Answer:
3. Oxidation-Reduction Reaction
Explanation:
Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
Fe(s) -2e- ----> Fe2+(aq) oxidation
Cl2(aq) + 2e- -----> 2Cl-(aq) reduction
The given unbalanced chemical reaction is the oxidation-reduction reaction. Therefore, option (3) is correct.
What is an oxidation-reduction reaction?Redox reactions can be defined as oxidation-reduction chemical reactions in which the reactants of the reaction undergo a change in their oxidation states. All the redox reactions are further broken down into two different processes: a reduction process and an oxidation process.
The oxidation and reduction reactions take place simultaneously in an Oxidation-Reduction reaction. The substance that is getting reduced in a reaction is known as the oxidizing agent, while a substance that is getting oxidized is the reducing agent.
The given chemical reaction is:
[tex]Fe(s) + Cl_2(aq) \longrightarrow Fe^{2+}(aq) + Cl^-(aq)[/tex]
The oxidation reaction for this reaction is: Fe (s) → Fe²⁺ (aq) + 2e⁻
The reduction reaction: Cl₂ (g) + 2e⁻ → 2Cl⁻ (aq)
Therefore, the given reaction between the iron and chlorine gas is the oxidation-reduction reaction or redox reaction.
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Which process is a physical change
Answer:
a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.
Example:
When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.
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(If you're referring to a question with these answers)
A. iron rusting
B. milk turning to curd
C. water boiling
D. paper burning
E. hard water staining pipes
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Answer:
C. Water Boiling
(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)
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A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.
When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.
Answer:
ΔErxn[tex]= -3.90*10^3KJ[/tex]
Explanation:
Given from the question
T1 = 25.87∘C
T2= 38.13∘C.
C= 5.73Kj/C
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
Calculate Keq for these reactions and predict if the equilibrium will lie to the right or to the left as written. (You may enter your answer in scientific notation, e.g. 1.0*10^-9. Enter your answer to two significant figures.) Reaction 1: + + pKa = 9 pKa = 38 Keq = Equilibrium position = _______ Reaction 2: + + pKa = 35 pKa = 25 Keq = Equilibrium position = _______
Complete Question
The complete question is shown on the first uploaded image
Answer:
For reaction 1
[tex]K_{eq} = 10^{29}[/tex]
The equilibrium position is to the right
For reaction 2
[tex]K_{eq} = 10^{-6.66}[/tex]The equilibrium position is to the left
Explanation:
Generally [tex]pKa[/tex] is mathematically evaluated as
[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]
And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]
From the question we are told that
For reaction 1
[tex]pKa_\ {left}} \ = 9[/tex]
[tex]pKa_\ {right }} \ = 38[/tex]
So
[tex]pKa = 9-38[/tex]
[tex]pKa =-29[/tex]
So [tex]K_{eq} = 10^{-(-29)}[/tex]
[tex]K_a = 10^{29}[/tex]
This implies that the equilibrium position is to the right
For reaction 2
[tex]pKa_\ {left}} \ = 15.9[/tex]
[tex]pKa_\ {right }} \ = 9.24[/tex]
So
[tex]pKa = 15.9-9.24[/tex]
[tex]pKa = 6.66[/tex]
So [tex]K_{eq} = 10^{-(6.66)}[/tex]
[tex]K_{eq} = 10^{-6.66}[/tex]
This implies that the equilibrium position is to the left
A chemistry student is given 600. mL of a clear aqueous solution at 37° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg. Using only the information from above, can you calculate the solubility of X at 21° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.
Answer:
The solubility is [tex]S = 0.0014 \ g[/tex]
Explanation:
From the question we are told that
The volume of the solution is [tex]V = 600 mL[/tex]
The initial temperature is [tex]T_i = 37 ^oC[/tex]
The final temperature is [tex]T_f = 21^oC[/tex]
The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]
Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent (for solubility of a solute to be calculated the solute must be able to saturate the solvent)
now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away
The solubility at 21 ° C is mathematically represented as
[tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]
Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as
[tex]m_w = V * \rho_w[/tex]
Where [tex]\rho = 1 \frac{g}{mL}[/tex]
So
[tex]m_w =600 * 1[/tex]
[tex]m_w =600g[/tex]
So
[tex]S = \frac{84}{600 * 100 g \ of water }[/tex]
[tex]S = 0.0014 \ g[/tex]
Calculate the percent saturated fat in the total fat in butter
Calculate the pH of this solution 0.0043 M of H2SO4=
Answer:
pH = - log [concentration]
pH = - log (0.0043M)
pH = 2.37
what is the equation for "acid dissociation constant" of "carbonic acid"
Answer:
H2CO3 = 2H+ + CO3-
Explanation:
It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.
uses of sodium chloride in daily life
Answer:
sodium chloride can be used as salt
extraction sodium metal by electrolysis
a common chemical in laboratory experiments
Answer:
sodium chloride can be used as preservatives,
in preserving foods.
Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Answer:
1. PEP is a feedback inhibitor of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Explanation:
Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.
Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.
The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.
PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.
Therefore, the true statements from the options above are 1, 4, 6.
Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.
Which of the following best describes isotopes?
An element with the same number of neutrons, but a different number of protons.
An element with the same number of protons, but a different number of electrons.
An element with the same number of electrons, but a different number of neutrons
An element with the same number of protons, but a different number of neutrons
Answer: An element with the same number of protons, but a different number of neutrons
Explanation:
The # of protons in an atom is what determines what atom it is (hydrogen has 1 proton, helium has 2 protons, etc ...). You cannot change the number of protons in an atom without changing what element the atom is.
The number of electrons in atoms varies greatly because electrons are constantly gained, lost, and shared during chemical reactions.
An isotope is a variation of the same element (so they must have the same # of protons) that have different masses (and therefore a different number of neutrons).
The answer is the fourth choice, "An element with the same number of protons, but a different number of neutrons"
The isotopes refer to an element that consists of a similar number of protons but have a distinct no of neutrons.
What are isotopes:It is considered to be the members of the family with respect to the elements that consist of a similar number of protons but have a distinct no of neutrons. The no of protons in the nucleus measured the atomic number of elements based on the periodic table.
Therefore, the fourth option is correct.
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Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]
Explanation:
P(aq)+Q(aq)⇌3R(aq)
This problem involves applying LeChatelier's principle.
LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.
1) Increase [P]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
2) Increase [Q]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
3) Increase [R]
This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.
4) Decrease [P]
This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
5) Decrease [Q]
This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
6) Decrease [R]
This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.
7) Triple [P] and reduce [Q] to one third
No shift in the direction of the net reaction because both changes cancels each other.
8) Triple both [Q] and [R]
No shift in the direction of the net reaction because both changes cancels each other.
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
Answer:
0.031 m
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
Chemical change
Element
Explanation:
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³. What is the length of one side of the cube in cm?
Step 1: Convert the mass to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]
Step 2: Calculate the volume (V) of the cube
[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]
Step 3: Calculate the length (l) of one side of the cube
We will use the following expression.
[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The balanced chemical equation is:
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
This is a chemical change because new substances are formed.
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
Precision can be defined as the?
Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.
Explanation:
Why is tape attracted to my skin? Give explanation
Answer:
Since the tape has extra electrons, it has a negative charge. When you move your finger close to the tape, electrons in your skin are repelled and move away. This makes the skin on your finger tip have a slight positive charge. Since positive and negative attract, the tape moves toward your finger.
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.
Answer:
The percentage yield is 50%
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10 °C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.
Answer:
The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).
Explanation:
The other options are true because:
A. Initial temp = 50 °C
Final temp = 270 °C
Differences in temp = 270 - 50 = 220°C
Rate = 10 °C/minute.
So, at 10 °C/minute,
total of 220°C /10 °C = number of minutes required to reach the final temp.
220/10 = 22 minutes
B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.
C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.
D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.
A new non-electrolyte molecule is discovered. When 241 mg of the molecule is dissolved in 250.0 mL of water, it has an osmotic pressure of 0.072 atm at 25 oC.What is the molar mass of the molecule
Answer:
327.89g/mol
Explanation:
Step 1:
The following data were obtained from the question:
Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)
Temperature (T) = 25°C = 25°C + 273 = 298K
Gas constant (R) = 0.0821 atm.L/Kmol
Mass of molecule = 241mg
Volume of water = 250mL
Molarity (M) =?
Osmotic pressure (Π) = 0.072 atm
Step 2:
Determination of the molarity of the molecule.
This can be obtained as follow:
Π = iMRT
0.072 = 1 x M x 0.0821 x 298
Divide both side by 0.0821 x 298
M = 0.072 / (0.0821 x 298)
M = 2.94×10¯³ mol/L
Step 3:
Determination of the number of mole of the molecule. This can be obtained as follow:
Molarity = 2.94×10¯³ mol/L
Volume of water = 250mL = 250/1000 = 0.25L
Mole of molecule =..?
Molarity = mole /Volume
2.94×10¯³ = mole / 0.25
Cross multiply
Mole of molecule = 2.94×10¯³ x 0.25
Mole of molecule = 7.35×10¯⁴ mole.
Step 4:
Determination of the molar mass of the molecule.
Mole of molecule = 7.35×10¯⁴ mole.
Mass of molecule = 241mg = 241×10¯³g
Molar mass of molecule =..?
Mole = Mass /Molar Mass
7.35×10¯⁴ = 241×10¯³/ Molar Mass
Cross multiply
7.35×10¯⁴ x molar mass = 241×10¯³
Divide both side by 7.35×10¯⁴
Molar Mass = 241×10¯³/7.35×10¯⁴
Molar Mass = 327.89g/mol
Therefore, the molar mass of the molecule is 327.89g/mol
3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm. The density of the metal is 5.30 g/cm3. Assume that 68% of the unit cell is occupied by Ba atoms. The molar mass of barium is 137.3 g/mol. Using this information, calculate Avogadro’s number. Show your calculation procedure that allows you to derive Avogadro’s number. Your answer must show six digits after the decimal point (i.e., 6.pppx1023) that is not necessarily the same as the known value. By showing your calculation-result down to six digits after the decimal point, you showcase that you did calculate the number, instead of simply adopting the known Avogadro’s number available in open resources.
Answer:
The Avogadro's number is [tex]N_A = 6.02289 *10^{23}[/tex]
Explanation:
From the question we are told that
The edge length is [tex]L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}[/tex]
The density of the metal is [tex]\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3[/tex]
The molar mass of Ba is [tex]Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol[/tex]
Generally the volume of a unit cell is
[tex]V = L^3[/tex]
substituting value
[tex]V = [5.02 *10^{-10}]^3[/tex]
[tex]V = 1.265*10^{-28}\ m^3[/tex]
From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),
The volume of barium atom is
[tex]V_a = \frac{V}{2} * 0.68[/tex]
substituting value
[tex]V_a = \frac{ 1.265*10^{-28}}{2} * 0.68[/tex]
[tex]V_a = 4.301 *10^{-29} \ m^3[/tex]
The Molar mass of barium is mathematically represented as
[tex]Z = N_A V_a * \rho[/tex]
Where [tex]N_A[/tex] is the Avogadro's number
So
[tex]N_A = \frac{ Z}{ V_a * \rho}[/tex]
substituting value
[tex]N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}[/tex]
[tex]N_A = 6.02289 *10^{23}[/tex]
How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?
Answer:
[tex]m_{CaF_2}0.375gCaF_2[/tex]
Explanation:
Hello,
In this case, for the studied reaction:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:
[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]
Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:
[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]
Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:
[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]
Best regards.
How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂
Answer:
0.032moles
Explanation:
2.10moles in 1000ml what about 15.25ml
(15.25×2.10)÷1000
0.032moles
Dry chemical hand warmers utilize the oxidation of iron to form iron oxide according to the following reaction: 4Fe(s)+3O2(g)→2Fe2O3(s) Standard thermodynamic quantities for selected substances at 25 ∘C Reactant or product ΔH∘f(kJ/mol) Fe(s) 0.0 O2(g) 0.0 Fe2O3(s) −824.2 Calculate ΔH∘rxn for this reaction.
Answer:
-1648.4 kJ/mol
Explanation:
Based on Hess's law:
ΔHr = ∑n×ΔH°f(products) - ∑n×ΔH°f(reactants)
In the reaction:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
ΔHr = 2 ΔH°f {Fe₂O₃} - (4ΔH°f {Fe(s)} + 3ΔH°f{O₂(g)}
As:
ΔH°f {Fe₂O₃} = -824.2kJ/mol
ΔH°f {Fe(s)} = 0.0kJ/mol
ΔH°f{O₂(g)} = 0.0kJ/mol.
Thus,
ΔHr = 2 ₓ -824.2kJ/mol =
-1648.4 kJ/molAnswer:
-1648.4 kJ
Explanation:
The product has the only nonzero heat of formation, so it is the only value needed to calculate the enthalpy of this reaction. Normally, you would want to express the enthalpy of a reaction with respect to one mole of a chemical species, whether it is a reactant or product. However, since the balanced chemical equation contains only coefficients greater than 1, you should consider how the enthalpy relates to one mole of each substance according to the coefficients. In other words, − 1648.4 kJ of heat is released when 4 mol of Fe reacts with 3 mol of O2 to produce 2 mol of Fe2O3 .
Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 60. g of hexane is mixed with 74.5 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.
Answer:
43.45g of water would be produced from the reaction.
Explanation:
Liquid became reacts with oxygen to produce carbon dioxide and water.
This type of reaction is known as combustion reaction between alkanes.
Equation of reaction.
Assuming the reaction occurs in an unlimited supply of oxygen,
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
From the above equation of reaction,
2 moles of C₆H₁₄ reacts with 19 moles of O₂ to produce 14 moles of H₂O.
To find the theoretical mass,
Number of moles = mass / molar mass
Molar mass of C₆H₁₄ = 86g/mol
Molar mass of O₂ = 16g/mol × 2 = 32g/mol
Molar mass of H₂O = 18g/mol
Mass of H₂O = number of moles × molar mass
Mass of H₂O = 14 × 18 = 252g
Mass of C₆H₁₄ = number of moles × molar mass
Mass of C₆H₁₄ = 2 × 86 = 172g
Mass of O₂ = number of moles × molar mass
Mass of O₂ = 19 × 32 = 608g
From the equation of reaction,
172g of C₆H₁₄ reacts with 608g of O₂ to produce 252g of H₂O
(172 + 608)g of reactants produce 252g of H₂O
780g of reactants produce 252g of H₂O
(60 + 75.5)g of reactants will produce a x g of H₂O
780g of reactants = 252g of H₂O
134.5g of reactants = x g of H₂O
X = (134.5 × 252) / 780
X = 43.45g of H₂O
Therefore, 43.45g of H₂O would be produced from 60g of hexane and 74.5g of oxygen
Answer:
[tex]m_{H_2O}=30.9gH_2O[/tex]
Explanation:
Hello,
In this case, the combustion of hexane is given by:
[tex]C_6H_{14}+\frac{19}{2} O_2\rightarrow 6CO_2+7H_2O[/tex]
The next step is to compute the reacting moles of hexane:
[tex]n_{C_6H_{14}}=60gC_6H_{14}*\frac{1molC_6H_{14}}{86gC_6H_{14}} =0.698molC_6H_{14}[/tex]
Then, the moles of hexane consumed by 74.5 g of oxygen using the molar ratio in the chemical reaction (1:19/2):
[tex]n_{C_6H_{14}}=74.5gO_2*\frac{1molO_2}{32gO_2} *\frac{1molC_6H_{14}}{19/2molO_2} =0.245molC_6H_{14}[/tex]
Therefore, as less moles of hexane are consumed by oxygen, it is in excess, so we compute the mass of water produced by the consumed 0.245 moles of hexane:
[tex]m_{H_2O}=0.245molC_6H_{14}*\frac{7molH_2O}{1molC_6H_{14}}*\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=30.9gH_2O[/tex]
Best regards.
Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.
Answer:
4.41
Explanation:
Step 1: Write the balanced equation
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
Step 2: Calculate the respective concentrations
[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]
[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]
[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]
Step 3: Make an ICE chart
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
I 0.100 0.300 0 0
C -x -3x +x +x
E 0.100-x 0.300-3x x x
Step 4: Find the value of x
Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396
Step 5: Find the concentrations at equilibrium
[CO] = 0.100-x = 0.100-0.0396 = 0.060 M
[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M
[CH₄] = x = 0.0396 M
[H₂O] = x = 0.0396 M
Step 6: Calculate the equilibrium constant (Kc)
[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.842 times that of the initial number of microstates
Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
Learn more about the Markovnikov rule here:
https://brainly.com/question/33423745
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Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
Answer:
Here's what I get
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
Your structure has two problems:
It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.Here's one way to draw a Lewis structure.
1. Draw a trial structure
Make F and O terminal atoms and give each one an octet (Fig. 1).
2. Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
3. Check the number of valence electrons available
1 S = 1 × 6 = 6 electrons
1 O = 1 × 6 = 6
4 F = 4 × 7 = 28
TOTAL = 40 electrons
The trial structure has the correct number of electrons.
4. Determine the formal charge on each atom.
To get the formal charges, we cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On S
VE = 6
BE = 5 bonding electrons = 5
FC = 6 - 5 = +1
(b) On O:
VE = 6
BE = 3 LP(six electrons) + 1 bonding electron = 7
FC = 6 - 7 = -1
(c) On F:
VE = 6
BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7
FC = 7 - 7 = 0
5. Minimize the formal charges
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
Move a lone pair from the O to make an S=O double bond (Fig. 2).
6. Recalculate the formal charges
(a) On S
VE = 6
BE = (3 bonding electrons) = 6
FC = 6 - 6 = 0
(b) On O:
VE = 6
BE = 2 LP(four electrons) + 2 bonding electrons = 6
FC = 6 - 6 = 0
Fig. 2 shows the Lewis structure in which all atoms have a formal charge of zero.
The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.
The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.
The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.
The lewis structure with the central S atom has been attached.
For more information about lewis structure, refer to the link:
https://brainly.com/question/4144781
If 200.4g of water is mixed with 101.42g of salt the mass of the final solution would be reported as
Answer:
301.8 g
Explanation:
We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.
m(solution) = m(solute) + m(solvent)
m(solution) = 200.4 g + 101.42 g
m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)
