Draw all of the monochlorination products that you might obtain from the free-radical chlorination of ethylcyclopropane.You do not have to consider stereochemistry.Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner.Separate multiple products using the + sign from the dropdown menu.
Answers
Answer:
-) (2-chloroethyl)cyclopropane
-) (1-chloroethyl)cyclopropane
-) 1-chloro-2-ethylcyclopropane
-) 1-chloro-1-ethylcyclopropane
Explanation:
The monochlorination are products in which Cl is added only 1 time. Now, in the ethylcyclopropane we will have 5 carbons, so we will have 5 options for monochlorination (see figure 1). But, for carbons 2 and 3 we will have equivalent structures, therefore at the end, we will have only 4 possible structures. In carbon 5, carbon 4, carbon 3 and carbon 1 (see figure 2).
Final answer:
Monochlorination of ethylcyclopropane can produce several isomers, based on the position where the chlorine atom is added - either on the cyclopropane ring or the ethyl group attached to it.
Explanation:
The subject of the question is the monochlorination products from the free-radical chlorination of ethylcyclopropane. During the free-radical chlorination, chlorine radicals can react with the different types of hydrogens in ethylcyclopropane - those on the cyclopropane ring as well as those on the ethyl group. Chlorination at the ring will yield two isomers due to two different types of hydrogens (geminal and vicinal), while chlorination on the ethyl group could occur at either one of the methylene hydrogens or the methyl group hydrogens. When considering the ethyl group, monochlorination would give a 1-chloroethylcyclopropane or a 2-chloroethylcyclopropane, depending on which hydrogen is substituted. With regard to the cyclopropane ring, substitution could lead to two different chlorocyclopropanes due to the different positions of hydrogen on the ring.
Related Questions
how many grams are in 0.5 moles of C10H16?
Please show your work for marked brainliest!
Answers
Answer:
68 g
Explanation:
Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =
= 136 g/mol
m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g
n(C10H16) - number of moles of C10H16
M(C10H16) - molar mass of C10H16
The ionization constant of water at a temperature above 25°C is 1.7 x 10-14. What is the pH of pure water at this temperature?
2H2O(l)--> H3O+(aq) + OH-(aq)
A) 7.88
B) 5.68
C) 6.88
D) 13.85
E) 7.00
Answers
Answer:
The pH of pure water at this temperature is 6.88.
Explanation:
Ionization constant of water = [tex]K_w=1.7 x 10^{-14}[/tex]
[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]
Water will dissociate into equal amount of hydronium and hydroxide ions
[tex][H_3O^+] = [OH^-]= x[/tex]
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
[tex]K_w=1.7 x 10^{-14}=[H_3O^+]\times [OH^-][/tex]
[tex]1.7 x 10^{-14}=x^2[/tex]
[tex]x=1.304\times 10^{-7}[/tex]
The pH of the solution is a negative logarithm of hydronium ions concentration.
The pH of pure water at this temperature:
[tex]pH=-\log [H_3O^+][/tex]
[tex]pH=-\log (1.304\times 10^{-7})=6.88[/tex]
The pH of pure water is:
C) 6.88.
Calculation for pH:Ionization constant of water = [tex]1.7 * 10^{-14}[/tex]
Chemical reaction for ionization of water:
[tex]2H_2O(l)---- > H_3O^+(aq) + OH^-(aq)[/tex]
Water dissociates into equal amount of hydronium and hydroxide ions
[tex][H_3O^+]=[OH^-]=x\\\\\K_w=[H_3O^+]*[OH^-]\\\\1.7*10^{-14}=x^2\\\\x=1.307*10^{-7}[/tex]
The pH of the solution is a negative logarithm of hydronium ions concentration.
The pH of pure water at this temperature:
[tex]pH=-log[H_3O^+]\\\\pH=-log(1.304*10^{-7})\\\\pH=6.88[/tex]
Thus, correct option is C.
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Which statements are true?
a. The evaporation of water is an exothermic process.
b. A combustion reaction is exothermic.
c. When energy is transferred as heat from the surroundings to the system, Deta H is negative.
d. When energy is transferred as heat from the system to the surroundings, Deta H is negative.
e. for an exothermic reaction Deta H is positive.
f. for an endothermic reaction Deta H is positive.
Answers
Explanation:
Evaporation is defined as a process in which liquid state of water is changing into vapor state.So, we need to break the bonds of liquid substance in order to convert it into vapor state. And, energy is absorbed for breaking of bonds which means that evaporation is an endothermic process.
Hence, the statement evaporation of water is an exothermic process is false.
When a hydrocarbon reacts with oxygen and leads to the formation of carbon dioxide and water then this type of reaction is known as combustion reaction.A combustion reaction will always release heat energy. Hence, combustion reaction is exothermic in nature.
When energy is transferred as heat from the surroundings to the system then it means energy is being absorbed by the system. And, absorption of heat is an endothermic process for which [tex]\Delta H[/tex] is positive.Whereas when energy is transferred from system to the surrounding then it means energy is released by the system which is an exothermic process.Hence, for an exothermic process value of [tex]\Delta H[/tex] is negative.
Thus, we can conclude that statements which are true are as follows.
A combustion reaction is exothermic.When energy is transferred as heat from the system to the surroundings, [tex]\Delta H[/tex] is negative.For an endothermic reaction Deta H is positive.At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 2Hg(l) + O2(g) Suppose 51.2 g of mercury(II) oxide is placed in a sealed 3.00-L vessel at 700 K. What is the partial
pressure of oxygen gas at equilibrium? (R = 0.0821 Lxatm/(Kxmol))
A) 0.075 atm
B) 0.0056 atm
C) 4.5 atm
D) 19 atm
E) 2.3 atm
Answers
Answer: Option (B) is the correct answer.
Explanation:
According to the given reaction equation, formula to calculate [tex]\Delta n[/tex] is as follows.
[tex]\Delta n[/tex] = coefficients of gaseous products - gaseous reactants
= 1 - 0
= 1
Also we know that,
[tex]K_{p} = K_{c} \times (RT)^{\Delta n}[/tex]
[tex]5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}[/tex]
[tex]K_{c} = 0.097 \times 10^{-3}[/tex]
For the equation, [tex]2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)[/tex]
Activity of solid and liquid = 1
As, [tex]K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}[/tex]
[tex]5.6 \times 10^{-3} = P_{O_{2}}[/tex]
Hence, [tex]P_{O_{2}}[/tex] = 0.0056 atm
Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.
Final answer:
To calculate the partial pressures of the three gases at equilibrium, we can use the equilibrium constant equation. Plugging in the given initial partial pressures into the equation gives a Kp value of 0.0135.
Explanation:
To solve this problem, we can use the equation for calculating the equilibrium constant (Kp):
Kp = ([NO]²)/([N₂O][O₂])
We are given the initial partial pressures of each gas, so we can plug these values into the equation. The initial partial pressures are as follows: [NO] = 0.08 atm, [N₂O] = 0.62 atm, and [O₂] = 0.24 atm. Plugging these values into the equation:
Kp = (0.08²)/(0.62·0.24) = 0.0135
Therefore, the equilibrium constant (Kp) for this reaction at the given temperatures and pressures is 0.0135.
The most common method for the synthesis of unsymmetrical ethers is the Williamson synthesis, a reaction (SN2) of an alkoxide ion with an alkyl halide. Two pathways are possible, but often one is preferred. Construct the preferred pathway for the synthesis of 2-propoxypropane from propene, with propene-derived alkyl halide and alkoxide intermediates, by dragging the appropriate intermediates and reagents into their bins. Not every given reagent or intermediate will be used.
Answers
Answer:
2- propanol treated with metal base NaH followed by hydrogen bromide to form 2- propoxypropane.
Explanation:
Formation of secondary halide:
Firstly propene reacts with sulfuric acid or water to form 2- Propanol.
Formation of primary halide:
Also, according to the Anti markovinikov rule propene undergoes addition to the hydrogen bromide in the presence of peroxide to form 1- bromopropane.
The secondary halide i.e, 2- propanol treated with metal base NaH to form sodium isopropoxide. Further treated with hydrogen bromide to form 2- Propoxypropane. This reaction follows [tex]S_{N}2[/tex] mechanism.
The overall reaction is as follows.
The preferred pathway for the synthesis of 2-propoxypropane from propene in the Williamson synthesis involves the reaction of propene with an alkyl halide and an alkoxide ion.
Explanation:The preferred pathway for the synthesis of 2-propoxypropane from propene in the Williamson synthesis involves the reaction of propene with an alkyl halide and an alkoxide ion. The alkoxide ion is formed by treating an alcohol with a strong base, such as sodium hydride (NaH). The alkyl halide is then added to the alkoxide ion, resulting in the formation of the desired product, 2-propoxypropane.
Alcohol + Strong Base (NaH) → Alkoxide Ion
Alkoxide Ion + Alkyl Halide → 2-Propoxypropane
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Two nonpolar organic liquids, hexane (C6H14) and heptane (C7H16), are mixed. Do you expect ΔHsoln to be a large positive number, a large negative number, or close to zero? Fill in the following sentence: ΔHsoln is determined by the relative magnitudes of the "old" solute-solute (ΔHsolute) and solvent-solvent (ΔHsolvent) interactions and the "new" solute-solvent (ΔHmix) interactions. In this process, the energy of mixing hexane and heptane (ΔHmix) is the energy of separating them individually (ΔHsolute + ΔHsolvent), so ΔHsoln is expected to be .
Answers
Answer:
ΔH of solution is expected to be close to zero.
Explanation:
When we mix two non polar organic liquids like hexane and heptane,the resulting mixture formed is an ideal solution.An ideal solution is formed when the force of attraction between the molecules of the two liquids is equal to the force of attraction between the molecules of the same type.
For instance if liquids A and B are mixed,
[tex]F_{A-A}[/tex] = [tex]F_{B-B}[/tex] = [tex]F_{A-B}[/tex]
Hence the condition before and after mixing remains unchanged.
Since enthalpy change is associated with inter molecular force of attraction the enthalpy change for ideal solution is zero.
More examples of ideal solutions are:
1. Ethanol and Methanol
2. Benzene and Toluene
3. Ethyl bromide and Ethyl iodide
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25�C. (The equation is balanced.)
3 Cl2(g) + 2 Fe(s) --> 6 Cl-(aq) + 2 Fe3+(aq)
Cl2(g) + 2 e- --> 2 Cl-(aq); E� = +1.36 V
Fe3+(aq) + 3 e- -->Fe(s); E� = -0.04 V
+1.32 V
-1.32 V
-1.40 V
+1.40 V
+4.16 V
Answers
Answer: + 1.40 V
Explanation:
The balanced chemical equation is:
[tex]3Cl_2(g)+2Fe(s)\rightarrow 6Cl^-(aq)+2Fe^{3+}(aq)[/tex]
Here Fe undergoes oxidation by loss of electrons, thus act as anode. Chlorine undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Fe^{3+}/Fe]}=-0.04V[/tex]
[tex]E^0_{[Cl_2/Cl^-]}=+1.36V[/tex]
[tex]E^0=E^0_{[Cl_2/Cl^-]}- E^0_{[Fe^{3+}/Fe]}[/tex]
[tex]E^0=+1.36-(-0.04V)=+1.40V[/tex]
The standard cell potential for the reaction is +1.40 V
A 24.201 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.053 g of water. A 13.774 g aliquot of this solution is then titrated with 0.1045 M HCl. It required 28.02 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Answers
Answer:
1.44%
Explanation:
Initially, there is 24.201g of the waste. Let this waste contain 'x' mols of ammonia([tex]NH_{3}[/tex]). Now, this waste is dissolved in 72.053g of water, thus making the weight of the solution to 24.201+72.053 = 96.254g. Now, from this, 13.774g is taken. Note that, when the aqueous waste is dissolved in water, the ammonia particles are uniformly distributed, i.e, the 'x' mols of ammonia is uniformly present in the 96.254g of the solution. Hence, when we take 13.774g of the solution, only a fraction of the ammonia particles is taken. This fraction is equal to [tex]\frac{13.774}{96.254}[/tex] of x, which is equal to 0.1431 times 'x'.
For the HCl solution, 28.02mL of 0.1045M solution contains 28.02x0.1045 = 2.9281mmols of HCl is present in it.
The basic titration reaction that occurs is : [tex]HCl + NH_{3}[/tex]→[tex]NH_{4}Cl[/tex] , i.e, one mol of ammonia requires one mol of HCl for neutralization. Therefore, for the above solution of HCl containing 2.9281mmols, same amount of ammonia, i.e, 2.9281mmols is required for complete neutralization.
Therefore, 0.1431x = 2.9281 mmols ⇒ x = 20.4619 mmols.
The molecular weight of ammonia is 17g/mol. Therefore, the weight of 20.4619mmols of ammonia has a weight(w) = 20.4619 x [tex]10^{-3}[/tex] x 17 = 0.3478 g.
Therefore the weight of ammonia in the initial aqueous waste is 0.3478 g. The total weight of the waste is 24.201g, hence, the percent weight of ammonia is given by [tex]\frac{0.3478}{24.201}[/tex]×100 = 1.44%
Final answer:
To calculate the weight percent of ammonia in the waste sample, the moles of HCl used in titration are converted to mass of NH3. This value is then adjusted for dilution and related to the original mass of the waste to find the weight percent of NH3, which is calculated to be 1.442%.
Explanation:
To calculate the weight percent of ammonia (NH3) in the aqueous waste sample from the fertilizer manufacturer, we follow a series of steps involving dilution and titration. The aliquot of the solution (13.774 g) titrated with 0.1045 M HCl required 28.02 mL to reach the endpoint, which indicates the amount of ammonia that was present in the aliquot.
We first need to calculate the number of moles of HCl that reacted with NH3 using the molarity of HCl and the volume used in the titration:
Moles of HCl = Molarity of HCl imes Volume of HCl in litersMoles of HCl = 0.1045 mol/L imes 0.02802 L = 0.00293 molSince each mole of NH3 reacts with one mole of HCl, the moles of NH3 in the aliquot will also be 0.00293 mol. To find the mass of NH3, we multiply the moles by the molar mass of NH3 (17.031 g/mol).
Mass of NH3 = moles of NH3 imes molar mass of NH3Mass of NH3 = 0.00293 mol imes 17.031 g/mol = 0.04988 gTo calculate the weight percent of NH3 in the original waste sample, we need to adjust the mass of NH3 found for the dilution and relate that to the original mass of the waste sample:
Total mass of the solution = mass of waste sample + mass of water addedTotal mass of the solution = 24.201 g + 72.053 g = 96.254 gWeight percent NH3 = (Mass of NH3 in original sample / mass of waste sample) imes 100%Since only a portion of the diluted sample was titrated, we calculate the proportion of the total solution that was titrated.Proportion of solution titrated = Mass of aliquot / Total mass of the solutionProportion of solution titrated = 13.774 g / 96.254 gMass of NH3 in original sample = Mass of NH3 found from titration / Proportion of solution titratedMass of NH3 in original sample = 0.04988 g / (13.774 g / 96.254 g)Mass of NH3 in original sample = 0.3489 gWeight percent NH3 = (0.3489 g / 24.201 g) imes 100% = 1.442%Therefore, the weight percent of ammonia in the initial aqueous waste sample is 1.442%.
The equilibrium: 2 NO2(g) \Longleftrightarrow⇔ N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forward (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?
Initial: forward rate < reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate
Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate
Answers
Answer:
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rateExplanation:
2NO₂(g) → N₂O₄(g) Kc=4.7
The definition of equilibrium is when the forward rate and the reverse rate are equal.
Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.
The work function of indium is 6.56 x 10^-19 J. What is the minimum frequency of a photon that will eject an electron from the indium metal surface?
1. 9.90 x 10^14 Hz
2. 2.97 x 10^23 Hz
3. 3.42 x 10^14 Hz
4. 4.57 x 10^26 Hz
5. 2.19 x 10^-27 Hz
Answers
Answer:
1. 9.90 x 10^14 Hz
Explanation:
The work function is the minimum energy required to eject electron's from the surface of the metal. The frequency of photons that are capable of doing this is called the threshold frequency. Given the work function, the threshold frequency can be calculated in the following way
Work function = Planck's constant * frequency
6.56 x 10^-19 J = 6.62607004 × 10-34 m²-kg/s * f
f = 9.90 x 10^14 Hz
The electron configuration of a neutral atom is 1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons. Name the element.
Answers
Answer:
The four quantum number for each electron will be:
[tex]1s^{2}[/tex]
[tex]n=1;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]2s^{2}[/tex]
[tex]n=2;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]2p^{6}[/tex]
[tex]n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]3s^{2}[/tex]
[tex]n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
Explanation:
As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.
Number of electrons =12
Atomic number = 12
Element : Magnesium
The principal shell is represented by "n"
i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0
The two electrons in s subshell will have either plus half or minus half spin quantum number
ii) for "p" subshell the value for l =1
thus m = 0 or +1 or -1
The two electrons in each orbital will have either plus half or minus half spin quantum number
Final answer:
The complete set of quantum numbers for each electron in the given electron configuration, which corresponds to the element magnesium (Mg), is derived from the configuration 1s²2s²2p⁶3s², by stating the values for the principal, angular momentum, magnetic, and spin quantum numbers for each electron.
Explanation:
The electron configuration of a neutral atom is 1s22s22p63s2. To write a complete set of quantum numbers for each of the electrons, we can consider the principles of quantum mechanics. Each electron in an atom is described by four quantum numbers: the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (ml), and spin quantum number (ms).
For the 1s subshell (2 electrons), the quantum numbers for each electron are (1, 0, 0, +1/2) and (1, 0, 0, -1/2). For the 2s subshell (2 electrons), the quantum numbers are similar, with the principal quantum number being 2: (2, 0, 0, +1/2) and (2, 0, 0, -1/2). For the 2p subshell (6 electrons), there are three magnetic quantum numbers (-1, 0, +1), and two spin states. Thus, the six sets of quantum numbers are (2, 1, -1, +1/2), (2, 1, -1, -1/2), (2, 1, 0, +1/2), (2, 1, 0, -1/2), (2, 1, +1, +1/2), and (2, 1, +1, -1/2). For the 3s subshell (2 electrons), the quantum numbers are (3, 0, 0, +1/2) and (3, 0, 0, -1/2).
Based on this configuration, the neutral atom would be magnesium (Mg), which has an atomic number of 12, corresponding to the 12 electrons described.
In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate the percent water in this hydrate?
What was the mass of water lost? What is the value of z (the number of waters in the formula)?
Answers
Answer:
1) The mass of water lost = 7.7892 grams
2) Z = 3: Cu(NO3)2*3H2O
Explanation:
Step 1: Data given
Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams
Mass of substance after heating = 27.0351 grams
Molar mass of Cu(NO3)2 = 187.56 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: Calculate mass of water
The mass of water is the mass lost after heating.
Mass water = 34.8243 - 27.0351 = 7.7892 grams of water
Step 3: Calculate moles of Cu(NO3)2
Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2
Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol
Moles Cu(NO3)2 = 0.144 moles
Step 4: Calculate moles of H2O
Moles H2O = 7.7892 grams / 18.02 g/mol
Moles H2O = 0.432 moles
Step 5: Calculate Z
z = moles H2O / moles Cu(NO3)2
Z = 0.432/0.144
Z = 3
This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O
Final answer:
To find the percent water in the hydrate, subtract the final mass after heating from the initial mass of the hydrate, then divide by the initial mass and multiply by 100. The mass of water lost is 7.7892g, and the percent water is 22.36%. To determine the value of z, calculate the moles of water and anhydrous Cu(NO3)2 to find z equals 3, making the formula Cu(NO3)2•3H2O.
Explanation:
The question involves determining the percent water in a hydrate, the mass of water lost during heating, and the value of z in the formula Cu(NO3)2•zH2O. Firstly, to calculate the mass of water lost, subtract the final mass after heating from the initial mass of the copper (II) nitrate hydrate. Then, to find the percent water, the mass of water lost is divided by the initial mass of the hydrate, multiplied by 100. Lastly, to determine the value of z, the moles of water lost are divided by the moles of anhydrous Cu(NO3)2.
To calculate the mass of water lost: 34.8243g (initial mass) - 27.0351g (final mass) = 7.7892g (mass of water lost).
To calculate the percent water in the hydrate: (7.7892g / 34.8243g) x 100 = 22.36%.
To find z, the number of waters, you need the molar mass of Cu(NO3)2 and H2O. Assume the molar mass of Cu(NO3)2 is approximately 187.55 g/mol. The mass of Cu(NO3)2 after heating is 27.0351g, which corresponds to 27.0351g / 187.55g/mol = 0.1441 mol. The moles of water lost is 7.7892g / 18.015g/mol = 0.4323 mol. Hence, z = 0.4323 mol / 0.1441 mol = 3, indicating the formula is Cu(NO3)2•3H2O.
Write a balanced chemical equation describing the oxidation of Cl2 gas by Cu3+ to form chlorate ion (ClO3-) and Cu2+ in an acidic aqueous solution. Use the smallest whole-number coefficients possible and indicate states of matter in your balanced reaction.
Answers
The detailed answer provides the balanced chemical equation for the oxidation of Cl2 by Cu3+ in an acidic solution, with states of matter indicated.
The balanced chemical equation describing the oxidation of Cl2 gas by Cu3+ to form chlorate ion (ClO3-) and Cu2+ in an acidic aqueous solution is:
3Cu^3+(aq) + 6Cl¯(aq) + 18H^+(aq) + Cl2(g) → 3Cu^2+(aq) + ClO3^-(aq) + 9H2O(l)
States of matter: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.
The oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] to form chlorate ion and [tex]Cu^{2+}[/tex] in an acidic aqueous solution is represented by the balanced chemical equation provided.
The given question concerns the concept of balancing chemical equations in chemistry and deriving correct chemical reactions for the mentioned reactants and products. Also, oxidation is mentioned which refers to addition of oxygen to an entity or removal of hydrogen from entity. The balanced chemical equation for the oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] in an acidic aqueous solution is:
[tex]2Cl_2(g) + 2Cu^{3+} (aq) + 4H^+ (aq) \rightarrow 2ClO^{3-} (aq) + 2Cu^{2+} (aq) + 2H_2O (l)[/tex]
This reaction results in the formation of chlorate ion ([tex]ClO^{3-}[/tex]) and [tex]Cu^{2+}[/tex] in the solution. The states of matter as gas, aqueous, and liquid are indicated in the equation.
Consider the following generic chemical equation. 2A+4B→3C What is the limiting reactant when each of the following amounts of A and B are allowed to react? Part A 2 molA; 5 molB 2 ; 5 B A SubmitRequest Answer Part B 1.8 molA; 4 molB 1.8 ; 4 B A SubmitRequest Answer Part C 3 molA; 4 molB 3 ; 4 B A SubmitRequest Answer Part D 22 molA; 40 molB 22 ; 40 B A
Answers
Answer:
A. Limiting reactant is A.
B. Limiting reactant is A.
C. Limiting reactant is B.
D. Limiting reactant is B.
Explanation:
It is possible to find the limiting reactant for a reaction taking the moles of a reactant that will react and using the chemical equation find the moles of the other reactant you will need.
For the reaction:
2A + 4B → 3C
A. 2 moles A requires:
2molA×[tex]\frac{4molB}{2molA}[/tex]= 4moles of B
As you have 5 moles of B, limiting reactant is A.
B. 1,8 moles A requires:
1,8 molA×[tex]\frac{4molB}{2molA}[/tex]= 3,6 moles of B
As you have 4 moles of B, limiting reactant is A.
C. 3 moles A requires:
3 molA×[tex]\frac{4molB}{2molA}[/tex]= 6 moles of B
As you have just 4 moles of B, limiting reactant is B.
D. 22 moles A requires:
22 molA×[tex]\frac{4molB}{2molA}[/tex]= 44moles of B
As you have just 40 moles of B, limiting reactant is B.
I hope it helps!
For each scenario, the limiting reactant is determined by comparing the moles used according to the stoichiometry of the reaction. In Parts A, B, and D, A is the limiting reactant; in Part C, B is limiting.
To determine the limiting reactant, compare the moles of each reactant to the stoichiometric coefficients in the balanced equation (2A + 4B → 3C).
1. Part A (2 mol A; 5 mol B):
- Moles of A used = 2 (coefficient in front of A)
- Moles of B used = [tex]\frac{5}{2}[/tex] (coefficient in front of B divided by 2)
- Since Moles of B used > Moles of A used, A is the limiting reactant.
2. Part B (1.8 mol A; 4 mol B):
- Moles of A used = 1.8 (coefficient in front of A)
- Moles of B used = [tex]\frac{4}{4}[/tex] (coefficient in front of B divided by 4)
- Since Moles of A used < Moles of B used, A is the limiting reactant.
3. Part C (3 mol A; 4 mol B):
- Moles of A used = 3 (coefficient in front of A)
- Moles of B used = [tex]\frac{4}{2}[/tex] (coefficient in front of B divided by 2)
- Since Moles of B used < Moles of A used, B is the limiting reactant.
4. Part D (22 mol A; 40 mol B):
- [tex]\( \text{Moles of A used} = \frac{22}{2} \)[/tex] (coefficient in front of A divided by 2)
- Moles of B used = 40 (coefficient in front of B)
- Since Moles of A used < Moles of B used, A is the limiting reactant.
In summary:
- Part A: Limiting reactant is A.
- Part B: Limiting reactant is A.
- Part C: Limiting reactant is B.
- Part D: Limiting reactant is A.
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A coal-burning power plant produces 500 megawatts of electricity while burning 225,000 kg of coal per hour (almost 250 tons each hour!). If the coal has an average energy content of 20 million Joules per kg, determine the efficiency of the power plant.
Answers
Answer:
40%
Explanation:
The efficiency of a plant is determined by
[tex]eff=\frac{E_{produced}}{E_{in}}[/tex]
Our produced energy is
[tex]500 \ MW[/tex]
While the input energy can be calculated
[tex]E_{in}=number\ of \ coal \ in \ * energy\ per\ coal\ unit\\ =\frac{225000}{3600}coal/ s\ *20\ MJ/coal\\ = 1250\ MW[/tex]
So, the efficiency of plant is
[tex]eff=\frac{500}{1250}}\\ 0.4[/tex]
or 40% efficient
Final answer:
The efficiency of the coal-burning power plant in question is 40%, calculated by comparing the electrical energy output to the energy input from burning coal.
Explanation:
To determine the efficiency of the coal-burning power plant, we compare the energy output in the form of electricity to the energy input from burning coal. First, calculate the total energy input per hour by multiplying the amount of coal burned by the average energy content of the coal. Then, convert the electrical energy output from megawatts to joules per hour. Finally, calculate efficiency using the formula: Efficiency (%) = (Energy output / Energy input) × 100.
Here are the calculations:
Energy input per hour = 225,000 kg/hour × 20× 106 J/kg = 4.5× 1012 J/hourEnergy output per hour = 500 MW × 3.6× 106 J/MWs = 1.8× 1012 J/hourEfficiency = (1.8× 1012 J/hour / 4.5× 1012 J/hour) × 100 = 40%Thus, the efficiency of the power plant is 40%.
Determine the molarity and mole fraction of a 1.15 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the final volume equals the sum of the volumes of acetone and ethanol.
Answers
Answer:
The molarity is 0.85M
The mole fraction = 0.05
Explanation:
Step 1: Data given
Molarity of acetone = 1.15 m
Density of acetone = 0.788 g/cm³
Density of ethanol = 0.789 g/cm³
Molar mass of acetone = 58.08 g/mol
Step 2: Calculate number of moles acetone
Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)
1.15 m = 1.15 moles acetone / 1 kg ethanol
Step 3: Calculate mass of acetone
1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone
Step 4: Calculate volume of acetone
66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone
Step 5: Calculate volume of ethanol
1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol
Step 6: Calculate total volume solution
Total solution volume = 84.76 + 1267.43 = 1352.2 mL = 1.3522L
Step 7: Calculate molarity
Molarity of acetone = moles acetone /volume solution = 1.15moles / 1.3522L
Molarity = 0.85 M
Step 8: Calculate moles ethanol
moles ethanol= mass/ molar mass = 1000g/ 46.0g/mol = 21.74 moles
Step 9: Calculate mole fraction
mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05
Consider the perchlorateanion.
(a) What is the central atom?
(b) How many lone pairs are around the central atom?
(c) What is the ideal angle between the chlorine-oxygen bonds?
(d) Compared to the ideal angle, you would expect the actual angle between the chlorine-oxygen bonds to be ___
Answers
Answer:
(a) Chlorine atom
(b) 0
(c) 109.5°
(d) about the same
Explanation:
Perchlorate is a negatively charged molecule that is composed of one carbon atom and four oxygen atoms. In this molecule, the central chlorine atom that is present in +7 oxidation state, is covalently bonded to four oxygen atoms.
Thus the number of bond pairs is 4 and the number of lone pairs on the central chlorine atom is 0.
So according to the VSEPR theory, perchlorate ion has a tetrahedral geometry. Therefore, the ideal bond angle for the perchlorate ion is 109.5°.
The actual bond angle of perchlorate is about the same as ideal bond angle. This is because the perchlorate is resonance stabilized and thus all the chlorine-oxygen bonds are equivalent.
In the perchlorate anion, chlorine is the central atom with zero lone pairs. The ideal angle between the chlorine-oxygen bonds is 109.5 degrees, though the actual angle is likely slightly less due to repulsion between bond pairs.
Explanation:Considering the perchlorate anion (ClO4−):
(a) The central atom in the perchlorate anion is chlorine.(b) There are zero lone pairs around the central chlorine atom in the perchlorate anion.(c) The ideal angle between the chlorine-oxygen bonds in the perchlorate anion is 109.5 degrees, which is characteristic of a tetrahedral electron geometry.(d) However, due to the repulsion between bond pairs, the actual angle between the chlorine-oxygen bonds is expected to be slightly less than 109.5 degrees.Learn more about Perchlorate Anion here:https://brainly.com/question/28296182
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What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:
½ N2(g) + ½ O2(g) ---> NO(g)
K1
2 NO(g) + O2(g) ---> N2O4(g)
K2
K1^2 + K2
2 K1 x K2
K1^2 x K2
K1 + K2
Answers
Answer : The correct expression will be:
[tex]K=(K_1)^2\times K_2[/tex]
Explanation :
The chemical reactions are :
(1) [tex]\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)[/tex] [tex]K_1[/tex]
(2) [tex]2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)[/tex] [tex]K_2[/tex]
The final chemical reaction is :
[tex]N_2(g)+2O_2(g)\rightleftharpoons N_2O_4[/tex] [tex]K=?[/tex]
Now we have to calculate the value of [tex]K[/tex] for the final reaction.
Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.
If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.
If the two equations are added then equilibrium constant will be multiplied.
Thus, the value of 'K' will be:
[tex]K=(K_1)^2\times K_2[/tex]
Chemistry: An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container, if the gas has a pressure of the 845mmHg at 22degrees Celsius?
Answers
Answer:
There are 29.4 grams of oxygen in the container
Explanation:
Step 1: Data given
Volume = 20.0 L
Pressure = 845 mmHg
Temperature = 22.0 °C
Molar mass of O2 = 32 g/mol
Step 2: Ideal gas law
p*V = n*R*T
⇒ p = the pressure of the gas = 845 mmHg = 1.11184
⇒ V = the volume of the gas = 20.0 L
⇒ n = the number of moles = TO BE DETERMINED
⇒R = the gasconstant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 22°C + 273 = 295 Kelvin
n = (p*V)/(R*T)
n = (1.11184*20.0)/(0.08206*295)
n = 0.9186 moles
Step 3: Calculate mass of NO2
Mass of O2 = Moles O2 * Molar mass O2
Mass of O2 = 0.9186 moles * 32 g/mol
Mass of O2 = 29.4 grams
There are 29.4 grams of oxygen in the container
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.163 M in sodium chloride and 0.019 M in magnesium sulfate? Assume complete dissociation for electrolytes.
Answers
Answer:
The minimum pressure should be 901.79 kPa
Explanation:
Step 1: Data given
Temperature = 25°C
Molarity of sodium chloride = 0.163 M
Molarity of magnesium sulfate = 0.019 M
Step 2: Calculate osmotic pressure
The formula for the osmotic pressure =
Π=MRT.
⇒ with M = the total molarity of all of the particles in the solution.
⇒ R = gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 25 °C = 298 K
NaCl→ Na+ + Cl-
MgSO4 → Mg^2+ + SO4^2-
M = 2(0.163) + 2(0.019 M)
M = 0.364 M
Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)
Π = 8.90 atm
(8.90 atm)(101.325 kPa/atm) = 901.79 kPa
The minimum pressure should be 901.79 kPa
A piston system absorbs 50.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.491 atm. The final volume of the system is 56.2 L. What was the initial volume of the system if the internal energy of the system decreased by 106.0 J? (To convert between L∙atm and J, use 1 L∙atm = 101.3 J.)
Answers
Answer:
The initial volume of the system: V₁ = 53.06 L
Explanation:
Given: heat absorbed by system: q = 50.5 J, Pressure: P = 0.491 atm, Final volume: V₂ = 56.2 L, The change in the internal energy: ΔE = -106.0 J
Initial volume: V₁ = ? L
According to the First Law of Thermodynamics:
ΔE = q - PΔV
⇒ PΔV = q - ΔE = 50.5 J - (-106.0 J) = 156.5 J
As, 1 L∙atm = 101.3 J ⇒ 1 J = (1 ÷ 101.3) L∙atm
⇒ PΔV = 156.5 J = (156.5 ÷ 101.3) L∙atm = 1.54 L∙atm
So,
ΔV = 1.54 L∙atm ÷ P = 1.54 L∙atm ÷ 0.491 atm = 3.14 L
∵ ΔV = V₂ - V₁ = 3.14 L
⇒ V₁ = V₂ - 3.14 L = 56.2 L - 3.14 L = 53.06 L
Therefore, the initial volume of the system: V₁ = 53.06 L
w Available Hint(s) Check all that apply. When a solid is converted to a liquid, heat is absorbed. When heat is applied to a solid, the molecular motion decreases as the temperature increases. The average kinetic energy of the system changes while all of a solid is converted to a liquid. The temperature increases while all of a liquid is converted to a gas. Temperature remains constant while all of a solid is converted to a liquid.
Answers
Answer:
Here's what I get
Explanation:
When a solid is converted to a liquid, heat is absorbed. TRUE.
Temperature remains constant while all of a solid is converted to a liquid. TRUE.
When heat is applied to a solid, the molecular motion decreases as the temperature increases. False. Molecular motion increases as the temperature increases.
The average kinetic energy of the system changes while all of a solid is converted to a liquid. False. The temperature remains constant during a phase change, so the average kinetic energy of the system does not change.
The temperature increases while all of a liquid is converted to a gas. False.
Elemental phosphorus reacts with chlorine gas according to the equation: P4(s)+6Cl2(g)❝4PCl3(l) A reaction mixture initially contains 45.55g P4 and 131.9g Cl2.
Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
Answers
Answer:
So after reaction completion. 7.11 g of P₄ will be left over.
Explanation:
So for number of moles of P₄ = Given mass/ molar mass = 45.55/ 123.88
= 0.367 moles
Number of moles of Cl₂ = Given mass/ molar mass = 131.9/70 = 1.884 moles
We will divide the values by dividing excess of reactants with their coefficient for finding excess of reactants.
P₄ = 0.367/ coefficient 1 = 0.367
Cl₂ = 1.884/ 6= 0.314
Phosphorus is thus the excess reactant.
We need to use the mole value of the limiting reactant in order to find the mass of P₄ used. Using dimensional analysis,
1.884 mol Cl₂ ( 1mole P₄/6 mol Cl₂ ) ( 123.88 g P₄/ 1 mol P₄) = 38.44 g P₄
This is the amount which is used up. So by subtracting this amount from the Initial amount
45.55 – 38.44 = 7.11 g of P₄
So after reaction completion. 7.11 g of P₄ will be left over.
Final answer:
To find the mass of the excess reactant, we need to determine the limiting reactant first. By comparing the moles of each reactant, we can find that the excess reactant is chlorine (Cl2) and the mass left is 127.7 g.
Explanation:
To determine the mass of the excess reactant, we need to find the limiting reactant first. We can do this by comparing the moles of each reactant to their stoichiometric coefficients. Firstly, we need to convert the given masses of P4 and Cl2 to moles using their molar masses: 123.9 g/mol for P4 and 70.9 g/mol for Cl2. This gives us 0.368 mol of P4 and 1.86 mol of Cl2. From the balanced equation, we see that 1 mole of P4 reacts with 6 moles of Cl2, so we need 1/6th of the moles of Cl2 to react with all of the P4. This means that we need 0.368/6 = 0.061 mol of Cl2. The excess moles of Cl2 is then 1.86 - 0.061 = 1.799 mol. Finally, we convert the excess moles to grams using the molar mass of Cl2: 1.799 mol * 70.9 g/mol = 127.7 g. Therefore, the mass of the excess reactant (Cl2) is 127.7 g.
Lets say the reaction H2SO4 + 2KOH -> K2SO4 + 2H2O, is known to proceed by a 20.85% yield via a particular procedure.
How many grams of KOH must be reacted with an excess of H2SO4, in order to collect 67.5g of H2O?
Answers
Answer:
There should react 1008.3 grams KOH
Explanation:
Step 1: Data given
yield = 20.85 %
actual yield of H2O = 67.5 grams
Molar mass of H2SO4 = 98.08 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of KOH = 56.11 g/mol
Step 2: The balanced equation
H2SO4 + 2KOH ⇒ K2SO4 + 2H2O
Step 3: Calculate moles of H2O
Moles H2O = Mass H2O / Molar mass H2O
Moles H2O = 67.5 grams / 18.02 g/mol
Moles H2O = 3.746 moles
Step 4: Calculate theoretical yield
% yield= (actual yield/theoretical yield)*100%
Theoretical yield= (Actual yield/Percent yield) * 100%
Theoretical yield = (3.746 moles /20.85) *100%
Theoretical yield = 17.97 moles of H2O
Step 5: Calculate moles of KOH
For 2 moles of H2O produced, we need 2 moles of KOH
For 17.97 moles of H2O produced, we need 17.97 moles of KOH
Step 6: Calculate mass of KOH
Mass KOH =moles KOH * Molar mass KOH
Mass KOH = 17.97 moles * 56.11 g/mol
Mass KOH = 1008.3 grams
There should react 1008.3 grams KOH
A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is the pressure of the final sample (in atm)?A.) 1.4atmB.) 2.8atmC.) 3.2atmD.) 4.3atm
Answers
Answer:
[tex]\large \boxed{\text{B.) 2.8 atm}}[/tex]
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
[tex]\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}[/tex]
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
[tex]\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\[/tex]
(c) Convert the pressure to atmospheres
[tex]p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}[/tex]
The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l )
Answers
Answer:
The entropy decreases.
Explanation:
The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where
Δngas = n(gaseous products) - n(gaseous reactants)
If Δngas > 0, the entropy increasesIf Δngas < 0, the entropy decreases.If Δngas = 0, there is little or no change in the entropy.Let's consider the following reaction.
2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)
Δngas = 0 - 3 = -3, so the entropy decreases.
If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.0% of its original value? View Available Hint(s) If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.0% of its original value? 40.1 days 16.5 days 7.15 days 0.0163 days
Answers
Answer:
16.5 days
Explanation:
Given that:
Half life = 26.5 days
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{26.5}\ days^{-1}[/tex]
The rate constant, k = 0.02616 days⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
35.0 % is decomposed which means that 0.35 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.35 = 0.65
t = 7.8 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.65=e^{-0.02616\times t}[/tex]
t = 16.5 days.
The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, assign the energy (infrared, ultraviolet, or visible), the nr value, and all possible n, values up to 7 Series Balmer Brackett Paschen Lyman Energy nf ni visible) infrared ultraviolet 1 2 3 6 7
Answers
Answer:
Hydrogen spectrum
Explanation:
Balmer series - Observed in the visible region
Brackett series - Observed in the infrared region
Paschen series - Observed in the infrared region
Lyman series - Observe in the Ultraviolet region.
There are four sets of lines seen in the hydrogen emission spectra are Lyman series in the UV region, Balmer in the visible region and bracket and paschen in the IR region.
What is emission spectra?Emission spectrum is set of radiations emitted by an atom in the order of increasing frequency or decreasing wavelengths. Electrons in atoms transit between various energy levels.
Electrons excites from the lower energy level to the higher energy level by the absorption of light and return back to the ground state by emitting a complementary light.
The lyman series in the hydrogen emission spectrum is starting from the energy level n= 1. It is seen in the UV-region. Balmer series is found in visible region and their n value starts from n = 2.
Similarly Bracket starts from n= 3 and paschen series starts from n= 4. Both are overlapping lines together and they are seen in IR region.
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Calculate the number of Carbon atoms in 1.50g of C6H12O6
Answers
Answer:
3.01 ·10↑22
Explanation:
First you want to convert the grams of Glucose to moles of Glucose.
[tex]\frac{1.5 grams of glucose}{180.15588grams/molesglucose} =.008326 Moles of Glucose[/tex]
Next find the formula units of glucose.
.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =
5.01 ·10↑21 Formula Units of Glucose
Now multiply the formula units of glucose by the amount of each element in the molecule.
So for Carbon:
6carbon · 5.01 · 10↑21 = 3.01 · 10↑22
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Answer:
Number of C-atoms = 3.01 *10^22
Explanation:
Step 1: Data given
Mass of glucose = 1.50 grams
Molar mass of Glucose = 180.156 g/mol
Molar mass of carbon = 12.01 g/mol
Step 2: Calculate number of moles glucose
Number of moles glucose = mass glucose / molar mass glucose
Moles glucose = 1.50 grams / 180.156 g/mol
Moles glucose = 0.00833 moles
Step 3: Calculate formula units of glucose
0.00833 moles glucose * 6.02 *10^23 = 5.01 *10^21
Step 4: Calculate number of carbon atoms
Glucose has 6 C'atoms in it's molecular formule
Number of C-atoms = 5.01 *10^21 * 6 = 3.01 *10^22
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm? Please report your answer two points past the decimal with the unit J/molK. ∆H˚fus = 3.17 kJ/mol.
Answers
Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :
[tex]\Delta S=\frac{\Delta H_{freezing}}{T_f}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H_{fus}[/tex] = change in enthalpy of fusion = 3.17 kJ/mol
As we know that:
[tex]\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol[/tex]
[tex]T_f[/tex] = freezing point temperature = [tex]-97.6^oC=273+(-97.6)=175.4K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H_{freezing}}{T_m}[/tex]
[tex]\Delta S=\frac{-3170J/mol}{175.4K}[/tex]
[tex]\Delta S=-18.07J/mol.K[/tex]
Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
Answer:
Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.K
Explanation:
The entropy change can be calculated by the given formula:
[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]
in which,
[tex]\Delta \text S[/tex] = change in entropy
[tex]\Delta \text H_{\text {fusion}} &=[/tex] change in enthalpy of fusion that is -3.17 kJ/mol (as given).
By the formula of enthalpy change,
[tex]\begin{aligned} \Delta \text H_{\text {fusion}} &= - \Delta \text H_{\text {freezing}} &= -3.17 {\text {kJ/mol}} &= -3170 \text {J/mol}\end{aligned}[/tex]
Now,
[tex]{\text T_{f}}[/tex] = Freezing Point Temperature = [tex]-97.6^{o} \text C[/tex] = 273 + (-97.6) = -175.4 K
Substituting the values in above equation of entropy change, we get,
[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]
[tex]\begin{aligned} \Delta \text S &= \frac{-3170 \text {J/mol}}{175.4 \text {K}} \\\\Delta \text S& = -18.07 \text {J/mol.K}\end{aligned}[/tex]
Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
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