Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260}{28}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%[/tex]
Therefore, the percent yield is, 91.8%
Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
1. Define the Law of Conservation of Mass (via text). Now that you’ve defined this law, explain what it means in your own words using an example.
Explanation:
The law of conservation of mass states that mass can neither be created nor be destroyed.
Explanation in own words = this means that in this universe no one can create or destroy mass.
No physical or chemical force.
Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if a 1.75 L sample at 23.0*C were to have a final volume of 3.50 L.
A. 46.0*C
B. 89.5*C
C. 169*C
D. 319*C
E. 592*C
Answer:
592 K or 319° C
Explanation:
From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;
V1/T1= V2/T2
Initial volume V1 = 1.75 L
Initial temperature T1= 23.0 +273 = 296 K
Final volume V2= 3.50 L
Final temperature T2 = the unknown
T2= V2T1/V1= 3.50 × 296 / 1.75
T2 = 592 K or 319° C
Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Answer:
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Explanation:
Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature, K = Constant
You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:
[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]
In this case:
P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 KReplacing:
[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]
Solving:
[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]
P2=1.44 atm
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?
Write a balanced chemical equation to support your answer.
Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
What is the empirical formula for the compound: C8H8S2?
Answer:
Empirical formula = C4H4SExplanation:
The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.
Calculate the mass percent of .485g of H, which reacts with O to form 2.32g H2O?
Answer:
53.1% of hydrogen reacts
Explanation:
The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.
The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:
Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water
1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:
0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H
In mass:
0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water
As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:
(Mass that reacts / Mass added) × 100
(0.260g / 0.485g) × 100 =
53.1% of hydrogen reactsHow many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation:
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that
[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]
where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula
[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]
After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added
Answer:
The correct answer is 1.60.
Explanation:
Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,
Moles = volume * concentration of HCl
= 25/1000*0.10 = 0.0025 moles
Similarly the moles of NaOH added will be determined by using the formula,
Moles of NaOH added = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
= 0.001 / 0.040 = 0.025 M
pH = -log[H+]
= -log[0.025]
= 1.60
The pH after 15 ml of NaOH has been volume is 1.60.
Calculation of Concentration of HCl Moles
It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,
Moles is = volume * concentration of HCl
Then is = 25/1000*0.10 = 0.0025 moles
Besides, the moles of NaOH added will be determined by using the formula,
When the Moles of NaOH added is = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
When The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
After that = 0.001 / 0.040 = 0.025 M
pH = -log[H+]
Then = -log[0.025]
Therefore, = 1.60
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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Round your answer to significant digit.
Answer:
Molarity Cu²⁺ = 0.423M Cu²⁺
Explanation:
40.8g of copper (II) acetate into 200mL of a 0.700M sodium chromate
The reaction of copper acetate with sodium chromate occurs as follows:
Cu(CH₃COO)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2CH₃COONa
In water, the Copper(II) acetate dissociates in Cu²⁺ cation.
To know final molarity of Cu²⁺ we need to calculate the moles of Cu²⁺ that don't react with chromate ion, thus:
Moles of 40.8g of copper(II) acetate (Molar mass: 181.63g/mol) are:
40.8g × (1mol / 181.63g) = 0.2246 moles of Copper(II) acetate
Moles of sodium chromate are:
0.200L ₓ (0.700mol / L) = 0.140 moles of sodium chromate.
As 1 mole of Copper(II) acetate reacts per mole of sodium chromate, moles of Copper(II) acetate = Moles of Cu²⁺ that remains after the reaction are:
0.2246mol - 0.140moles = 0.0846 moles of Cu²⁺
Molarity is ratio between moles of solute (Moles Cu²⁺) and volume in liters of solution (200mL = 0.200L):
Molarity Cu²⁺ = 0.0846 moles / 0.200L
Molarity Cu²⁺ = 0.423M Cu²⁺For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.
1. a snowman melts on a spring day
2. a document goes through a paper shredder
3. a water bottle cools down in a refrigerator
4. silver tarnishes
5. dissolved sigar precipitates out of water to form rock candy
A. Entropy is increasing
B. Entropy is decreasing
Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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For the reaction system, 2SO2(g) + O2(g) <--> 2SO3(g), the equilibrium concentrations are: SO3: 0.120M SO2: 0.860M O2: 0.330M Calculate the value of Kc for this reaction.
Answer:
0.0590 M⁻¹
Explanation:
Kc represents the equilibrium constant. It is given as;
Kc = [products] / [reactants]
For the reaction; 2SO2(g) + O2(g) <--> 2SO3
Products = SO3
Reactants = SO2 and O2
Kc is given as;
Kc = [SO3]² / [SO2]² [O2]
Kc = 0.120² / (0.860)² (0.330)
Kc = 0.0144 / 0.2440 = 0.0590 M⁻¹
Find the [OH−] of a 0.32 M methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10−4.) Express your answer to two significant figures and include the appropriate units.
Answer:
[tex][OH^-]=0.01165M[/tex]
Explanation:
Hello,
In this case, for the dissociation of methylamine:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
We can write the basic dissociation constant as:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
That in terms of the reaction extent [tex]x[/tex], turns out:
[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]
[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]
That has the following solution for [tex]x[/tex]:
[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]
Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:
[tex][OH^-]=0.01165M[/tex]
Best regards.
2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g
Answer: The mass of sodium chloride in 219 g solution is 32.9 g
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]
To find mass of sodium chloride in solution:
[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]
Mass percent of sodium chloride= 15.0 %
Mass of solution = 219g
[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]
[tex]{\text{Mass of sodium chloride}=32.9g[/tex]
Thus mass of sodium chloride in 219 g solution is 32.9 g
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
Decide which element probably has a density most and least similar to the density of lithium.
Answer:
Helium and potassium
Explanation:
The density of helium is 0.18 and potassium 0.86, while lithium is 0.53
Which types of electron orbitals will have higher energy than a 4d orbital?
A) 4p
B) 3s
C) 5s
D) 4f
Answer:
D) 4f
Explanation:
To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.
1s
2s 2p
3s 3p 3d 3f
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d 6f
7s 7p 7d 7f
In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.
From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.
Therefore, 4f is the correct answer.
Answer:
D) 4f
Explanation:
Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the Li2+ ion.
Answer: E = - 19.611×[tex]10^{-18}[/tex] J
Explanation: The lowest possible energy can be calculated using the formula:
[tex]E_{n} = - Z^{2}.\frac{k}{n^{2}}[/tex]
where:
Z is atomic number of the atom;
k is a constant which contains other constants and is 2.179×[tex]10^{-18}[/tex] J
n is a layer;
For the lowest possible, n=1.
Atom of Lithium has atomic number of Z=3
Substituing:
[tex]E_{1} = - 3^{2}.\frac{2.179.10^{-18}}{1}[/tex]
[tex]E_{1} =[/tex] [tex]-19.611.10^{-18}[/tex] J
The energy for the electron in the [tex]Li^{+2}[/tex] ion is - 19.611 joules
The lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion is equal to [tex]1.96\times 10^{-17}\; Joules[/tex]
To determine the lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion, we would use the Bohr model:
Mathematically, Bohr's model is given by the equation:
[tex]Energy = -Z^2 \frac{k}{n^2}[/tex]
Where:
Z is the atomic number of an atom.n is the number of energy level.k is Rydberg constant.We know that the atomic number of lithium (Li) is equal to 3.
Also, at the lowest possible energy, n = 1.
Rydberg constant = [tex]2.179 \times 10^{-18}[/tex]
Substituting the parameters into the equation, we have;
[tex]E_1 = -3^2 \times \frac{2.179 \times 10^{-18}}{1^2} \\\\E_1 =9 \times 2.179 \times 10^{-18}\\\\E_1 =1.96\times 10^{-17}\; Joules[/tex]
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If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)
Answer:
B
Explanation:
The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.
A chemist adds of a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemist has added to the flask.
Answer:
3.383x10⁻³ micromoles of HgCl
Explanation:
The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.
The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:
0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.
Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:
3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =
3.383x10⁻³ micromoles of HgClSample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
Question 14 of 25
What type of reaction is BaCl2 + Na,504 → 2NaCl + Baso,?
A. Single-replacement
B. Synthesis
C. Double-replacement
D. Decomposition
double displacement
bcoz each of the reactants combines with other reactants to obtain the product
Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 CReplacing:
Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C
Solving:
Q=3,294.9 J
The heat absorbed by the sample of water is 3,294.9 J
Determine which complex of the electron transport chain (respiratory chain) each phrase describes. (Coenzyme Q is also called ubiquinone or ubiquinol, depending on whether it is in oxidized or reduced form.)
Complex I:
Complex II:
Complex III:
Complex IV:
Here are the choices that need to be put in the correct complex:
1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase)
2) Coenzyme Q-cytochrome c oxidoreductase
3) Electron transfer from succinate to ubiquinone (coenzyme Q)
4) Electron transfer from cytochrome c to O2
5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
6) Cytochrome c oxidase
7) Electron transfer from ubiquinol (QH2) to cytochrom c
8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Answer:
Complex I: (1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase), (8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Complex II: (3) Electron transfer from succinate to ubiquinone (coenzyme Q) (5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
Complex III: (2) Coenzyme Q-cytochrome c oxidoreductase, (7) Electron transfer from ubiquinol (QH2) to cytochrome c
Complex IV: (4) Electron transfer from cytochrome c to O2, (6) Cytochrome c oxidase
Explanation:
The electron transport chain (ETC) in the mitochondria provides a pathway by which electrons are transferred from NADH and FADH₂ through a series of membrane-bound carriers to molecular oxygen reducing it to water.
The electron transport chain electron carriers are organized into four complexes, Complexes I - IV.
Complex I : It is also called NADH:ubiquinone reductase. It transfers electrons from NADH to ubiquinone (also known as coenzyme Q)
Complex II : It is also called succinate dehydrogenase. It functions to tranfer electrons from succinate to FAD and then to ubiquinone.
Complex III : It is also called ubiquinone:cytochrome c oxidoreductase. It functions to transfer electrons from ubiquinol (reduced ubiquinone) to cytochrome c.
Complex IV : It is also called cytochrome oxidase. It functions to transfer electrons from cytochrome c to molecular oxygen reducing it to water.
The electron transporter chain is a series of enzymatic reactions to produce and store energy for the organism’s correct functioning. Complex I: 1 and 8. Complex II: 3 and 5. Complex III: 2 and 7. Complex IV: 4 and 6.
---------------------------------
Electron transporter chain
The electron transporter chain is located in the internal mitochondrial membrane. It constitutes a series of enzymatic reactions to release and save energy for the organism’s correct functioning.
Along the chain, there are four proteinic complexes in the membrane, I, II, III, and IV, that contain the electrons transporters and the enzymes necessary to catalyze the electrons' transference from one complex to the other.
Different redox reactions occur to pass electrons along the chain.
Released energy creates a proton concentration gradient used to synthesize ATP.
1) NADH provides electrons to the first complex, Complex I (NADH-
ubiquinone or NADH-coenzyme Q oxidoreductase).
From there, electrons go to the coenzyme Q (Ubiquinone) that carries them to complex II and III. Meanwhile, complex I pomp four protons to the intermembrane space.
2) Complex II (succinate-dehydrogenase) receives electrons from CoQ and also receives electrons from FADH2. Electrons are sent from complex II to ubiquinone Q that carries these electrons to complex III.
3) Complex III (Cytochrome C-reductase) receives electrons from ubiquinone Q and pomps protons to the intermembrane space.
Electrons are transferred to Cytochrome c.
Electrons travel from cytochrome c to complex IV.
4) Complex IV (Cytochrome C-oxidase) is the last complex that pomps protons to the intermembrane space. It takes electrons from cytochrome C and sends them to oxygen.
5) Electrons are sent to O₂ molecules, which also receive protons in the matrix to create water molecules.
Four electrons are needed to produce two water molecules from one O₂ molecule.
The proton gradient is used to produce ATP molecules.
Now, we can join the complexes with the phrases.
Complex I:
1) NADH-ubiquinone (NADH-coenzyme Q oxidoreductase)
8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Complex II:
3) Electron transfer from succinate to ubiquinone (coenzyme Q)
5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
Complex III:
2) Coenzyme Q - cytochrome c oxidoreductase
7) Electron transfer from ubiquinol (QH₂) to cytochrom c
Complex IV:
6) Cytochrome C oxidase
4) Electron transfer from cytochrome c to O₂
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The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____
Answer:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
The reactant that is reduced is Q.
Explanation:
The complete equation for the reaction is such that:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently, [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].
From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation: