(NH4)2SO
Explanation:
Aluminium sulfate react with ammonium hydroxide to produce aluminium hydroxide and ammonium sulfate.
If you are looking for the equation
Compound B, C6H12O2, was found to be optically active, and it was oxidized to an optically active carboxylic acid A, by Ag (aka, Tollens Reagent). Oxidation of B by PCC gave an optically inactive compound X that reacted with Zn amalgam/HCl to give 3-methylpentane. With Na2Cr2O7/H2SO4, compound B was oxidized to an optically inactive dicarboxylic acid C, C6H10O4. Provide the structures of A, B, and C (ignore specific configuration of any stereocenters).
Answer:
Check the explanation
Explanation:
Acidipic Acid (which is an essential dicarboxylic acid for manufacturing purposes with about 2. 5 billion kilograms produced per year. It is mainly used for the production of nylon and its related materials.)
Going by the question, since the [tex]H_{2} CrO_{4}[/tex] is comparatively mild oxidizing agent than the [tex]CrO_{3}[/tex], it only oxidizes the carbon group
Kindly check the attached image below for the full explanation to the question above.
3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cyclohexene (density = 0.811 g/mL, Molecular weight = 82.143 g/mol). What is the theoretical yield in Moles of cyclohexene? (hint: reaction scheme same as problem 1, 1:1 mole ratio) (10 pts)
Answer:
[tex]n_{C_6H_{10}}=0.03molC_6H_{10}[/tex]
Explanation:
Hello,
In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:
[tex]n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}[/tex]
Besides, the mass could be computed as well by using the molar mass of cyclohexene:
[tex]m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}[/tex]
Even thought, the volume could be also computed by using its density:
[tex]V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}[/tex]
Best regards.
Answer:
The theoretical yield of moles cyclohexene is 0.0288 moles
Explanation:
Step 1: Data given
Volume of cyclohexanol = 3 mL
Density cyclohexanol = 0.9624 g/mL
Molar mass of cyclohexanol = 100.158 g/mol
Sulfuric acid is in excess
Density of cyclohexene = 0.811 g/mL
Molar mass of cyclohexene = 82.143 g/mol
Step 2: The balanced equation
C6H12O + H2SO4 → C6H10 + H3O + HSO4
Step 3: Calculate mass cyclohexanol
Mass cyclohexanol = density cyclohexane * volume
Mass cyclohexanol = 0.9624 g/mL * 3mL
Mass cyclohexanol = 2.8872 grams
Step 4: Calculate moles cyclohexanol
Moles cyclohexanol = mass cyclohexanol / molar mass
Moles cyclohexanol = 2.8872 grams / 100.158 g/mol
Moles cyclohexanol = 0.0288 moles
Step 5: Calculate moles cyclohexene
For 1 mol cyclohexanol we need 1 mol H2SO4 to produce 1 mol cyclohexene
For 0.0288 moles cyclohexanol we'll have 0.0288 moles cyclohexene
The theoretical yield of moles cyclohexene is 0.0288 moles
How many molecules are contained in 0.500 moles of hydrogen gas?
A.
1.20 x 1024 molecules
B.
6.02 x 10-23 molecules
C.
3.01 x 10-23 molecules
D.
3.01 x 1023 molecules
Answer:
C= 3.01 x 10-23 molecules
Explanation:
n = N/NA
0.5 = N/6.02×10^23
N= 3.01×10^23
You make dilutions of curcumin stock solution and measure the absorbance of each dilution to obtain the following data: Conc. (M) Abs. 1.60E-05 1.93 1.21E-05 1.62 8.09E-06 1.08 4.04E-06 0.44 2.02E-06 0.13 Which data points should you keep when making your calibration curve? Choose all that apply. (Hint: Plot the data.)
Answer:
The following data point should be kept when making your calibration.
1.21E-05 M
1.60E-05 M
8.09E-06 M
4.04 E-06 M
Explanation:
From the graph it is obvious that the reading for the concentration 2.02E-06 M does not fall on the straight line. So this point should not be taken to construct the calibration plot.
Find attached of the graph.
Why do some distances have negative values?
Answer:
distance with a negative value just means that it's going in the opposite direction.
Explanation:
example: -50m East is the just 50m west
In physics, negative values in a calculation often refer to displacement, indicating movement in the opposite direction of the chosen positive direction.
Distance itself remains a positive value as it does not account for direction.
In the context of physics, distances don't inherently have negative values, but displacement can. The term displacement refers to the change in position of an object and takes direction into account. For example, if we start at a reference point and consider east as positive, then a movement 100 meters west would be represented as -100 meters of displacement.When calculating displacement, we choose a frame of reference and a positive direction. The negative sign indicates a movement in the opposite direction to what we defined as positive.On the other hand, distance refers to the total path length traveled and is always a positive value, since it does not take direction into account.