Answer:
Second reaction
NO2 + F -------> NO2F
Rate of reaction:
k1 [NO2] [F2]
Explanation:
NO2 + F2 -----> NO2F + F slow step1
NO2 + F -------> NO2F fast. Step 2
Since the first step is the slowest step, it is the rate determining step of the reaction
Hence:
rate = k1 [NO2] [F2]
Answer:
Cl(g)+NO_2Cl(g)-->NO_2(g)+Cl_2(g)
Explanation:
COLLEGE CHEMISTRY 35 POINTS
Determine the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen according to the following balanced equation:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)
Please show your work.
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 %
In a Grignard reagent, the carbon bonded to the magnesium has a partial ________ charge, because carbon is ________ electronegative than magnesium. This makes this carbon of the Grignard ________. In a Grignard reagent, the carbon bonded to the magnesium has a partial ________ charge, because carbon is ________ electronegative than magnesium. This makes this carbon of the Grignard ________. negative, less, electrophilic positive, less, electrophilic positive, more, electrophilic negative, more, nucleophilic positive, less, nucleophilic
Answer:
In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic.
Explanation:
Organometallic compounds have covalent bonds between carbon atoms and metallic atoms. Organometallic reagents are useful because they have nucleophilic carbon atoms, in contrast to the electrophilic carbon atoms of the alkyl halides. The majority of metals are more electropositive than carbon, and the bond is biased with a positive partial charge on the metal and a negative partial charge on the carbon.
Answer:
In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic In a Grignard reagent, the carbon bonded to the magnesium has a partial positive charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard less electrophilic.
Explanation:
The Grignard reagent is a highly reactive organomagnesium compound formed by the reaction of a haloalkane with magnesium in an ether solvent. The carbon atom of a Grignard reagent has a partial negative charge. Hence a Grignard reagent will have a general formula RMgX. Where R must contain a carbanion.
Grignard reagents are versatile reagents used in many applications in organic chemistry.
After treating cyclohexanone with LDA, he then added tert-butyl bromide to the same reaction vessel, which one would normally do for a one-pot two-step reaction sequence. However, none of the desired compound formed. Instead, he isolated three organic compounds in a 1:1:1 ratio. What is the structure of each of these organic compounds?
Answer:
2-Tert-butyl-cyclohexanone, 2,6-Di-Tert-Butylcyclohexanone and Di-isopropyl-amine.
Explanation:
Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 125 mL of a buffer that is 0.360 M in both acetic acid ( CH 3 COOH ) and its conjugate base ( CH 3 COO − ) . Calculate the maximum volume of 0.300 M HCl that can be added to the buffer before its buffering capacity is lost.
Answer:
The maximum volume is 122.73 mL
Explanation:
125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.
pKa of acetic acid = 4.76
Using Henderson Hasselbalch equation.
pH =pKa + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
= 4.76 + [tex]log \frac{0.3600}{0.3600}[/tex]
= 4.76
Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.
When HCl is added , it reacts with CH₃COO⁻ to give CH₃COOH.
However, [CH₃COOH] increases and the log term results in a negative value.
Let assume, the new pH is less than 4.76
Let say 3.76; calculating the concentration when the pH is 3.76; we have
3.76 = 4.76 + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 4.76 - 3.76
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 1.00
[tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 0.1
Let number of moles of acid be x (i.e change in moles be x); &
moles of acetic acid and conjugate base present be = Molarity × Volume
= 0.360 M × 125 mL
= 45 mmol
replacing initial concentrations and change in the above expression; we have;
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1[/tex]
[tex]\frac{45-x}{45+x} =0.1[/tex]
0.1(45 + x) = 45 - x
4.5 + 0.1 x = 45 -x
0.1 x + x = 45 -4.5
1.1 x = 40.5
x = 40.5/1.1
x = 36.82
So, moles of acid added = 36.82 mmol
Molarity = 0.300 M
So, volume of acid = [tex]\frac{moles}{molarity }[/tex] = [tex]\frac{36.82 \ mmol}{0.300}[/tex]
= 122.73 mL
A chemist must prepare 800mL of potassium hydroxide solution with a pH of 13.90 at 25C.?
He will do this in three steps.
1. fill a 800mL volumetric flask about halfway with distilled water.
2. weigh out a small amount of solid potassium hydroxide and add it to the flask.
3. fill the flask to the mark with distilled water.
Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step.
Answer:
35.6 g
Explanation:
Step 1: Calculate the [H⁺]
We use the following expression
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -13.90 = 1.259 × 10⁻¹⁴ M
Step 2: Calculate the [OH⁻]
We use the ionic product of water (Kw).
Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]
[OH⁻] = 0.7943 M
Step 3: Calculate the moles of OH⁻
[tex]\frac{0.7943mol}{L} \times 0.800 L = 0.635 mol[/tex]
Step 4: Calculate the required moles of KOH
KOH is a strong base that dissociates according to the following equation.
KOH → K⁺ + OH⁻
The molar ratio of KOH to OH⁻ is 1:1. Then, the required moles of KOH are 0.635 moles.
Step 5: Calculate the mass of KOH
The molar mass of KOH is 56.11 g/mol.
[tex]0.635 mol \times \frac{56.11g}{mol} =35.6 g[/tex]
g A mass spectrum. The peak at mass 100 has a 8% relative abundance. THe peak with mass 85 has a 40% abundance. The peak at 71 has a 3% abundance. The peak at 57 has a 30% abundance. The peak at mass 43 has a 100% abundance. The peak at mass 29 has a 18% abundance. In this mass spectrum, which peak most likely represents the molecular ion
Answer:
The peak at mass 100 with a 8% relative abundance is the molecular ion peak
Explanation:
Molecular ion peak has the highest charge to mass ratio,
Mass of 100 is same as mass to charge ratio =100
The reaction:L-glutarmate + pyruvate → α-ketoglutarate + L-alanine is catalyzed by the enzyme L-glutamate-pyruvate aminotransferase. At 300 K, the equilibrium constant for the reaction is 1.1. Predict whether the forward reaction (left to right) will occur spontaneous if the concentrations of the reactants and products are [L-glutarmate] = 3.0 x 10-5 M, [pyruvate] = 3.3 x 10-4 M,[α-ketoglutarate] = 1.6 x 10-2 M, and [L-alanine] = 6.25 x 10-3 M.
Answer:
Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}
or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4
Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.
Explanation:
The phosphorylation of glucose to glucose 6-phosphate Group of answer choices is so strongly exergonic that it does not require a catalyst. is an exergonic reaction not coupled to any other reaction. is an endergonic reaction that takes place because it is coupled to the exergonic hydrolysis of ATP. is an exergonic reaction that is coupled to the endergonic hydrolysis of ATP.
Answer:
The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.
Explanation:
In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.
Treatment of ethyl acetoacetate with NaOEt (2 equiv) and BrCH2CH2Br forms compound X. This reaction is the first step in the synthesis of illudin-S, an antitumor substance isolated from the jack-o'-lantern, a saffron-colored mushroom. What is the structure of X
Answer:
See explanation below
Explanation:
In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.
In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.
The mechanism of this reaction to reach X is shown in the attached picture.
6. Air at 300°C and 130kPa ows through a horizontal 7-cm ID pipe at a velocity of 42.0m/s. (a) Calculate _ Ek W, assuming ideal-gas behavior. (b) If the air is heated to 400°C at constant pressure, what is Δ _ Ek _ Ek 400°C _ Ek 300°C? (c) Why would it be incorrect to say that the rate of transfer of heat to the gas in Part (b) must equal the rate of change of kinetic energy?
Answer:
Check the explanation
Explanation:
kindly check the attached image below to see the answer to question A and B
(c) The energy balance equation if the pressure is constant (ΔE=0) is,
ΔE= ΔU + Δ[tex]E_{k}[/tex]
From the above relation, it is clear that some heat energy used to raises the temperature of the air.
Hence, the internal energy is not equal to zero. Therefore, the rate of transfer of heat to air is not equal to the rate of thane in kinetic energy of the air.
That is ΔE ≠ Δ[tex]E_{k}[/tex]
Answer:
a) [tex]\delta E_k[/tex] = 112.164 W
b) [tex]\delta E_k[/tex] = 42.567 W
c) From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
Explanation:
The given data include:
Inlet diameter [tex](d_1)[/tex] = 7 cm = 0.7 m
Inlet velocity [tex](v_1)[/tex] = 42 m/s
Inlet pressure [tex](P_1)[/tex] = 130 KPa
Inlet temoerature [tex](T_1)[/tex] = 300°C = (300 + 273.15) = 573.15 K
a) Assuming Ideal gas behaviour
Inlet Volumetric flowrate [tex](V_1)[/tex] = Inlet velocity [tex](v_1)[/tex] × area of the tube
[tex]= v_1*(\frac{\pi}{4})0.07^2\\\\= 0.161635 \ m^3/s[/tex]
Using Ideal gas law at Inlet 1
[tex]P_1V_1 =nRT_1[/tex]
where ; n = molar flow rate of steam
making n the subject of the formula; we have:
[tex]n = \frac{P_1V_1}{RT_1}[/tex]
[tex]n = \frac{130*42}{8.314*573.15}[/tex]
[tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Moleular weight of air = 28.84 g/mol
The mass flow rate = molar flowrate × molecular weight of air
[tex]= 4.4096*10^{-3} \ * 28.84\\= 0.12717 \ kg/s[/tex]
Finally: the kinetic energy at Inlet [tex](\delta E_k) = \frac{1}{2}*mass \ flowrate *v_1^2[/tex]
= [tex]0.5*0.12717*42^2[/tex]
= 112.164 W
b) If the air is heated to 400°C;
Then temperature at 400°C = (400 + 273.15)K = 673.15 K
Thee pressure is also said to be constant ;
i.e [tex]P_1= P_2[/tex] = 130 KPa
Therefore; the mass flow rate is also the same ; so as the molar flow rate:
Thus; [tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Using Ideal gas law at Inlet 2
[tex]P_2V_2 = nRT_2[/tex]
making [tex]V_2[/tex] the subject of the formula; we have:
[tex]V_2 = \frac{nRT_2}{P_2}[/tex]
[tex]V_2 = \frac{4.4906*10^{-3}*8.314*673.15}{130}[/tex]
[tex]V_2 = 0.189836 \ m^3/s[/tex]
Assuming that the diameter is constant
[tex]d_1 = d_2 = 0.07 \ cm[/tex]
Now; the velocity at outlet = [tex]\frac{V_2}{\frac{\pi}{4}d_2^2}[/tex]
= [tex]\frac{0.0189836}{\frac{\pi}{4}(0.07)^2}[/tex]
= 49.33 m/s
Change in kinetic energy [tex]\delta E_k[/tex] = [tex]\frac{1}{2}*mass \ flowrate * \delta V[/tex]
= [tex]0.5*0.12717 *(49.33^2 -42^2)[/tex]
= 42.567 W
c).
From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
What parts of a circuit is a conductor?
Answer:
copper
Explanation:
These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load
Write a balanced equation for the decomposition reaction described, using the smallest possible integer coefficients. When calcium carbonate decomposes, calcium oxide and carbon dioxide are formed.
Answer:
1. The balanced equation is given below:
CaCO3 —> CaO + CO2
2. The coefficients are: 1, 1, 1
Explanation:
Step 1:
The word equation is given below:
calcium carbonate decomposes to produce calcium oxide and carbon dioxide.
Step 2:
The elemental equation. This is illustrated below:
Calcium carbonate => CaCO3
calcium oxide => CaO
carbon dioxide => CO2
CaCO3 —> CaO + CO2
A careful observation of the above equation shows that the equation is balanced.
The coefficients are: 1, 1, 1
Final answer:
The balanced equation for the decomposition reaction of calcium carbonate is CaCO3(s) → CaO(s) + CO2(g), representing the formation of calcium oxide and carbon dioxide gas from calcium carbonate.
Explanation:
The decomposition reaction of calcium carbonate forming calcium oxide and carbon dioxide can be represented by the following balanced chemical equation:
CaCO3(s) → CaO(s) + CO2(g)
This equation indicates that one mole of calcium carbonate (CaCO3) decomposes to form one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2) gas upon heating. The reactants and products are represented using their smallest possible integer coefficients, reflecting the principle of conservation of mass and the stoichiometry of the decomposition reaction.
An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm absolute. State a quick way to show thatthe system has two phases. Estimate (i) the molar compositions of each phase, (ii) thepercentage of the total moles in the vessel that are in the vapor phase, and (iii) thepercentage of the vessel volume occupied by the vapor phase.
The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
How many moles are in 12.0 grams of O2
Answer:
Moles = 0.375
Explanation:
Moles= m/M
= 12/32 = 0.375mol
There are approximately 0.375 moles of O₂ in 12.0 grams of O₂.
To find the number of moles in a given mass of a substance, one can use the formula:
[tex]\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]
The molar mass of oxygen (O₂) is approximately 32.0 g/mol, since one oxygen atom has an atomic mass of approximately 16.0 g/mol and O₂ has two oxygen atoms.
Given the mass of O₂ is 12.0 g, we can calculate the moles as follows:
[tex]\[ \text{moles of \ } {O_2} = \frac{12.0 \text{ g}}{32.0 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of} \ {O_2} = 0.375 \text{ mol} \][/tex]
Therefore, there are 0.375 moles of O₂ in 12.0 grams of O₂.
Calculate the amount of energy in kilojoules needed to change 369 gg of water ice at −−10 ∘C∘C to steam at 125 ∘C∘C. The following constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C)Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40 J/(mol⋅∘C)Cm (water)=75.40 J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C)Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/molΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/molΔHvap=+40.67 kJ/mol Express your answer with the appropriate units.
Answer:
We need 1136.4 kJ energy
Explanation:
Step 1: Data given
The mass of water = 369 grams
the initial temperature = -10°C
The finaltemperature = 125 °C
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6010 J/mol
ΔHvap=+40670 J/mol
Step 2: :Calculate the energy needed to heat ice from -10 °C to 0°C
Q = n*C*ΔT
⇒Q = the energy needed to heat ice to 0°C
⇒with n = the moles of ice = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of ice = 36.57 / J/mol°C
⇒ ΔT = the change of temperature = 10°C
Q = 20.48 moles * 36.57 J/mol°C * 10°C
Q = 7489.5 J = 7.490 kJ
Step 3: calculate the energy needed to melt ice to water at 0°C
Q = n* ΔHfus
Q = 20.48 moles * 6010 J/mol
Q = 123084.8 J = 123.08 kJ
Step 4: Calculate energy needed to heat water from 0°C to 100 °C
Q = n*C*ΔT
⇒Q = the energy needed to heat water from 0°C to 100 °C
⇒with n = the moles of water = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of water =75.40 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 100°C
Q = 20.48 moles * 75.40 J/mol°C* 100°C
Q = 154419.2 J = 154.419 kJ
Step 5: Calculate energy needed to vapourize water to steam at 100°C
Q = 20.48 moles * 40670 J/mol
Q = 832921.6 J = 832.922 kJ
Step 6: Calculate the energy to heat steam from 100 °C to 125 °C
⇒Q = the energy needed to heat steam from 100 °C to 125 °C
⇒with n = the moles of steam = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of steam = 36.04 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 25 °C
Q = 20.48 moles * 36.04 J/mol*°C * 25°C
Q = 18452.5 J = 18.453 kJ
Step 7: Calculate total energy needed
Q = 7489.5 J + 123084.8 J + 154419.2 J + 832921.6 J + 18452.5 J
Q = 1136367.6 J
Q = 1136.4 kJ
We need 1136.4 kJ energy
The amount of energy needed is [tex]\( \\1120 \text{ kJ}} \).[/tex]
To calculate the amount of energy required to change 369 g of water from ice at[tex]\(-10^\circ \text{C}\)[/tex] to steam at [tex]\(125^\circ \text{C}\),[/tex] we need to consider the different phases of water and the energy required for each phase change and temperature change.
Let's break down the calculation step by step.
1. Energy to heat ice from [tex]\(-10^\circ \text{C}\) to \(0^\circ \text{C}\).[/tex]
First, calculate the energy required to heat the ice initially at[tex]\(-10^\circ \text{C}\)[/tex] to its melting point [tex]( \(0^\circ \text{C}\) ).[/tex]
Specific heat capacity of ice [tex](\(C_m\)): \(36.57 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Molar mass of water (H₂O). [tex]\(18.015 \text{ g/mol}\).[/tex]
Amount of ice in moles:
[tex]\[ \text{moles of ice} = \frac{369 \text{ g}}{18.015 \text{ g/mol}} \approx 20.49 \text{ mol} \][/tex]
Energy required to heat ice.
[tex]\[ Q_1 = n \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.57 \text{ J/(mol} \cdot \text{°C)} \times (0^\circ \text{C} - (-10^\circ \text{C})) = 7485.39 \text{ J} \][/tex]
[tex]\( Q_1 = 7.48539 \text{ kJ} \)[/tex]
2. Energy to melt the ice at [tex]\(0^\circ \text{C}\).[/tex]
Latent heat of fusion [tex](\(\Delta H_{\text{fus}}\)) for ice: \(6.01 \text{ kJ/mol}\)[/tex].
Energy required to melt ice:
[tex]\[ Q_2 = \text{moles of ice} \times \Delta H_{\text{fus}} = 20.49 \text{ mol} \times 6.01 \text{ kJ/mol} = 123.05 \text{ kJ} \][/tex]
[tex]\( Q_2 = 123.05 \text{ kJ} \)[/tex]
3.Energy to heat water from[tex]\(0^\circ \text{C}\) to \(100^\circ \text{C}\)[/tex].
Specific heat capacity of water[tex](\(C_m\))[/tex]: [tex]\(75.40 \text{ J/(mol} \cdot \text{°C)}\)[/tex].
Energy required to heat water:
[tex]\[ Q_3 = \text{moles of water} \times C_m \times \Delta T = 20.49 \text{ mol} \times 75.40 \text{ J/(mol} \cdot \text{°C)} \times (100^\circ \text{C} - 0^\circ \text{C}) = 155,297.80 \text{ J} \][/tex] [tex]\( Q_3 = 155.2978 \text{ kJ} \)[/tex]
4. Energy to vaporize water at [tex]\(100^\circ \text{C}\)[/tex]:
Latent heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) for water: \(40.67 \text{ kJ/mol}\).[/tex]
Energy required to vaporize water:
[tex]\[ Q_4 = \text{moles of water} \times \Delta H_{\text{vap}} = 20.49 \text{ mol} \times 40.67 \text{ kJ/mol} = 833.34 \text{ kJ} \][/tex]
[tex]\( Q_4 = 833.34 \text{ kJ} \)[/tex]
5. Energy to heat steam from [tex]\(100^\circ \text{C}\) to \(125^\circ \text{C}\):[/tex]
Specific heat capacity of steam [tex](\(C_m\)): \(36.04 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Energy required to heat steam:
[tex]\[ Q_5 = \text{moles of steam} \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.04 \text{ J/(mol} \cdot \text{°C)} \times (125^\circ \text{C} - 100^\circ \text{C}) = 1831.79 \text{ J} \][/tex]
[tex]\( Q_5 = 1.83179 \text{ kJ} \)[/tex]
6. Total energy required.
Summing up all the energy contributions.[tex]\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 7.48539 \text{ kJ} + 123.05 \text{ kJ} + 155.2978 \text{ kJ} + 833.34 \text{ kJ} + 1.83179 \text{ kJ} = 1120.00598 \text{ kJ} \][/tex]
Suppose the formation of nitryl fluoride proceeds by the following mechanism:
Step. Elementary reaction. Rate constant
1. NO2(g)+F2(g)=NO2F(g)+F(g). K1
2. F(g)+NO2(g)=NO2F(g). K2
Suppose also k1<
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally observable rate law for the overall chemical reaction:
Note your answer should not contain concentrations or any intermediates.
Express the rate constant k for the overall chemical reaction in terms of k1, k2, and If necessary the rate constants k-1 and k-2 for the reverse of two elementary reactions in the mechanism.
The balanced chemical equation for the overall chemical reaction is 2NO2(g) + F2(g) → 2NO2F(g). The experimentally observable rate law for the overall chemical reaction is Rate = k [NO2]^2 [F2]. The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as k = k1 (k2/k-1).
Explanation:The balanced chemical equation for the overall chemical reaction is:
2NO2(g) + F2(g) → 2NO2F(g)
The experimentally observable rate law for the overall chemical reaction is:
Rate = k [NO2]2 [F2]
The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as:
k = k1 (k2/k-1)
a sample of a radioactive isotope initially contains 20 x 10^10 atoms. After 16 days, 5 x 10^10 atoms remain. what is the half-life of the isotope?
A)12 days
B)20 days
C)10 days
D)16 days
E)8 days
The half-life of the radioactive isotope is 8 days because after 16 days (which is two half-lives), the sample is reduced from 20 x 10^10 atoms to 5 x 10^10 atoms.
Explanation:The question deals with the concept of the half-life of a radioactive isotope, which is the time required for half of the radioactive atoms in a sample to decay. According to the problem statement, an initial sample containing 20 x 1010 atoms decays to 5 x 1010 atoms in 16 days. To find the half-life (t1/2), we can use the fact that after one half-life, the number of atoms would be halved. Starting with 20 x 1010 atoms, after one half-life, there would be 10 x 1010 atoms, and after two half-lives, there would be 5 x 1010. This indicates that two half-lives have passed in 16 days, making the half-life 8 days.
The correct answer is 8 days, option E.
A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) 4730 L 5378 L 3525 L 742.2 L
Answer:
742.2 L
Explanation:
First we must find the number of moles of nitroglycerine reacted.
Molar mass of nitroglycerine= 227.0865 g/mol
Mass of nitroglycerine involved = 1×10^3 g
Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol
n= 4.40361 moles
T= 1985°C + 273= 2258K
P= 1.100atm
R= 0.082atmLmol-1K-1
Using the ideal gas equation:
PV= nRT
V= nRT/P
V= 4.40361× 0.082× 2258/1.1
V= 742 L
Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.
The balanced reaction is:
4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₃H₅N₃O₉(s): 4 moles O₂(g): 1 moles CO₂(g): 12 moles H₂O(g): 10 molesN₂(g): 6 molesThen 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]
Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:
[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?
[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]
amount of moles of gas= 31.93625 moles
On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
P= 1.1 atmV= ?n= 31.93625 molesR= 0.082[tex]\frac{atmL}{molK}[/tex]T= 1985 C= 2258 K (being 0 C=273 K)Replacing:
1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K
Solving:
V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm
V= 5375.625 L
Finally, the volume of gas produced is 5375.626 L.
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Answer:
Problem 1 => 8778 joules
Problem 2 => 14,630 joules
Explanation:
Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.
For the listed problems...
Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.
Note the temperature change in the problem context => use q = m·c·ΔT ...
q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)
Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.
Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...
q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules
Hope this helps. Doc
2. A mixture of 60 wt% benzene, 35% toluene, and 5% naphthalene is being distilled. The distillate product should be 99.5 wt% benzene. Also, 99% of benzene fed should be recovered in the distillate, Determine/calculate: a. Distillate and bottom product flow rates (D and B) per 1 kg of feed b. compositions of distillate and bottom product c. fraction of toluene fed that is recovered in the bottom product
Answer:
(a) Distillate (D) = 0.597 kg
Bottom product (B) = 0.403 kg
(b) Compositions of distillate product:
Benzene = 0.995
Toluene = 0.05
Compositions of bottom product:
Benzene = 0.0149
Naphthalene = 0.1241
Toluene = 0.8610
(c) Fraction of toluene fed that is recovered in the bottom product = 99.14%
Explanation:
See the attached file for the calculation
Determine the ground-state electron configuration and bond order for each of the Period 2 diatomic molecules. Specify which MO is the HOMO and which is the LUMO. List the molecules in order of increasing dissociation and list them in order of bond length. You may need to THINK a bit to get this correct
Answer:
Explanation:
check the attachment below for correct explanations.
For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each oscillating lattice site contributes 3kT of energy, by equipartition), as well as a contribution from the mobile electrons (if it's a metal). At room temperature the latter contribution is typically much smaller than the former, so we will ignore it here. In other words, you can reasonably estimate the specific heat simply by counting the number of atoms! Use this fact to estimate the specific heat of copper (atomic mass = 63.6), given that the specific heat of aluminum (atomic mass = 27.0) is 900 J/kg-K.
Answer:
The specific heat of copper is [tex]C= 392 J/kg\cdot ^o K[/tex]
Explanation:
From the question we are told that
The amount of energy contributed by each oscillating lattice site is [tex]E =3 kT[/tex]
The atomic mass of copper is [tex]M = 63.6 g/mol[/tex]
The atomic mass of aluminum is [tex]m_a = 27.0g/mol[/tex]
The specific heat of aluminum is [tex]c_a = 900 J/kg-K[/tex]
The objective of this solution is to obtain the specific heat of copper
Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by [tex]1 ^o K[/tex]
The general equation for specific heat is
[tex]C = \frac{dU}{dT}[/tex]
Where [tex]dT[/tex] is the change in temperature
[tex]dU[/tex] is the change in internal energy
The internal energy is mathematically evaluated as
[tex]U = 3nk_BT[/tex]
Where [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]
T is the room temperature
n is the number of atoms in a substance
Generally number of atoms in mass of an element can be obtained using the mathematical operation
[tex]n = \frac{m}{M} * N_A[/tex]
Where [tex]N_A[/tex] is the Avogadro's number with a constant value of [tex]6.022*10^{23} / mol[/tex]
M is the atomic mass of the element
m actual mass of the element
So the number of atoms in 1 kg of copper is evaluated as
[tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]
The number of atom is
[tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]
[tex]= 9.46*10^{24} \ atoms[/tex]
Now substituting the equation for internal energy into the equation for specific heat
[tex]C = \frac{d}{dT} (3 n k_B T)[/tex]
[tex]=3nk_B[/tex]
Substituting values
[tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]
[tex]C= 392 J/kg\cdot ^o K[/tex]
Consider the reaction: 2N2(g) + O2(g)2N2O(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.19 moles of N2(g) react at standard conditions. S°system = J/K
Answer:
ΔS = -114.296 J/K
Explanation:
Step 1: Data given
Temperature = 298 K
Number of moles N2 = 2.19 moles
S°(N2) = 191.6 J/K*mol
S°(O2)= 161.1 J/k*mol
S°(N2O) = 219.96 J/K*mol
Step 2: The balanced equation
2N2(g) + O2(g) ⇆ 2N2O(g)
Step 3: Calculate ΔSrxn
ΔSrxn = ∑S°products - ∑S°reactants
ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol
ΔSrxn = -104.38 J/K
Step 4: Calculate ΔS for 2.19 moles
The reaction is for 2 moles N2
ΔS = -114.296 J/K
Final answer:
The entropy change for a chemical reaction can be determined using standard absolute entropies. In this case, the given reaction of N2(g) and O2(g) to form N2O(g) can be analyzed using these principles.
Explanation:
The entropy change for the given reaction can be calculated using the standard absolute entropies at 298K. The formula to calculate the entropy change is ΔS = ΣnS(products) - ΣnS(reactants). In this case, for the reaction 2N2(g) + O2(g) → 2N2O(g), the entropy change can be calculated using the standard absolute entropies of each component.
Step 1: Data given
Temperature = 298 K
Number of moles N2 = 2.19 moles
S°(N2) = 191.6 J/K*mol
S°(O2)= 161.1 J/k*mol
S°(N2O) = 219.96 J/K*mol
Step 2: The balanced equation
2N2(g) + O2(g) ⇆ 2N2O(g)
Step 3: Calculate ΔSrxn
ΔSrxn = ∑S°products - ∑S°reactants
ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol
ΔSrxn = -104.38 J/K
Step 4: Calculate ΔS for 2.19 moles
The reaction is for 2 moles N2
ΔS = -114.296 J/K
A 900.0 mLmL sample of 0.18 MHClO4MHClO4 is titrated with 0.27 MLiOHMLiOH. Determine the pHpH of the solution after the addition of 600.0 mLmL of LiOHLiOH (this is the equivalence point). A 900.0 sample of 0.18 is titrated with 0.27 . Determine the of the solution after the addition of 600.0 of (this is the equivalence point). 11.24 13.03 2.76 0.97 7.00
Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.
Explanation:
The given data is as follows.
Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,
Molarity of [tex]HClO_{4}[/tex] = 0.18 M,
Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.
No. of moles = Molarity × Volume
= 0.18 M × 0.9 L
= 0.162 moles
Volume of NaOH = 600.0 ml = 0.6 L
Molarity of NaOH = 0.27 M
No. of moles of NaOH = Molarity × Volume
= 0.27 M × 0.6 L
= 0.162 moles
This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.
Also we know that,
[tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]
As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.
This means that pH = 7.
Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.
2. Calculate the mass of 3.47x1023 gold atoms.
3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams.
Explanation:
Data given:
number of atoms of gold = 3.47 x [tex]10^{23}[/tex]
mass of the gold in given number of atoms = ?
atomic mass of gold =196.96 grams/mole
Avagadro's number = 6.022 X [tex]10^{23}[/tex]
from the relation,
1 mole of element contains 6.022 x [tex]10^{23}[/tex] atoms.
so no of moles of gold given = [tex]\frac{3.47 X 10^{23} }{6.022 X 10^{23} }[/tex]
0.57 moles of gold.
from the relation:
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
rearranging the equation,
mass = number of moles x atomic mass
mass = 0.57 x 196.96
mass = 113.44 grams
thus, 3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams
The mass of 3.47 x 10^23 gold atoms calculated using Avogadro's number and the molar mass of gold equals approximately 113.52 grams. To calculate this, the given number of atoms was first converted into moles and the number of moles were then multiplied by the molar mass.
Explanation:In chemistry, to calculate the mass of a certain number of atoms, we use the concept of a mole and Avogadro's number (6.022 x 1023). For gold atoms, the molar mass is 197 g/mol. Given that the number of gold atoms is 3.47 x 1023, we can use the relationship that 1 mole of gold atoms = 6.022 x 1023 gold atoms. Therefore, the mass of 3.47 x 1023 gold atoms can be calculated as follows:
First, we convert the number of atoms to moles using Avogadro's number: 3.47 x 1023 atoms / 6.022 x 1023 atoms/mol = 0.576 moles of gold.
Then, we multiply this moles by the molar mass of gold to get the mass in grams: 0.576 moles * 197 g/mol = 113.52 g.
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In this chemical Formula for Ammonia, the Subscripts indicate what?
A. When added together, we know that Nitrogen and Hydrogen combined create four molecules of Ammonia
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
C. There are two molecules of Nitrogen and two molecules of Hydrogen combined to make Ammonia
D. The compound N2 and the compound H2 make Ammonia
Answer:
B
Explanation:
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
Draw the curved arrow(s) to depict the formation of the keto form of an enolate ion via a strong base, B. Complete the resonance structures of the enolate anion\'s keto and enolate forms with bonds, charges, and nonbonding electron pairs. Use curved arrows to show how the keto form resonates to the enolate form.
The enolate ion is formed from a carbonyl function via a strong base, leading to resonance structures between keto and enolate forms. The resonance involves shifting of electron pairs, with the enolate form being more prevalent under basic conditions.
Explanation:An enolate ion is formed when a strong base, B, abstracts a proton α to a carbonyl function. The negatively charged oxygen donates a pair of electrons, pushing the electrons from the double bond into the alpha carbon to form the enolate form. The curved arrow represents the movement of a pair of electrons.
The resonance of the keto form to the enolate form involves shifting of electron pairs, with the electron pair on the carbon moving to form a C=O bond and the π bond between carbon and oxygen moving to the oxygen, forming a p-orbital with a lone pair of electrons and making the oxygen negatively charged.
By this mechanism, the keto and enolate forms are interconvertible with the enolate form being more prevalent under basic conditions. The full resonance structures for both the keto and enolate forms would consist of all their bonds, charges, and nonbonding electron pairs.
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What is the daily cost of incandescent lamp
Draw the correct structure for (3S,4S)-3,4-dimethylhexane. Show stereochemistry clearly. To ensure proper grading, explicitly draw all four groups, including wedge/dash bonds, around a chirality center. Indicate whether the compound could exist in an optically active form.
Find the figure in the attachment
Construct a simulated 1H NMR spectrum for 1-chloropropane by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Follow these steps: identify the different types of hydrogen atoms, determine the chemical shifts, assign integration values, and drag and drop splitting patterns.
Explanation:To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Here's a step-by-step guide:
Identify the different types of hydrogen atoms in 1-chloropropane. In this molecule, there are three types: H on the methyl group, H on the second carbon, and H on the third carbon.Determine the chemical shifts of the hydrogen atoms. The chemical shift values for H on the methyl group, H on the second carbon, and H on the third carbon are typically around 0.9-1.2 ppm, 1.2-1.4 ppm, and 2.5-3.0 ppm, respectively.Assign integration values to the signals. The integration values represent the relative number of hydrogen atoms in each type. For example, the H on the methyl group typically has an integration value of 3, indicating that there are three hydrogen atoms in the methyl group.Drag and drop the appropriate splitting patterns into the boxes on the chemical shift baseline. The splitting pattern depends on the neighboring hydrogen atoms and follows the n+1 rule. For example, if a hydrogen atom has two neighboring hydrogen atoms, it will show a triplet splitting pattern.Learn more about simulated 1H NMR spectrum for 1-chloropropane here:https://brainly.com/question/9812005
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