6.81% is the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen.
Explanation:
Data given:
mass of carbon dioxide formed = 12 grams (actual yield)
atomic mass of CO2 = 44.01 grams/mole
moles of [tex]C_{8} H_{18}[/tex] = 0.5
Balanced chemical reaction:
2 [tex]C_{8} H_{18}[/tex] + 25 [tex]O_{2}[/tex] ⇒ 16 C[tex]O_{2}[/tex] + 18 [tex]H_{2} _O{}[/tex]
number of moles of carbon dioxide given is
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
number of moles= [tex]\frac{12}{44}[/tex]
number of moles of carbon dioxide gas = 0.27 moles
from the reaction 2 moles of [tex]C_{8} H_{18}[/tex] reacts to produce 16 moles of C[tex]O_{2}[/tex]
So, when 0.5 moles reacted it produces x moles
[tex]\frac{16}{2}[/tex] = [tex]\frac{x}{0.5}[/tex]
x = 4
4 moles of carbon dioxide formed, mass from it will give theoretical yield.
mass = number of moles x molar mass
mass = 4 x 44.01
= 176.04 grams
percent yield =[tex]\frac{actual yield}{theoretical yield}[/tex] x 100
percent yield = [tex]\frac{12}{176.04}[/tex] x 100
percent yield = 6.81 %
Calcium hydride reacts with water to form hydrogen gas according to the unbalanced equation below: CaH2(s) + H2O(l) --> Ca(OH)2(aq) + H2(g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of calcium hydride are needed to generate 15.0 L of hydrogen gas at 25 degrees C and 825 torr of pressure?
Answer:
28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.
Explanation:
CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)
Using ideal gas law, PV = nRT
=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)
From stoichiometry of given equation,
=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)
Converting moles to grams, multiply by formula weight,
=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s) (2 sig. figs.)
A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.9 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
A flammable gas made up of only carbon and hydrogen is found to effuse through a porous barrier in 1.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be.
Answer:
A. The molar mass of the unknown gas is 16g/mol
B. Compound is CH4 i.e methane
Explanation:
A. Step 1:
Representation:
Let t1 be time for unknown gas
Let t2 be the time for bromine vapor
Let M1 be molar mass of the unknown gas
Let M2 be the molar mass of bromine vapor
A. Step 2 :
Data obtained from the question.
Time for the unknown gas (t1) = 1.50 min
Time for Br2 (t2) = 4.73 min
Molar Mass of unknown gas (M1) =?
Molar Mass of Br2 (M2) = 80 x 2 = 160g/mol
A. Step 3:
Determination of the molar mass of the unknown gas.
Applying the equation:
t2/t1 = √(M2/M1)
The molar mass of can be obtained as follow:
t2/t1 = √(M2/M1)
4.73/1.5 = √(160/M1)
Take the square of both side
(4.73/1.5)^2 = 160/M1
9.94 = 160/M1
Cross multiply to express in linear form.
9.94 x M1 = 160
Divide both side by 9.94
M1 = 160/9.94
M1 = 16g/mol
Therefore, the molar mass of the unknown gas is 16g/mol
B. Identification of the gas.
The gas contains C and H only. From the calculations made above, the molar mass of the unknown gas is 16g/mol.
We know also that the molar mass of carbon is 12g/mol and that of Hydrogen is 1g/mol
Therefore,
C + H = 16
There would be only 1 atom of C in the compound since the molar mass of the compound is 16g/mol. With these understanding, let us determine the number of H atom in the compound. This is illustrated below :
C + H = 16
12 + H = 16
H = 16 - 12
H = 4
Divide by the molar mass of H i.e 1
H = 4/1 = 4
There are 4 atoms of H in the compound. Therefore, the compound is CH4 i.e methane
The molar mass of the unknown gas is 16 g/mol hence the unknown gas is methane.
We must note that the time taken for a gas to diffuse is directly proportional to the molar mass of the gas.
Hence;
t1/t2 = √M1/M2
Let t1 = time taken for the flammable gas to diffuse = 1.50 min
Let t2 = time taken for the bromine vapor to diffuse = 4.73 min
M1 = molar mass of the flammable gas = ?
M2 = molar mass of the bromine vapor = 160 g/mol
Substituting values;
1.50/4.73 = √M1/160
(1.50/4.73)^2 = M1/160
0.1006 = M1/160
M1 = 0.1006 × 160
M1 = 16 g/mol
The unknown gas is methane.
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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
There are three competing factors at play here: 1. The effective nuclear charge 2. The size of the atom and the force of attraction according to Coulomb's law 3. A pair of electrons in an orbital The first ionization energy is the energy required to completely remove the first electron from the atom. The higher in energy an electron is to start with, the less additinal energy will be required to remove it, which translates to a lower ionization energy. By the same token, an electron arrangement which is lower in energy will require more energy to remove an electron.
Answer:
These three factors are required for ionization potential or ionization energy.
Explanation:
Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.
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.
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]
The charges and sizes of the ions in an ionic compound affect the strength of the electrostatic interaction between the ions and thus the strength of the lattice energy of the ionic compound. Arrange the compounds according to the magnitudes of their lattice energies based on the relative ion charges and sizes.- MgS
- NaCl
- MgCl_2
- KBr
The charges and sizes of ions in an ionic compound determine the strength of its lattice energy.
Explanation:The strength of the lattice energy in an ionic compound is determined by the charges and sizes of the ions. The larger the charges and the smaller the ion sizes, the stronger the electrostatic interaction and the higher the lattice energy. Based on this, the compounds can be arranged in order of their magnitudes of lattice energies:
MgSKBrNaClMgCl2MgS has the highest lattice energy because of the high charge and small size of both the Mg2+ and S2- ions.
So the order of lattice energies would be: MgS > MgCl_2 > NaCl > KBr
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At 5000 K and 1.000 atm, 83.00% of the oxygen molecules in a sample have dissociated to atomic oxygen. At what pressure will 95.0% of the molecules dissociate at this temperature
Answer:
At [tex]P_{total} = 0.240\ atm[/tex]; 95.0% of the molecules will dissociate at this temperature
Explanation:
The chemical reaction of this dissociation is:
[tex]O_2 \leftrightarrow 2O_g[/tex]
The ICE table is as follows:
[tex]O_2 \ \ \ \ \ \ \ \leftrightarrow \ \ \ \ \ \ \ \ 2O_g[/tex]
Initial 100 0
Change -83 +166
Equilibrium 17 166
The mole fractions of each constituent is now calculated as:
[tex]Mole \ fraction \ of \ O(X_o) = \frac{166}{183}[/tex] = 0.9071
[tex]Mole \ fraction \ of \ O_2(X_o_2_}}) = \frac{17}{183}[/tex] = 0.0929
Given that the total pressure [tex]P_{total}[/tex] = 1.000 atm ; the partial pressure of each gas is calculated by using Raoult's Law.
[tex]Partial \ Pressure \ of \ O (P_o) = X_oP_{total}\\\\ = 0.9071 \ atm[/tex]
[tex]Partial \ Pressure \ of \ O_2 (P_o_2) = X_oP_{total}\\\\ = 0.0929 \ atm[/tex]
Now; we proceed to determine the equilibrium constant [tex]K_c[/tex]; which is illustrated as:
[tex]K_c = \frac{Po^2}{Po_2} \\ \\ K_c = \frac{(0.9072)^2}{0.0929} \\ \\ =8.86 \ atm[/tex]
Let assume that the partial pressure of [tex]O_2[/tex] be x ;&
the change in pressure of [tex]O_2[/tex] be y ; then
we can write that the following as the changes in concentration of species :
[tex]O_2 \ \ \ \ \ \ \ \leftrightarrow \ \ \ \ \ \ \ \ 2O_g[/tex]
Initial x 0
Change -y +2 y
Equilibrium x - y 2 y
From above; we can rewrite our equilibrium constant as:
[tex]K_c = \frac{(2y)^2}{x-y} \\ \\ 8.86 = \frac{(2y)^2}{x-y} ----- equation (1)[/tex]
From the question; we are told that provided that 95% of the molecules dissociate at this temperature. Therefore, we have:
[tex]\frac{y}{x}*100 = 95[/tex]% -------- equation (2)
Solving and equating equation 1 and 2 ;
x = 0.123 atm
y = 0.117 atm
Thus, the pressure required can be calculated as :
[tex]P_{total} = (x-y) +2y \\ \\ P_{total} = (0.123- 0.117)+ 2(0.123) \\ \\ \\ P_{total} = 0.240 \ atm[/tex]
It is difficult to prepare an amide from a carboxylic acid and an amine directly, since an acid-base reaction occurs which renders the amine nitrogen non-nucleophilic. Typically, in such an amide synthesis, the carboxylic acid OH group is first transformed into a better, nonacidic leaving group. In practice, amides are often prepared by treating the carboxylic acid with dicyclohexylcarbodiimide (DCC). The amine is then added and nucleophilic acyl substitution occurs easily because dicyclohexylurea is a good leaving group. This method of amide bond formation is a key step in the laboratory synthesis of peptide bonds (amide bonds) between protected amino acids. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution.
A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both species 1 and 2. The pressure of this two- phase system is 1 bar; the temperature is 25°C. At this temperature, 1 = 200 bar and P2sat = 0.10 bar. Determine good estimates of x1 and y1. State and justify all assumptions.
Answer:
x1= 4.5 × 10^-3, y1= 0.9
Explanation:
A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2
Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:
1. The vapor phase is ideal at pressure of 1 bar
2. Henry's law apply to dilute solution only.
3. Raoult's law apply to concentrated solution only.
Where,
Henry's constant for species 1 H= 200bar
Saturation vapor pressure of species 2, P2sat= 0.10bar
Temperature = 25°C= 298.15k
Apply Henry's law for species 1
y1P= H1x1...... equation 1
y1= mole fraction of species 1 in vapor phase.
P= Total pressure of the system
x1= mole fraction of species 1 in liquid phase.
Apply Raoult's law for species 2
y2P= P2satx2...... equation 2
From the 2 equations above
P=H1x1 + P2satx2
200bar= H1
0.10= P2sat
1 bar= P
Hence,
P=H1x1 + (1 - x1) P2sat
1bar= 200bar × x1 + (1 - x1) 0.10bar
x1= 4.5 × 10^-3
The mole fraction of species 1 in liquid phase is 4.5 × 10^-3
To get y, substitute x1=4.5 × 10^-3 in equation 1
y × 1 bar = 200bar × 4.5 × 10^-3
y1= 0.9
The mole fraction of species 1 in vapor phase is 0.9
What is Binary solution?
A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2
If we consider the pressure of the 2 phase system is 1 bar.
The assumption are as follows:
The vapor phase is ideal at pressure of 1 bar. Henry's law apply to dilute solution only. Raoult's law apply to concentrated solution only.Where, these values are given:
Henry's constant for species 1 H= 200bar
Temperature = 25°C= 298.15K
P₂sat= 0.10 bar
Apply Henry's law for species 1
y₁P= H₁x₁.......... (i)
where y₁= mole fraction of species 1 in vapor phase, P= Total pressure of the system ,x₁= mole fraction of species 1 in liquid phase.
Apply Raoult's law for species 2
y₂P= P₂sat. x₂...........(ii)
From (i) and (ii)
P=H₁x₁ + P₂sat. x₂
200bar= H₁
0.10= P₂sat
1 bar= P
Hence,
P=H₁x₁ + (1 - x₁) P₂sat
1bar= 200bar × x₁ + (1 - x₁) 0.10bar
x₁= [tex]4.5 * 10^{-3}[/tex]
The mole fraction of species 1 in liquid phase is [tex]4.5 * 10^{-3}[/tex]
To get y, substitute x₁=[tex]4.5 * 10^{-3}[/tex] in (i)
y × 1 bar = 200bar × [tex]4.5 * 10^{-3}[/tex]
y₁= 0.9
The mole fraction of species 1 in vapor phase is 0.9.
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Measurements show that the enthalpy of a mixture of gaseous reactants increases by 215. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -155. kJ of work is done on the mixture during the reaction.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 2 significant digits.Is the reaction exothermic or endothermic?
Answer: The change in energy of the gas mixture during the reaction is 60 kJ. The reaction is endothermic.
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done on the system
w = -155 kJ
q = +215kJ {Heat absorbed by the system is positive}
[tex]\Delta E=+215+(-155)kJ=60kJ[/tex]
As the heat is absorbed and enthalpy increases, the reaction is endothermic.
Draw the organic product(s) of the reaction of p-methylbenzoic acid with CH3MgBr in dry ether, then H3O+ in the window below. If no new products are formed, tell OWL by drawing ethane, CH3CH3.
Answer:
Methane is produced as a new product.
Explanation:
[tex]CH_{3}MgBr[/tex] acts as a base toward -COOH group in p-methylbenzoic acid.
Hence an acid-base reaction occurs between p-methylbenzoic acid and [tex]CH_{3}MgBr[/tex] to produce methane and p-methylbenzoate.
[tex]H_{3}O^{+}[/tex]addition will convert p-methylbenzoate back to p-methylbenzoic acid.
Hence, methane is produce as a new product.
Reaction sequences are given below.
Mark each of the following statements as either True or False: The rate law is deduced directly from the coefficients of the overall reaction. The rate equation for this mechanism is rate = k [O][ClO]. ClO(g) is an intermediate formed in this reaction mechanism. Step 2 of the mechanism is a bimolecular. The sum of the two steps is equal to the overall reaction: O(g) + O3(g) → 2O2(g). Cl(g) is a catalyst in this reaction mechanism.
Answer:
Details of true/false statements are given below.
Explanation:
The rate law is deduced directly from the coefficients of the overall reaction. False
The rate equation for this mechanism is rate = k [O][ClO]. ClO(g) is an intermediate formed in this reaction mechanism. True
Step 2 of the mechanism is a bimolecular. True
The sum of the two steps is equal to the overall reaction: O(g) + O3(g) → 2O2(g). True
Cl(g) is a catalyst in this reaction mechanism. True
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.
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.
Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlorite (NaOCl). What will be the pH of this buffer solution after adding 0.341 mol of HCl? The Ka of hypochlorous acid is 2.9 x 10-8 . Assume the change in volume of the buffer solution is negligible
Answer:
pH = 6.82
Explanation:
To solve this problem we can use the Henderson-Hasselbach equation:
pH = pKa + log[tex]\frac{[NaOCl]}{[HOCl]}[/tex]We're given all the required data to calculate the original pH of the buffer before 0.341 mol of HCl are added:
pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54[HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 MpH = 7.54 + log [tex]\frac{4}{4}[/tex]pH = 7.54By adding HCl, we simultaneously increase the number of HOCl and decrease NaOCl:
pH = 7.54 + log[tex]\frac{[NaOCl-HCl]}{[HOCl+HCl]}[/tex]pH = 7.54 + log [tex]\frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}[/tex]pH = 6.82Question 3After creating a Beer's Law plot using standard solutions of Q, you determined the slope of Beer's Law to be 0.515 M-1. Your unknown solution of Q tested in Part B of the experiment had an absorbance of 0.145. Determine the concentration (in molarity) of the unknown solution Q from Part B.Question 4Refer to the procedure stated in the manual pages for Part A to answer the following question.Using the equation editor embedded in this question, show a sample calculation determining the original concentration of the provided unknown Q in Part A from the diluted concentration calculated in question 3 above.Lab ManualYou have been provided with a solution of unknown Q, the actual molar concentration is listed as the unknown number. Dilute 15.00 mL of the provided solution to a final volume of 50.00 mL. You may only use the equipment and reagents listed above. Be sure to record your unknown number in your notebook. After making your dilution, calculate the concentration of your diluted solution.
The concentration of the unknown solution 'Q' from Part B is determined to be 0.282 M using Beer's Law. The original concentration of the unknown 'Q' from Part A, before dilution, is calculated to be 0.94 M.
Explanation:The concentration of the unknown solution Q can be determined using the Beer's Law plot. Beer's Law or the Beer-Lambert law connects the absorbance of a solution to its concentration through the following equation: A = εcl, where 'A' is the absorbance, 'ε' is the molar absorptivity, 'c' is the concentration and 'l' is the path length. The slope of the Beer's Law plot corresponds to the product, 'εl'. So, the concentration can be calculated as: c=A/(εl), which gives c= 0.145/0.515 M-1 = 0.282 M
Regarding the dilution of the solution in Part A, we are given that 15.00 mL of the solution was diluted to a final volume of 50.00 mL. This can be used to calculate the original concentration. The formula used in this case is C1V1 = C2V2, where C1 is the original concentration, V1 is the original volume, C2 is the diluted concentration, and V2 is the final volume. Substituting C2 with the concentration we just calculated, the original concentration C1 can be calculated as: C1 = (C2V2)/V1 = (0.282 M × 50.00 mL)/ 15.00 mL = 0.94 M.
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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.
If the standard reduction potential of a half-cell is positive, which redox reaction is spontaneous when paired with a hydrogen electrode?
A. oxidation
B. both reduction and oxidation
C. reduction
D. neither reduction nor oxidation
If the standard reduction potential of a half-cell is positive, the reduction reaction is spontaneous when paired with a hydrogen electrode
Explanation:
The relative standard reduction potential of the half-cell in which reduction occurs; more positive than the other half-cell.If the standard reduction potential of a half-cell is positive, the reduction reaction is spontaneous when paired with a hydrogen electrode.The reduction is a chemical process in which electrons are added to an atom or an ion; it always occurs accompanied by oxidation of the reducing agent.The reduction to happen the electrons gained by the material that is being decreased must be transported from the atoms of ions of a different material.Answer:
Reduction
Explanation:
An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part E An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part F An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part G Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Question:
Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.
Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answer:
D) 85 J/K
E) - 50 J/K
F) 62.5 J/K
G) 12.5 J/K
Explanation:
Let's make use of the entropy equation: ΔS = [tex] \frac{Q}{T} [/tex]
Part D)
Given:
T = 20°C = 20 +273 = 293K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{293} [/tex]
= 85 J/K
Part E)
Given:
T = 500K
Q = -25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{-25*1000}{500} [/tex]
= - 50 J/K
Part F)
Given:
T = 400K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{400} [/tex]
= 62.5 J/K
Part G:
Given:
T1 = 400K
T2 = 500K
Q = 25.0 kJ
The net entropy change will be:
ΔS = [tex] (\frac{25*1000}{400}) + (\frac{-25*1000}{500}[/tex]
= 12.5 J/K
Answer:
A) The change in entropy [tex]\delta S = 0.085J/K[/tex]
B) The change in entropy [tex]\delta S = -50J/K[/tex]
C) The change in entropy [tex]\delta S = 62.5J/K[/tex]
D) The net change in entropy [tex]\delta S = 12.5J/K[/tex]
Explanation:
A)
[tex]T = 20^oC + 273k\\\\T = 293k[/tex]
expression for change in entropy,
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25}{293}\\\\\delta S = 0.085J/K[/tex]
B) [tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{-25*10^3}{500}\\\\\delta S = -50J/K[/tex]
The negative sign indicates the heat lost into the surrounding
C)
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25*10^3}{400}\\\\\delta S = 62.5J/K[/tex]
The entropy remains constant in the adiabatic process because no heat is given to the system in this process.
D)
[tex]\delta S_1 = \frac{\delta Q_1}{T_1}\\\\\delta S_1 = \frac{25*10^3}{400}\\\\\delta S_1 = 62.5J/K[/tex]
similarly,
[tex]\delta S_2 = \frac{\delta Q_2}{T_2}\\\\\delta S_2 = \frac{25*10^3}{500}\\\\\delta S_2 = 50J/K[/tex]
Therefore the net change in entropy is,
[tex]\delta S = \delta S_1 - \delta S_2\\\\\delta S = 62.5 - 50\\\\\delta S = 12.5J/K[/tex]
Entropy is not a conserved quantity because it can be created but cannot be destroyed.
The net change in entropy is calculated by difference of the change in entropy at two different temperatures.
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uestion 174 Which sequence of reactions is expected to produce the product below as the final, and major, organic product? I 1. HC≡CH, NaNH2; 2. (CH3)2CHCH2Br 3. Aqueous H2SO4, HgSO4 II 1. CH3C≡CH, NaNH2; 2. (CH3)2CHBr; 3. Disisamylborane; 4. H2O2, NaOH III 1. (CH3)2CHBr, NaNH2; 2. CH3C≡CH; 3. O3; 4. H2O IV 1. CH3C≡CH, NaNH2: 2. (CH3)2CHCH2Br; 3. BH3·THF; 4. H2O2, NaOH V 1. (CH3)2CHCH2Br, NaNH2; 2. HC≡CH; 3. 9-BBN; 4. H2O2, NaOH I II III IV V
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correction option is is [tex]I[/tex]
Explanation:
The mechanism of the reaction is show on the second uploaded image
Three elements have the electron configurations 1s22s22p63s1, 1s22s22p63s2, and 1s22s22p5. The first ionization energies of these elements (not in the same order) are 1.681, 0.738, and 0.496 MJ/mol. The atomic radii are 160, 186, and 64 pm. Identify the three elements, and match the appropriate values of ionization energy and atomic radius to each configuration.
Answer:
Explanation:
1s²2s²2p⁶3s² = magnesium ionization energy, 0.738 MJ/mol atomic radius 160 pm
1s²2s²2p⁶3s¹ = Sodium, ionization energy, 0.496 MJ/mol, atomic radius 186 pm
1s²2s²2p⁵ = Florine ionization energy, 1.681 MJ/mol, atomic radius 64 pm
Ionization energy is the minimum amount of energy needed to remove an electron from its orbital around the atom from the influence of the atom while atomic radius is one-half the distance between the nuclei of identical atoms that are bonded together.
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
To a flask, 15.0 mL of 1.25 M hydrofluoric acid is added. Then, 3.05 M KOH is used to titrate the acid sample. Write the balanced net ionic equation for the acid-base reaction.
Answer:
HF + OH- = F- + H2O
Explanation:
Since hydrofluoric acid does not ionize in aqueous solution, the fluoride ion is still present as part of the product
Answer:
H+ (aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
Volume of hydrofluoric acid = 15.0 mL = 0.015 L
Molarity = 1.25 M
Molarity of KOH = 3.05 M
Step 2: The unbalanced equation
HF(aq) + KOH(aq) → KF(aq) + H2O(l)
This equation is already balanced
Step 3: The net ionic equation
The net ionic equation shows only those elements, compounds, and ions that are directly involved in the chemical reaction.
The elements, compounds, and ions that do not take part in the chemical reaction are called spector ions.
H+ (aq) + F-(aq) + K+(aq) + OH-(aq)→ K+(aq) +F-(aq) + H2O(l)
We'll remove all the spector ions.
H+ (aq) + OH-(aq) → H2O(l)
Hydrogen peroxide can act as either an oxidizing agent or a reducing agent depending on the species present in solution. Write the balanced half-reaction equations for each situation. Write the balanced half-reaction equation for when H2O2(aq) acts as an oxidizing agent in an acidic solution. Phases are optional. half-reaction equation: Write the balanced half-reaction equation for when H2O2(aq) acts as a reducing agent in an acidic solution. Phases are optional. half-reaction equation: A disproportionation reaction is one in which a single species oxidizes and reduces itself. Write the complete balanced equation for the disproportionation reaction of H2O2(aq) . Phases are optional. disproportionation reaction:
Answer:
1) H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O
2) H₂O₂ → 2H⁺ + 2e⁻ O₂
3) 2H₂O₂ → 2H₂O + O₂
Explanation:
Half-reaction equation for when H₂O₂(aq) acts as an oxidizing agent in an acidic solution (this means H₂O₂ is reduced):
H₂O₂ + 2H⁺ + 2e⁻ → 2H₂OIt is a reduction because the oxidation number of O changes from -1 to -2.
Half-reaction equation for when H₂O₂(aq) acts as a reducing agent in an acidic solution (this means H₂O₂ is oxidized):
H₂O₂ → 2H⁺ + 2e⁻ O₂
It is an oxidation because the oxidation number of O changes from -1 to 0.
Disproportionation reaction of H₂O₂(aq):
2H₂O₂ → 2H₂O + O₂
Hydrogen peroxide can act as both an oxidizing and a reducing agent in an acidic solution. It becomes oxidized when acting as a reducing agent and gets reduced when it acts as an oxidizing agent. In a disproportionation reaction, it can both oxidize and reduce itself.
Explanation:Hydrogen peroxide (H2O2) can indeed act as either an oxidizing or reducing agent depending on the species present in solution. When
H2O2
acts as an oxidizing agent in an acidic solution, the balanced half-reaction is:
H2O2 + 2H+ + 2e- → 2H2O
This reaction shows H2O2 being reduced, thus it is acting as an oxidizer because it causes the oxidation of other substances by accepting electrons. In contrast, when
H2O2 acts as a reducing agent in an acidic solution, the balanced half-reaction is:
H2O2 → O2 + 2H+ + 2e-
In this case, H2O2 is being oxidized to O2 so it acts as a reducer because it donates electrons which allows the reduction of other substances.
Finally, in a disproportionation reaction, H2O2 can act as both the oxidizing and reducing agent, showing the same substance functioning as an oxidant and a reductant. The complete balanced equation for this disproportionation reaction of H2O2 is:
2H2O2 → 2H2O + O2
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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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What is the correct formula for the compound formed by CA2+ and NO2-
Answer:
Ca(NO2)2
Explanation:
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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