Answer:J.K^-1. kg^-1
Explanation:
Heat capacity= H
Mass=m
Specific heat capacity= c
H=mc
J.K^-1 = c ×kg
c= J.K^-1/kg
=J.K^-1.kg^-1
What is the standard entropy change for the reaction below? 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) S o (CO(g)) = 197.7 J/(mol·K) S o (CO2(g)) = 213.8 J/(mol·K) S o (NO(g)) = 210.8 J/(mol·K) S o (N2(g)) = 191.6 J/(mol·K)
Answer:
The standard entropy change for the reaction is -197.8 J/mol*K
Explanation:
Step 1: Data given
S°(CO(g)) = 197.7 J/(mol*K)
S°(CO2(g)) = 213.8 J/(mol*K)
S°(NO(g)) = 210.8 J/(mol*K)
S°(N2(g)) = 191.6 J/(mol·K)
Step 2: The balanced equation
2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)
Step 3: Calculate ΔS°
ΔS° = ∑S°(products) - ∑S°(reactants)
ΔS° = (191.6 + 2*213.8) - (2*210.8+2*197.7) J/mol*K
ΔS° = 619.2 J/mol*K - 817.0 J/mol *K
ΔS° = -197.8 J/mol* K
The standard entropy change for the reaction is -197.8 J/mol*K
Calculate the concentration of buffer components present in 287.00 mL of a buffer solution that contains 0.310 M NH4Cl and 0.310 M NH3 immediately after the addition of 1.50 mL of 6.00 M HNO3.
Final answer:
To determine the new concentrations of NH3 and NH4+ after the addition of HNO3, the moles of HNO3 added are calculated and the amounts of NH3 converted to NH4+ are accounted for. The final concentrations are found to be approximately 0.277 M for NH3 and 0.339 M for NH4+, after considering the change in total volume.
Explanation:
The student is asking to calculate the concentration of buffer components in a solution after the addition of a strong acid. To solve this, we have to determine how much the strong acid (HNO3) will neutralize the NH3 component of the buffer, and how it will change the concentrations of NH3 and NH4+ (the buffer components).
First, calculate the number of moles of HNO3 added:
1.50 mL × 6.00 M = 0.009 moles of HNO3Since NH3 and NH4+ are in a 1:1 molar ratio in the solution and they react with HNO3 in a 1:1 ratio, the added HNO3 will react with the NH3:
0.310 M × 0.28700 L = 0.08887 moles of NH3 (initial)0.08887 moles - 0.009 moles = 0.07987 moles of NH3 (after reaction with HNO3)And NH4+ will increase by 0.009 moles (since each mole of NH3 reacts to form one mole of NH4+):
0.310 M × 0.28700 L = 0.08887 moles of NH4+ (initial)0.08887 moles + 0.009 moles = 0.09787 moles of NH4+ (after reaction with HNO3)The new volume of the solution would be the initial volume of buffer solution plus the volume of HNO3 added, which totals 288.50 mL or 0.28850 L.
The new concentrations are therefore:
NH3 concentration = 0.07987 moles / 0.28850 L = approximately 0.277 MNH4+ concentration = 0.09787 moles / 0.28850 L = approximately 0.339 MThis change in concentrations due to the addition of HNO3 means the buffer will have adjusted to these new concentrations of NH3 and NH4+.
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 5.61 kJ·mol−1 at 2000. K and −52.80 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature. K at 2000. K= K at 3000. K= Assuming that ΔH∘rxn is independent of temperature, determine the value of ΔH∘rxn from this data. ΔH∘rxn=
Answer:
The equilibrium constant at 2000 K is 0.7139
The equilibrium constant at 3000 K is 8.306
ΔH = 122.2 kJ/mol
Explanation:
Step 1: Data given
the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K
the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K
Step 2: The equation
1/2X2(g)⟶X(g)
Step 3: Determine K at 2000 K
ΔG = -RT ln K
⇒R = 8.314 J/mol *K
⇒T = 2000 K
⇒K is the equilibrium constant
5610 J/mol = -8.314 J/molK * 2000 * ln K
ln K = -0.337
K = e^-0.337
K = 0.7139
The equilibrium constant at 2000 K is 0.7139
Step 4: Determine K at 3000 K
ΔG = -RT ln K
⇒R = 8.314 J/mol *K
⇒T = 3000 K
⇒K is the equilibrium constant
-52800 J/mol = -8.314 J/molK * 3000 * ln K
ln K = 2.117
K = e^2.117
K = 8.306
The equilibrium constant at 3000 K is 8.306
Step 5: Determine the value of ΔH∘rxn
ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)
ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)
2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)
2.455 = -ΔH/8.314 * (-1.67*10^-4)
-14700= -ΔH/8.314
-ΔH = -122200 J/mol
ΔH = 122.2 kJ/mol
When The constant at 2000 K is 0.7139
Then The constant at 3000 K is 8.306
ΔH = 122.2 kJ/mol
What is Equilibrium?Step 1: Data is given
When the quality molar Gibbs effectiveness of the formation of X(g) is 5.61 kJ/mol at 2000 K
When the quality molar Gibbs effectiveness of the formation of X(g) is -52.80 kJ/mol at 3000 K
Step 2: The equation is:
[tex]1/2X2(g)⟶X(g)[/tex]
Step 3: Then Determine K at 2000 K
ΔG = -RT ln K
[tex]⇒R = 8.314 J/mol *K[/tex]
⇒[tex]T = 2000 K[/tex]
⇒ at that time K is that the constant
Then 5610 J/mol = [tex]-8.314 J/molK * 2000 * ln K[/tex]
ln K is = [tex]-0.337[/tex]
K is =[tex]e^-0.337[/tex]
K is = [tex]0.7139[/tex]
When The constant at 2000 K is 0.7139
Step 4: Then Determine K at 3000 K
ΔG = -RT ln K
⇒[tex]R = 8.314 J/mol *K[/tex]
⇒[tex]T = 3000 K[/tex]
⇒K is that the constant
[tex]-52800 J/mol = -8.314 J/mol K * 3000 * ln K[/tex]
ln [tex]K = 2.117[/tex]
K = e^2.117
K = 8.306
The constant at 3000 K is 8.306
Step 5: at the moment Determine the worth of ΔH∘rxn
ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)
ln 8.306 /0.713 = -Δ[tex]H/8.314 * (1/3000 - 1/2000)[/tex]
2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)
2.455 = -ΔH/8.314 * [tex](-1.67*10^-4)[/tex]
-14700= -ΔH/8.314
-ΔH = [tex]-122200 J/mol[/tex]
Then ΔH = [tex]122.2 kJ/mol[/tex]
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Draw the mechanism arrows for both propagation steps for the radical addition of HBr to the alkene. When drawing single-headed radical arrows, this software requires that they meet at one atom (not in space between atoms like you may do in class). In the second box you will need to draw the first product and another reactant. In the last box you will need to draw an additional product.
Answer:
See the attached file for the structure
Explanation:
See the attached file for the explanation
A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt bridge. F2(g) + 2I-(aq) 2F-(aq) + I2(s)
The anode reaction is: + +
The cathode reaction is: + +
In the external circuit, electrons migrate the F-|F2 electrode the I-|I2 electrode. In the salt bridge, anions migrate the F-|F2 compartment the I-|I2 compartment.
Answer:
See explanation below
Explanation:
The anode reaction is :
2I^-(aq) -------> I2(g) +2e
Cathode reaction
F2(g) + 2e------> 2F^-(aq)
In the external circuit, electrons migrate from the I-|I2 electrode (anode) to the F-|F2 electrode (cathode)
In the salt bridge, anions migrate from the F-|F2
A voltaic cell is a type of cell in which a spontaneous redox reaction generates an electric current. This current is generated by the migration of electrons from the anode to the cathode in the external circuit, and the migration of anions in the salt bridge maintains electrical neutrality.
Explanation:A voltaic cell is a type of electrochemical cell where a spontaneous redox reaction generates an electric current. In this particular voltaic cell, the anode reaction is 2I-(aq) -> I2(s) + 2e-, where iodide ions are oxidized to solid iodine (losing electrons). The cathode reaction is F2(g) + 2e- -> 2F-(aq), where gaseous fluorine is reduced to fluoride ions (gaining electrons).
In the external circuit, electrons migrate from the anode (I-|I2 electrode) to the cathode (F-|F2 electrode). This migration of electrons generates the electric current. Lastly, in the salt bridge, anions migrate from the anode compartment (I-|I2) to the cathode compartment (F-|F2), allowing the cell to maintain electrical neutrality throughout the redox reaction.
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Which of these statements are true for a neutral, aqueous solution at 25 °C? pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00 Which of these statements are true for a neutral, aqueous solution regardless of temperature? [ H + ] = [ OH − ] pH = 7.00 pOH = 7.00
The statement which are true for a neutral, aqueous solution at 25 °C are:
pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00The statement which is/are true for a neutral, aqueous solution regardless of temperature is;
[ H + ] = [ OH − ]We must know that at standard temperature, 25°C, a neutral, aqueous solution has it's pH = pOH = 7. Additionally, the hydrogen ion concentration, [H+] is equal to its hydroxide ion concentration, [OH-]
pH is slightly affected by change in temperature as it decreases with increase in temperature. In a neutral aqueous solution, there are always the same concentration of hydrogen ions,[H+] and hydroxide ions, [OH-] and hence, the solution is still neutral (even if its pH changes).Ultimately, the pH of a solution decreases with increase in temperature and vice versa.
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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL volumetric flask and diluting to 25.00 mL prior to removing the 10.00 mL sample used above. What was the molar concentration of phosphoric acid in the original stock solution?
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]
10 ml 17.50 ml
(x) M 0.200 M
Molarity = [tex]\frac{0.2*17.5}{1000}[/tex]
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= [tex]0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}[/tex]
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = [tex]\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }[/tex]
Molar Concentration = [tex]\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000[/tex]
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
Which of the following statements is true? a. At equilibrium BOTH the rate of the forward reaction equals that of the reverse reaction AND the rate constant for the forward reaction equals that of the reverse. b. The equilibrium state is dynamic even though there is no change in concentrations. c. The equilibrium constant for a particular reaction is constant under all conditions. d. Starting with different initial concentrations will yield different individual equilibrium concentrations and a different relationship of equilibrium concentrations. e. None of these is true.
Answer:
a) True
Explanation:
a) From the definition of the equilibrium. When a reversible reaction is carried out in a closed vessel, a stage is reached when the forward and the backward reactions proceed with the same speed. This stage is known as chemical equilibrium.
The equipment used for aviation communications emits high‑frequency radiowave energy with a wavelength of 0.250 km. What is the energy of exactly one photon of this radiowave radiation?
Explanation:
Given that,
The wavelength of high‑frequency radio wave, [tex]\lambda=0.25\ km=250\ m[/tex]
We need to find the energy of exactly one photon of this radio wave radiation. It is given by :
[tex]E=\dfrac{nhc}{\lambda}[/tex]
Here, n = 1
[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{250}\\\\E=7.95\times 10^{-28}\ J[/tex]
So, the energy of exactly one photon of this radio wave radiation is [tex]7.95\times 10^{-28}\ J[/tex].
The energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
Given that,
The equipment used for aviation communications emits high‑frequency radio wave energy with a wavelength of 0.250 km i.e. = 250 m.Based on the above information, the calculation is as follows:
We know that
[tex]E = nhc \div \lambda\\\\= (6.63 \times 10^{-34} \times 3 \times 10^8) \div 250[/tex]
= [tex]7.95 \times 10^{-28}J[/tex]
Therefore we can conclude that the energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
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Huan wants to enter the science fair at his school. He has a list of ideas for his project. Which questions could be
answered through scientific investigation? Check all that apply.
Does pressure have an effect on the volume of a gas?
Which physicist was the smartest?
Is the information on the periodic table difficult to understand?
Which brand of soap is the best for cleaning grease off dishes?
Which laboratory experiment is the most fun to perform?
Correct answer:
•Does pressure have an effect on the volume of a gas?
•Which brand of soap is the best for cleaning grease off dishes?
The questions that can be answered by science must be empirical, that is they must be answerable by experiments.
The question that can be answered by science is; Which brand of soap is the best for cleaning grease off dishes?
This question can be answered by using various brands of soap to clean the same type of dish with the same type of grease and comparing the results. The answer to this question is pure empirical.
The other questions listed can not be answered by experiment. Their answer may vary from person to person therefore they are not scientific questions.
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Any help would be appreciated. Confused.
Answer:
q(problem 1) = 25,050 joules; q(problem 2) = 4.52 x 10⁶ joules
Explanation:
To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...
=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:
Heat of fusion (ΔHₓ) = 80 cal/gram
Heat of vaporization (ΔHv) = 540 cal/gram
specific heat of ice [c(i)] = 0.50 cal/gram·°C
specific heat of water [c(w)] = 1.00 cal/gram·°C
specific heat of steam [c(s)] = 0.48 cal/gram·°C
Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ... Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.
q(warming ice) = m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal
q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal
q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal
q(evaporation of water) = m·ΔHv = (10g)(540cal/g) = 5400 cal
q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal
Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.
Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).
Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.
Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.
Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).
Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.
Problems containing a temperature change:
NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve. Good luck on your efforts. Doc :-)
What pressure is needed to reduce the volume of gases in a car’s cylinder from 48.0 cm3 at 102 kPa to 5.20 cm3?
Answer: The pressure that is needed is around 405kPa-755kPa
To summarize into 405PPM-755PPM
Explanation: In which 405 PPM-755 PPM which is about the same amount of pressure that water pressurises a car in water or a better example is that it's the same amount of pressure as if a penny was dropped from the sky towards a person holding a square piece of cardboard in which then the penny would directly go straight through the piece of cardboard.
Which statement describes solctices
Answer:funk
hot dog cat
Explanation:
uhughuhuhuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
Answer:
They occur when the sun reaches its highest or lowest point in the sky
The concentration of a mixture can be increased in which of the following ways?
Answer:
C. adding more powder solute
Explanation:
The concentration of a mixture can be increased by removing solvent.
The concentration of a mixture can be increased in several ways. If we consider a distillery wanting to achieve a higher concentration of alcohol, they need to remove solvent, which in this case is water, from their product. This can be achieved through processes such as distillation, which separates alcohol from water due to their different boiling points. Another approach is the addition of a substance that reacts with the water, effectively removing water content from the mixture.
To increase the concentration of a solution in general, one can also add more solute (the substance being dissolved) into the solution or remove solvent (the substance dissolving the solute). It is important not to confuse the process of dilution, which involves adding solvent and decreases solute concentration, with the process of concentrating a solution, which involves removing solvent and increases solute concentration. Therefore, methods such as evaporation, reverse osmosis, or chemical reaction can be employed to increase the concentration of a solute in a solution.
What is the density of N2 gas (molar mass: 28 g/mol), at 400 K and 2 atm?
Answer:
Density= 1.7g/dm3
Explanation:
Applying
P×M= D×R×T
P= 2atm, Mm= 28, D=? R= 0.082, T= 400K
2×28= D×0.082×400
D= (2×28)/(0.082×400)
D= 1.7g/dm3
The procedure for testing your unknown solution in this week's lab is identical to the procedure which you conducted in Week 1. The only difference is, of course, your Unknown Solution may or may not contain all of the ions which you tested for in Week 1. With that being said, please consider the following scenario: You enter the lab and obtain an Unknown Solution from the Stockroom. You begin testing the solution through the steps outlined in the flowchart on p. 9 of the Exp 22 document. You first add HCl, and centrifuge your mixture. You observe the formation of a white precipitate in the bottom of the test tube. After pouring off the supernatant liquid, you add hot water to the white precipitate. Upon addition of the hot water, you still have some white precipitate in the bottom of the test tube. You add ammonia, NH3, to the test tube and observe the formation of a gray-black precipitate. Which of the following is the best conclusion to draw at this point
A. The unknown solution definitely has Ag+ present.
B. The unknown solution could have Agt present, or Hg22+ present, or BOTH.
C. The unknown solution definitely has Hg22+ present.
D.The Unknown Solution definitely has Pb2+ present.
Answer:
The correct answer is option C. The unknown solution definitely has Hg22+ present.
Explanation:
In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.
Balanced chemical equation of the reation is -
Hg2Cl2 + 2NH3 ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl
Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.
Answer:
C- the unknown solution definitely has Hg22+ present.
Explanation:,
The box below to the left represents ions in a certain volume of 0.10MHCl(aq) . In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq)
Complete question:
The student performs a second titration using the 0.10MNaOH(aq) solution again as the titrant, but this time with a 20.mL sample of 0.20MHCl(aq) instead of 0.10MHCl(aq).
1. The box below to the left represents ions in a certain volume of 0.10MHCl(aq). In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq). (Do not include any water molecules in your drawing.)
Answer:
The concentration of H+ and Cl- will be doubled.
Explanation:
See the attached photo for the representation of ions in the same volume.
Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in of hot water (). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature.It is likely that less rock candy will be formed in batch A. It is likely that no rock candy will be formed in either batch. I need more information to predict which batch is more likely to form rock candy.
Rock candy is formed through a chemical process known as crystallization which requires a supersaturated sugar solution. In the scenario, batch A prepared with hot water is more likely to form rock candy as it can dissolve more sugar creating a supersaturated solution. Batch B prepared at room temperature may not form as much rock candy due to lesser sugar dissolution.
Explanation:The question relates to the chemical process of crystallization, particularly in the formation of rock candy. When making rock candy, a supersaturated solution of sugar and water is required. This condition is achieved when as much sugar as possible is dissolved in hot water. Once cooled, the oversaturated solution starts to crystallize and forms rock candy. So in the given scenario, batch A is more likely to produce rock candy because it involves the preparation of a supersaturated solution through dissolving sugar in hot water. Batch B, prepared at room temperature, may not dissolve as much sugar as batch A, and thus, less or no rock candy might be formed.
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Batch A, which dissolved sugar in hot water that was then cooled, is more likely to form rock candy as the cooling process can lead to the crystallization of sugar from a supersaturated solution.
The process of making rock candy involves dissolving sugar in water to create a saturated solution from which sugar crystals can form upon cooling or evaporation of the solvent. The solubility of sugar increases with temperature, which means hot water can dissolve more sugar than room temperature water. Therefore, for batch A, dissolving sugar in hot water likely supersaturates the solution, and as it cools to room temperature, excess sugar will crystallize out.
Batch B, on the other hand, dissolves sugar at room temperature, potentially creating a saturated solution, but without the temperature change, it is less likely to form a supersaturated environment and thus may yield less crystallization compared to batch A. In summary, batch A is more likely to form rock candy due to the temperature-dependent solubility of sugar, and the cooling process allows crystals to form from the supersaturated solution.
Equation: 3Cu(s) 8HNO3(aq) --> 2NO(g) 3Cu(NO3)2(aq) 4H2O(l) In the above reaction, the element oxidized is ______, the reducing agent is ______ and the number of electrons transferred from reducing to oxidizing agent in the equation, as written, is ______.
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
In the given chemical reaction, the element oxidized and the reducing agent is Copper (Cu). The number of electrons transferred from the reducing agent to the oxidizing agent is six.
Explanation:In the reaction equation, 3Cu(s) + 8HNO3(aq) --> 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l), the element oxidized is Cu (Copper), as Copper goes from an oxidation state of 0 in Cu(s) to +2 in Cu(NO3)2. Hence, the reducing agent is also Copper (Cu). Oxidation is the process of losing electrons, and in this chemical reaction, each copper atom loses 2 electrons (going from 0 to +2). Since the equation shows the reaction of 3 copper atoms, therefore, the total number of electrons transferred from reducing to oxidizing agent in the equation, as written, is 6.
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Complete the electron pushing mechanism for the formation of the following cyclic acetal under acidic conditions by adding any missing atoms, bonds, charges, non-bonding electrons, and curved arrows. Note the use of a generic base B: that represents any basic molecule in solution, in this case another ethylene glycol.
Question:
The question is incomplete. See the attached file for the complete question and answer.
Explanation:
Find attached for explanation.
The first two pages is the additional question while the 3rd and last page is the answer .
Which statements are true about balancing chemical reactions?
Check all that apply
A. Atoms that are in only one of the reactants and only one of the
products should be done last.
B. Single atoms should be done last.
C. Balancing reactions involves trial and error
D. The final coefficients should be the biggest numbers possible
Final answer:
Balancing chemical reactions involves trial and error, with the aim of achieving the same number of atoms for each element on both sides of the equation. Atoms that are in only one of the reactants and only one of the products should be done last, and the final coefficients should be the biggest numbers possible.
Explanation:
There are several statements that are true about balancing chemical reactions:Atoms that are in only one of the reactants and only one of the products should be done last. This is because it is easier to balance elements that appear in multiple reactants and products first.The final coefficients should be the biggest numbers possible. This is because coefficient values should be integers and the aim is to attain the simplest whole number ratio.Balancing reactions involves trial and error. It requires adjusting the coefficients of the reactants and products until the number of atoms of each element on either side of the equation is balanced.The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation. Which compounds donate electrons to the electron transport chain? H 2 O H2O FADH 2 FADH2 ADP ADP NADH NADH NAD + NAD+ O 2 O2 FAD FAD ATP ATP Which compound is the final electron acceptor? NADH NADH ATP ATP H 2 O H2O NAD + NAD+ FADH 2 FADH2 FAD FAD O 2 O2 ADP ADP Which compounds are the final products of the electron transport chain and oxidative phosphorylation? ADP ADP NADH NADH O 2 O2 NAD + NAD+ FADH 2 FADH2 ATP ATP H 2 O H2O FAD
Answer:
The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation.
a)The compounds that donate electrons to the electron transport chain are NADH and . FADH2
b) O2 is the final electron acceptor.
c) The final products of the electron transport chain and oxidative phosphorylation are NAD+, H2O, ATP and FAD
Explanation:
119. In analytical chemistry, bases used for titrations must often be standardized; that is, their concentration must be precisely determined. Standardization of sodium hydroxide solutions can be accomplished by titrating potassium hydrogen phthalate (KHC8H4O4), also known as KHP, with the NaOH solution to be standardized. a. Write an equation for the reaction between NaOH and KHP. b. The titration of 0.5527 g of KHP required 25.87 mL of an NaOH solution to reach the equivalence point. What is the concentration of the NaOH solution
Answer:
0.1046M NaOH solution
Explanation:
a. KHP is a salt used as primary standard because allows direct standarization of bases solutions. The reaction of KHP with NaOH is:
KHP + NaOH → H₂O + KP⁻ + Na⁺
As you can see, KHP has 1 acid proton that reacts with NaOH.
Molar mass of KHP is 204.22g/mol; 0.5527g of KHP contains:
0.5527g KHP × (1mol / 204.22g) = 2.706x10⁻³moles of KHP. As 1 mole of KHP reacts per mole of NaOH, at equivalence point you must add 2.706x10⁻³moles of NaOH
As you spent 25.87mL of the solution, molarity of the solution is:
2.706x10⁻³moles of NaOH / 0.02587L = 0.1046M NaOH solution
Answer:
NaOH + KHC8H4O4 → NaKC8H4O4 + H2O
Concentration NaOH = 0.105 M
Explanation:
Step 1: Data given
Mass of KHP = 0.5527 grams
Molar mass KHP = 204.22 g/mol
Volume of NaOH = 25.87 mL
Step 2: The balanced equation
NaOH + KHC8H4O4 → NaKC8H4O4 + H2O
Step 3: Calculate moles KHP
Moles KHP = mass KHP / molar mass KHP
Moles KHP = 0.5527 grams / 204.22 g/mol
Moles KHP = 0.002706 moles
Step 4: Calculate moles NaOH
For 1 mol NaOH we need 1 mol KHP to react
For 0.002706 moles KHP we need 0.002706 moles NaOH
Step 5: Calculate concentration NaOH
Concentration = moles / volume
Concentration NaOH = 0.002706 moles / 0.02587 L
Concentration NaOH = 0.105 M
a cell contains 180 m^3 of gas at 7400 Pa and a machine. The machine is turned on remotely and expands the box. During this process, the machine also gives off 260 kJ of heat to the gas, and the internal energy is determined to be -69kJ. What is the final volume of the cell? Assume pressure stays constant.
Please Show All Work
Answer:
[tex]224.5 m^3[/tex]
Explanation:
By using the first law of thermodynamics, we can find the work done by the gas:
[tex]\Delta U=Q-W[/tex]
where in this problem:
[tex]\Delta U=-69 kJ[/tex] is the change in internal energy of the gas
[tex]Q=+260 kJ[/tex] is the heat absorbed by the gas
W is the work done by the gas (positive if done by the gas, negative otherwise)
Therefore, solving for W,
[tex]W=Q-\Delta U=+260-(-69)=+329 kJ = +3.29\cdot 10^5 J[/tex]
So, the gas has done positive work: it means it is expanding.
Then we can rewrite the work done by the gas as
[tex]W=p(V_f-V_i)[/tex]
where:
[tex]p=7400 Pa[/tex] is the pressure of the gas
[tex]V_i=180 m^3[/tex] is the initial volume of the gas
[tex]V_f[/tex] is the final volume
And solving for Vf, we find
[tex]V_f=V_i+\frac{W}{p}=180+\frac{3.29\cdot 10^5}{7400}=224.5 m^3[/tex]
Final answer:
To determine the final volume of the gas, the first law of thermodynamics was used, considering the process as isobaric due to constant pressure. The final volume was calculated to be approximately 45045.9 m³ after applying the formula and accounting for the work done and heat supplied.
Explanation:
To find the final volume of the gas, we can use the first law of thermodynamics, which is given by ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the question states the pressure remains constant (isobaric process), the work done by the gas can be found by W = pΔV, where p is the pressure and ΔV is the change in volume.
From the given data, we have:
Initial internal energy change (ΔU) = -69 kJ (since the value is negative, this indicates that the gas loses energy)Heat added to the system (Q) = 260 kJInitial volume (V₁) = 180 m³Pressure (p) = 7400 PaUsing ΔU = Q - W and substituting W with pΔV, we get:
-69 kJ = 260 kJ - (7400 Pa × ΔV)
Convert kJ to J by multiplying by 1000 (since 1 kJ = 1000 J):
-69000 J = 260000 J - (7400 Pa × ΔV)
We can then solve for ΔV:
ΔV = (260000 J + 69000 J) / 7400 Pa
ΔV = 44865.9 m³ (rounded to one decimal place)
To find the final volume (V₂), we add the change in volume to the initial volume:
V₂ = V₁ + ΔV
V₂ = 180 m³ + 44865.9 m³
V₂ = 45045.9 m³
So, the final volume of the gas after the process is approximately 45045.9 m³.
Use the Ideal Gas Law to calculate the number of moles (n) of helium in a 4000 Liter weather balloon near the top of Mt. Rainier with a pressure of 0.6 atm and temperature of 260K.
Answer: 112.5moles
Explanation:
P= 0.6atm, V= 4000L, R= 0.082, T= 260K
Applying PV = nRT
0.6×4000= n×0.082×260
Simplify n= (0.6× 4000)/(0.082×260)
n= 112.5moles
Using the Ideal Gas Law, we calculated that the weather balloon near the top of Mt. Rainier, with given conditions, contains approximately 113.1 moles of helium.
The Ideal Gas Law formula is PV = nRT, where:
P = pressure (in atm)V = volume (in liters)n = number of molesR = ideal gas constant (0.0821 L·atm/mol·K)T = temperature (in Kelvin)We can use this formula to calculate the number of moles of helium in the weather balloon.
Given:
Pressure (P) = 0.6 atmVolume (V) = 4000 LTemperature (T) = 260KUsing the Ideal Gas Law:
PV = nRT[tex]n = \frac{PV}{RT}[/tex]Plugging in the given values:
[tex]n = \frac{0.6 \, \text{atm} \times 4000 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \times 260 \, \text{K}}[/tex]n ≈ 113.1 molesThus, the number of moles of helium in the weather balloon is approximately 113.1 moles.
Labels of many food products have expiration dates, at which point they are typically removed from the supermarket shelves. A particular natural yogurt degrades (reacts) with a half-life of 45 days. The manufacturer of the yogurt wants unsold product pulled from the shelves when it degrades to no more than 80% (0.8) of its original quality. Assume the degradation process is first order. What should be the "best if used before" date on the container with respect to the date the yogurt was packaged?
Answer:
Explanation:
Since it first order, we use order rate equation
In ( [tex]\frac{A1}{A0}[/tex]) = -kt where A1 is the final quality = 0.8 (80%), A0 is the initial quality = 1 ( 100%)
also, t half life = [tex]\frac{In2}{k}[/tex] where k is rate constant
k = [tex]\frac{In 2}{45 days}[/tex] = 0.0154
In ( [tex]\frac{0.8}{1}[/tex]) = - 0.0154 t
-0.223 / -0.0154 = t
t = 14.49 approx 14.5 days from the date the yogurt was packaged
You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = g
Answer:
[tex]m_{ice} = 65.336\,g[/tex]
Explanation:
Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:
[tex]Q_{water} = Q_{ice}[/tex]
[tex](100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)[/tex]
The amount of ice that is melt is:
[tex]m_{ice} = 65.336\,g[/tex]
Draw the line‑bond structure of oleic acid (cis‑9‑octadecenoic acid), CH 3 ( CH 2 ) 7 CH = CH ( CH 2 ) 7 COOH , at physiological pH. Hydrogen atoms attached to carbon atoms do not need to be drawn.
Answer : The line-bonds structure of oleic acid (cis‑9‑octadecenoic acid) is shown below.
Explanation :
In line-bonds formula, the organic structures of the compound are represented in such a way that covalent bonds are represented by line and frame the structure by zig-zag straight lines which emits all the hydrogen atom.
The terminals of the line and vertex represents carbon atoms whose valences are satisfied by formation of single bonds with H atom.
The given compound is, (cis‑9‑octadecenoic acid)
In this compound, the parent chain is 18 membered and the carboxylic acid functional group is attached to it.
A) Al3 (aq) is a stronger oxidizing agent than I2(s), and I-(aq) is a stronger reducing agent than Al(s).B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).C) Al(s) is a stronger oxidizing agent than I-(aq), and Al3 (aq) is a stronger reducing agent than I2(s).D) I-(aq) is a stronger oxidizing agent than Al(s), and I2(s) is a stronger reducing agent than Al3 (aq).
Answer:
B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).
Explanation:
An oxidizing agent accepts electrons in a redox reaction and become reduced while a reducing agent looses electrons and become oxidized.
Hence in a redox reaction, oxidizing agents are reduced while reducing agents are oxidized.
Looking at I2 and Al3+, I2 is a better oxidizing agent since it has a reduction potential of +0.54V compared to -1.66V for Al3+.
Given the statements above, the converse must be true, that is; All is a better reducing agent compared to I-
Benzene was treated with a set of unknown reagents. Using the spectral data below, draw the most likely product and select the most likely set of reagents. 1H NMR: See spectra below. 13C NMR: 142 ppm, 128 ppm, 128 ppm, 125 ppm, 38 ppm, 24 ppm, and 13 ppm. MS: There is no detectable M+2 peak. IR: Peaks observed at 3100 cm-1 and at 2900 cm-1. There is NO absorbance near 1700 cm-1.
Answer:
CHECK THE ATTACHMENT TO SEE THE STRUCTURAL DIAGRAM
Explanation:
The spectral data suggests the product is ethylbenzene, which can be synthesized from benzene using ethylchloride and a catalyst like AlCl3 via a Friedel-Crafts alkylation reaction.
Explanation:
The spectral data provided corresponds to the molecule ethylbenzene. The presence of a 13C NMR signal at 142 ppm indicates a carbon directly invested in a aromatic system. The signals at 38 ppm, 24 ppm and 13 ppm are indicative of the ethyl side chain. The IR peaks at 3100 cm-1 and 2900 cm-1 indicate C-H stretching of aromatic and aliphatic hydrogens respectively. The absence of an absorbance near 1700 cm-1 indicates absence of carbonyl group. The lack of a M+2 peak in MS data suggests no halogens are part of the structure. To obtain ethylbenzene from benzene, the reagents used would include ethylchloride and aluminum chloride (AlCl3) in a Friedel-Crafts alkylation reaction.Benzene, ethylchloride and AlCl3 are the key components to this reaction.
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