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.
Answers
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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Related Questions
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
Answers
Answer
The transport of the substance into the cell is reduced or reversed in the absence of ATP. I think........
Explanation:
Chose the statements below that are TRUE about bases. Group of answer choices Metals in the second group make bases of the form M(OH)2 Ammonium is an example of a "Nitrogen with 3 groups" base Metals in the first group make bases of the form M(OH)2 A nitrogen with three groups and a lone pair is a base since it can accept a proton HNO3 has N as the central atom so it is a base
Answers
Answer:
Metals in the second group make bases of the form M(OH)₂A nitrogen with three groups and a lone pair is a base since it can accept a proton.Explanation:
The answer choices are:
Metals in the second group make bases of the form M(OH)₂ Ammonium is an example of a "Nitrogen with 3 groups" base Metals in the first group make bases of the form M(OH)₂ A nitrogen with three groups and a lone pair is a base since it can accept a proton HNO₃ has N as the central atom so it is a baseSolution1. Metals in the second group make bases of the form M(OH)₂
TRUE
The metals in the second group have two valence electrons, thus they can lose them to form cation with +2 oxidation state and bond with two (OH)⁻ groups, forming bases with the general formula M(OH)₂. For instance, Mg(OH)₂ and Ca(OH)₂.
2. Ammonium is an example of a "Nitrogen with 3 groups" base
FALSE
Ammonium is NH₃. The "nitrogen with 3 groups" base are the compounds where the three hidrogens are replaced with R-groups, where R are alkyl groups.
3. Metals in the first group make bases of the form M(OH)₂
FALSE
Since metals in first group of the periodic table have one valence electrons they make bases of the form MOH. For instance, NaOH and KOH.
4. A nitrogen with three groups and a lone pair is a base since it can accept a proton
TRUE
Since nitrogen has five valence electrons, when it bonds to three groups, it will retain a lone pair of electrons which make it a base under the Lewis concept and can accept a proton.
5. HNO₃ has N as the central atom so it is a base
FALSE
It is true that N is the central atom in HNO₃ but this compound is an acid as it releases a proton in aqueous solution.
completethe mass in grams of sample of carbon containing 68 atoms?
Answers
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.
Select the statements below that are false and think about why they are not true.
a) If a chemical reaction is spontaneous then the ?Ssys must always be positive.
b)If a chemical reaction with a negative ?Ssys is spontaneous then ?Ssurr must be positive.
c) If a chemical reaction is spontaneous then the ?Euniv must be increasing.
d) We don\'t need to worry about spontaneous reactions because they are rare.
Answers
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.
what do you think could be happening to the sand eels
Answers
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:
Cocaine has been prepared by a sequence beginning with a Mannich reaction between dimethyl acetonedicarboxylate, methylamine, and butanedial. The mechanism involves the following steps: 1. Following initial protonation of the carbonyl oxygen, nucleophilic attack by the amine forms carbinolamine 1; 2. Proton transfer and elimination of water forms iminium ion 2; 3. The enol form of the dicarboxylate ester attacks the iminium ion to form adduct 3; 4. Adduct 3 tautomerizes to form enol 4; 5. Following protonation of the aldehyde, cyclization occurs as protonated carbinolamine 5 is formed; 6. Proton transfer and elimination of water lead to iminium ion 6; 7. The enol attacks the iminium ion in a second cyclization reaction to form bicyclic ion 7; 8. Deprotonation leads to the final product. Write out the mechanism on a separate sheet of paper and then draw the structure of protonated carbinolamine 1.
Answers
Answer:
Explanation:
check the attached file for answers. Thanks
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
Answers
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.
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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
Answers
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
Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH. Which statement is most likely to be true?
a. All three titrations require the same volume of NaOH to reach their first equivalence point.
b. All three titrations have the same initial pH.
c. All three titrations have the same pH at their first equivalence point.
Answers
c. All three titrations have the same pH at their first equivalence point.
Explanation:
- In the titration of a monoprotic weak acid with NaOH (i), the first equivalence point corresponds to the complete neutralization of the weak acid with NaOH. The pH at this point is determined by the salt formed, and for a monoprotic weak acid, it will be neutral (pH 7).
- In the titration of a diprotic weak acid with NaOH (ii), the first equivalence point also corresponds to the neutralization of the weak acid with NaOH. The pH at this point is determined by the salt formed, and it will also be neutral (pH 7).
- In the titration of a strong acid with NaOH (iii), the first equivalence point is when the strong acid is completely neutralized by NaOH. The resulting solution will be basic, with a pH greater than 7.
Therefore, the statement "All three titrations have the same pH at their first equivalence point" is most likely to be true.
A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.
Answers
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can use the formula C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
47.2 mL * 150 mM = 24.0 mM * V₂V₂ = 295 mLAnd converting into L becomes:
295 mL * [tex]\frac{1 L}{1000mL}[/tex] = 0.295 LThe 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.
Answers
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]
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.
Answers
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
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.
Answers
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.
How many atoms does 32 grams of sulfur contain?
Answers
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 .
Answers
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.
Aqueous aluminum sulfate is mixed with aqueous ammonium hydroxide
Answers
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
The temperature of a sample of liquid water changes from 50°C to 30°C. Which statement best explains the change that must
happen at the molecular level?
A)The molecules move more quickly and their average kinetic energy increases
B)The molecules move more slowly and their average kinetic energy decreases
C)The molecules move more quickly and their average kinetic energy decreases
D)The molecules move more slowly and their average kinetic energy increases
Answers
Answer:
The statement which best explains the change that must happen at the molecular level is below:
Explanation:
The molecules move more slowly and their average kinetic energy decreases
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.)
Answers
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.
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?
Answers
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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https://brainly.com/question/4222261
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)
Answers
Answer:
1.34 KJ
Explanation:
To calculate the reaction free energy for the chemical reaction, find the solution below.
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
Answers
Answer:
C= 3.01 x 10-23 molecules
Explanation:
n = N/NA
0.5 = N/6.02×10^23
N= 3.01×10^23
"Phenylacetic acid, C7H7COOH, builds up in the blood of people with phenylketonuria. The pKa of phenylacetic acid is 4.32 at 25 ºC. What is the pH of a 0.15 M aqueous solution of calcium phenylacetate, Ca(C7H7COO)2, at 25 ºC?"
Answers
Answer:
pH = 8.9
Explanation:
To solve this problem, we need to write the overall reaction of the aqueous solution of Ca(C₇H₇COO)₂:
Ca(C₇H₇COO)₂ <----------> Ca²⁺ + 2C₇H₇COO⁻
This means that the concentration of the phenyl acetate is 2 times the concentration of the calcium phenylacetate:
[C₇H₇COO⁻] = 2 * 0.15 = 0.30 M
We have the concentration of the phenylacetate, this is a conjugate base of the phenylacetic acid, therefore, we need the Kb of the conjugate. This can be calculated using the following expression:
Kw = Kb * Ka -----> Kb = Kw/Ka
And Ka = 10^(-pKa)
Replacing the values (Assuming Kw = 1x10⁻¹⁴)
Ka = 10⁽⁻⁴°³²⁾ = 4.786x10⁻⁵
Kb = 1x10⁻¹⁴ / 4.786x10⁻⁵ = 2.089x10⁻¹⁰
Now that we have Kb, we can write the reaction of the phenylacetate in water, and an ICE chart
C₇H₇COO⁻ + H₂O <--------> C₇H₇COOH + OH⁻ Kb = 2.089x10⁻¹⁰
i) 0.30 0 0
e) 0.30-x x x
Kb = [C₇H₇COOH] [OH⁻] / [C₇H₇COO⁻] Replacing the above values
2.089x10⁻¹⁰ = x² / 0.30 - x
As Kb is very small, x would be a very small value too, so we can neglect 0.30 - x to 0.30 only:
2.089x10⁻¹⁰ = x² / 0.30
2.089x10⁻¹⁰ * 0.30 = x²
√6.27x10⁻¹¹ = x
x = 7.92x10⁻⁶ M
This is the concentration of the phenylacetate and OH⁻. With this value we can calculate the pOH and then, the pH:
pOH = -log[OH⁻]
pOH = -log(7.92x10⁻⁶)
pOH = 5.1
Finally the pH:
pH = 14 - pOH
pH = 14 - 5.1
pH = 8.9And this is the pH of the calcium phenylacetate solutionThe calcium phenylacetate, Ca(C7H7COO)2, at 25 ºC pH is = 8.9
Calculation of Calcium phenylacetate
Then To translate this situation, After That, we need to note the overall reaction of the aqueous resolution of Ca(C₇H₇COO)₂:
Then Ca(C₇H₇COO)₂ <----------> Ca²⁺ + 2C₇H₇COO⁻
This define that the concentration of the phenylacetate is 2 times the concentration of the calcium phenylacetate:
That is [C₇H₇COO⁻] = 2 * 0.15 is = 0.30 M
After that We have the concentration of the phenylacetate, which is a conjugate base of the phenylacetic acid, Thus, we need the Kb of the conjugate. Now, This can be calculated using the subsequent declaration:
Kw is = Kb * Ka -----> Kb is = Kw/Ka
And also Ka is = 10^(-pKa)
Then We are Replacing the values (Accepting Kw is = 1x10⁻¹⁴)
Ka is = 10⁽⁻⁴°³²⁾ = 4.786x10⁻⁵
Kb is = 1x10⁻¹⁴ / 4.786x10⁻⁵ is = 2.089x10⁻¹⁰
Now that we have Kb, then we can compose the reaction of the phenylacetate in water, and also an ICE chart
Then C₇H₇COO⁻ + H₂O <--------> C₇H₇COOH + OH⁻ Kb is = 2.089x10⁻¹⁰
i) 0.30 0 0
e) 0.30-x x x
Kb is = [C₇H₇COOH] [OH⁻] / [C₇H₇COO⁻] Then Replacing the above values
2.089x10⁻¹⁰ is = x² / 0.30 - x
When Kb is very small, x then they would be a very small value too, so we can neglect 0.30 - x to 0.30 only:
2.089x10⁻¹⁰ is = x² / 0.30
2.089x10⁻¹⁰ * 0.30 is = x²
√6.27x10⁻¹¹ is = x
x is = 7.92x10⁻⁶ M
Now This is the concentration of the phenylacetate and also OH⁻. With this value then we can calculate the pOH and also then, the pH:
pOH is = -log[OH⁻]
pOH is = -log(7.92x10⁻⁶)
pOH is = 5.1
Finally the pH is:
pH is = 14 - pOH
pH is = 14 - 5.1
Therefore pH is = 8.9
So, This is the pH of the calcium phenylacetate solution
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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.
Answers
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.
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)
Answers
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
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
Answers
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
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?
Answers
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
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)
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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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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
Answers
Answer:
the ans is option b(help navigate at night)
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.
Answers
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.
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
Answers
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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