Answer:
495nm
Explanation:
The energy of a photon could be obtained by using:
E = hc / λ
Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.
The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:
242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.
Replacing in the equation:
E = hc / λ
4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ
λ = 4.946x10⁻⁷m
Is maximum wavelength of light that could break a Cl-Cl bond.
Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:
4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
495nmHELPPP PLZ !
Which one of the following is an example of a difficulty caused by thermal
expansion?
A. butter melting too fast
B. falling through the ice while skating
C. sagging power lines on a cold day
D. cracking in the walls of a house
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⁺True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.
Answer:
Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.
For each row in the table below, decide whether the pair of elements will form a molecular compound held together by covalent chemical bonds
Answer:
1- yes
HBr--hydrogen bromide
2- no
BaBr₂----barium bromide
3- yes
NCl----- nitrogen chlorine
Hydrogen ,bromine and nitrogen , chlorine are the pair of elements which will form a molecular compound by covalent bond as they have 1, 7,5, 7 valence electrons respectively.
What is a covalent bond?Covalent bond is defined as a type of bond which is formed by the mutual sharing of electrons to form electron pairs between the two atoms.These electron pairs are called as bonding pairs or shared pair of electrons.
Due to the sharing of valence electrons , the atoms are able to achieve a stable electronic configuration . Covalent bonding involves many types of interactions like σ bonding,π bonding ,metal-to-metal bonding ,etc.
Sigma bonds are the strongest covalent bonds while the pi bonds are weaker covalent bonds .Covalent bonds are affected by electronegativities of the atoms present in the molecules.Compounds having covalent bonds have lower melting points as compared to those with ionic bonds.
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33. Hydrocarbons that release pleasant odors are called_________
hydrocarbons. (1 point)
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
Answer:
substituted hydrocarbons
Explanation:
i think
The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:
Answer:
The two-step mechanism is a slow mechanism and a fast mechanism. When we combine them, the result is
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
Explanation:
We know that the decomposition of hydrogen peroxide is catalyzed by iodide ion, which means that the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one:
H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction
IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction
If we cancel the same type of molecules and ions, the final result is:
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
The two-step mechanism represents the slow mechanism and a fast mechanism. At the time of combining them, the result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
Two-step mechanism:
The decomposition of hydrogen peroxide should be catalyzed by iodide ion, which represents the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one
Now
H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction
IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction
Now in case of cancelling, the same type of molecules and ions, the final result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
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Calculate the frequency (Hz) and wavelength (nm)
of the emitted photon when an electron drops from
the n = 4 to the n=2 level in a hydrogen atom
Answer:
wavelength, λ = 486.6 nm
frequency, f = 6.16 * 10¹⁴ Hz
Explanation:
a. Wavelength
Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)
Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s
1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)
1/λ = 2.055 * 10⁶ m
λ = 4.866 * 10⁻⁷ m
wavelength, λ = 486.6 nm
b. Frequency
Using f = c/λ
f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m
frequency, f = 6.16 * 10¹⁴ Hz
Which of the following statements about water is not true?
Answer:
Water has a low specific heat capacity and so large bodies of water moderate temperatures on Earth.
Explanation:
Water has a very high specific heat capacity, meaning that it has to absorb a lot of energy to raise the temperature by one degree. Because water has a high specific heat capacity, large bodies of water can moderate the temperature of nearby land.
Hope this helps.
Which of the following best describes the interaction of the alkali metals with water? Select the correct answer below: A. They all dissolve easily in water. B. They do not react or dissolve in water. C. They react strongly with water to produce an alkaline solution and hydrogen. D. They react strongly with water to produce an alkaline solution and oxygen.
Answer:
C. They react strongly with water to produce an alkaline solution and hydrogen
Explanation:
All alkali metals react vigorously with cold water. In the reaction, hydrogen gas is given off and the metal hydroxide is produced.
Hope that helps.
the equilibrium concentrations were found to be [H2O]=0.250 M , [H2]=0.600 M , and [O2]=0.800 M . What is the equilibrium constant for this reaction?
Answer:
Keq = 0.217
Explanation:
Let's determine the equilibrium reaction.
In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.
2H₂(g) + O₂(g) ⇄ 2H₂O (g) Keq
Let's make the expression for the equilibrium constant
Products / Reactants
We elevate the concentrations, to the stoichiometry coefficients.
Keq = [H₂O]² / [O₂] . [H₂]²
Keq = 0.250² / 0.8 . 0.6² = 0.217
The Haber Process is the main industrial procedure to produce ammonia. The reaction combines nitrogen from air with hydrogen mainly from natural gas (methane) and is reversible and exothermic. The enthalpy change for this reaction is - 92 kJ mol-1. In an experiment, 1.5 moles of N2 and 4.0 moles of H2 is mixed in a 1.50 dm3 reaction vessel at 450 °C. After reaching equilibrium, the mixture contained 0.9 mole of NH3.
A) With the above information, write the reaction equilibrium equation in the Haber process. t.
B) Calculate Kc for this reaction.
C) What is the equilibrium yield of ammonia in this reaction?
D) Referring to Le Chatelier's principle and above information, suggest two ways to increase the yield of ammonia in this reaction and explain.
Answer:
A) [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex].
B) [tex]Kc=0.0933[/tex].
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
B) For the calculation of Kc, we rate the equilibrium expression:
[tex]Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent [tex]x[/tex], we have:
[tex][NH_3]=0.6M=2*x[/tex]
[tex]x=\frac{0.6M}{2}=0.3M[/tex]
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
[tex][N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M[/tex]
[tex][H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M[/tex]
Therefore, the equilibrium constant is:
[tex]Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933[/tex]
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
odine atoms will combine to form I2 in liquid hexane solvent with a rate constant of 1.5×1010L/mol⋅s. The reaction is second order in I . Since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instantaneously, usually by photochemical decomposition of I2. Suppose a flash of light creates an initial [I] concentration of 2.00×10−2 M .
How long will it take for 94% of the newly created iodine atoms to recombine to form I2? Express your answer using two significant figures.
Answer:
The time taken is [tex]t = 1.11 *10^{-9} \ s[/tex]
Explanation:
From the question we are told that
The rate constant is [tex]k = 1.50 *10^{10} \ L /mol \cdot s[/tex]
The initial concentration of iodine atom is [tex][I] = 2.0*10^{-2} \ M[/tex]
Generally the integrated rate law for a second order reaction is mathematically represented as
[tex]\frac{1 }{[I_r]} = \frac{1}{[I]} * k * t[/tex]
Where [tex][I_r][/tex] is the concentration of the remaining iodine atom after the recombination which is mathematically evaluated as
[tex][I_r ] = [I_o ] *[/tex][100% - 94%]
The reason for the 94% is that we are told from the question that only 94% of the iodine atom recombined
=> [tex][I_r ] = [I_o ] *[/tex][6%]
=> [tex][I_r ] = [I_o ] *0.06[/tex]
substituting values
[tex][I_r ] = 2.0 *10^{-2}*0.06[/tex]
[tex][I_r ] = 1.2 *10^{-3}[/tex]
So
[tex]\frac{1 }{1.2 *10^{-3}} = \frac{1}{2.0 *10^{-2}} * 1.50*10^{10} * t[/tex]
[tex]t = 1.11 *10^{-9} \ s[/tex]
It will take "1.11 × 10⁻⁹ s".
Chemical reaction:A process wherein the two or even more compounds collide with both the proper orientation as well as enough effort to generate a new substance or the outcomes, is considered as Chemical reaction. This process involves the breaking as well as formation of atom connections.
According to the question,
Rate constant, k = 1.5 × 10¹⁰ L/mol.s
Initial concentration, [I] = 2.0 × 10⁻² M
By using the integrated rate law,
→ [tex]\frac{1}{[I]_r} = \frac{1}{[I]}[/tex] × k × t ...(Equation 1)
Now,
The concentration of remaining Iodine atom,
[[tex]I_r[/tex]] = [[tex]I_o[/tex]] × [100% - 94%]
= [[tex]I_o[/tex]] × 6%
= [[tex]I_o[/tex]] × 0.06
By substituting the values is above equation,
[[tex]I_r[/tex]] = 2.0 × 10⁻² × 0.06
= 1.2 × 10⁻³
hence,
The time taken will be:
[tex]\frac{1}{[I]_r} = \frac{1}{[I]}[/tex] × k × t
By substituting the values,
[tex]\frac{1}{1.2\times 10^{-3}} = \frac{1}{2.0\times 10^{-2}}[/tex] × 1.50 × 10 × t
t = 1.11 × 10⁻⁹ s
Thus the above answer is correct.
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Why is phosphorus unusual compared to other group 15 elements? Select the correct answer below: A. There are compounds that contain phosphorus-phosphorus bonds with uncommon oxidation states. B. Phosphorus is relatively unreactive. C. Phosphorus only forms compounds where the oxidation number of phosphorus is 5+. D. Phosphorus is the most electronegative of the group 15 elements.
Answer:
There are compounds that contain phosphorus-phosphorus bonds with uncommon oxidation states.
Explanation:
Phosphorus is a member of group 15 in the periodic table. Its common oxidation States are -3 and +5. Phosphorus is believed to firm some of its compounds by the participation of hybridized d-orbitals in bonding although this is also disputed by some scientists owing to the high energy of d - orbitals.
Phosphorus form compounds having phosphorus-phosphorus bonds in unusual oxidation states such as diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S hence the answer.
g A spontaneous process is one in which: A. releases a large amount of heat B. may happen (is possible) C. will rapidly approach equilibrium D. will happen quickly
Answer:
A. releases a large amount of heat
Explanation:
A reaction is said to be spontaneous if it can proceed on its own without the addition of external energy. A spontaneous reaction is not determined by the length of time, because some spontaneous reactions are completed after a long period of time. They are exothermic in nature. An example is the conversion of graphite to carbon which takes a long period of time to complete. Spontaneous reactions are known to increase entropy in a system. Entropy is the rate of disorder in a system.
In the combustion of fire, energy is released to the surroundings as there is a decrease in energy. This is an example of a spontaneous reaction because it is an exothermic reaction, which causes an increase in entropy and a decrease in energy.
The titration of 78.5 mL of an unknown concentration H3PO4 solution requires 134 mL of 0.224 M KOH solution. What is the concentration of the H3PO4 solution
Answer:
0.127 M.
Explanation:
The balanced equation for the reaction is given below:
H3PO4 + 3KOH —› K3PO4 + 3H2O
From the balanced equation above, we obtained the following data:
Mole ratio of acid, H3PO4 (nA) = 1
Mole ratio of base, KOH (nB) = 3
Data obtained from the question include:
Volume of acid, H3PO4 (Va) = 78.5 mL
Molarity of acid, H3PO4 (Ma) =...?
Volume of base, KOH (Vb) = 134 mL
Molarity of base, KOH (Mb) = 0.224 M
The concentration of the acid, H3PO4 can be obtained as follow:
MaVa / MbVb = nA/nB
Ma x 78.5 / 0.224 x 134 = 1/3
Cross multiply
Ma x 78.5 x 3 = 0.224 x 134 x 1
Divide both side by 78.5 x 3
Ma = (0.224 x 134 x 1) /(78.5 x 3)
Ma = 0.127 M
Therefore, the concentration of the acid, H3PO4 is 0.127 M.
A pressure cooker contains 5.68 liters of air at a temperature of 394 K. If the absolute pressure of the air in the pressure cooker is 205 pascals, how many moles of air are in the cooker? The cooker contains _______ moles of air. 1 SEE ANSWER
Answer:
Explanation:
We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )
Let it be V₂
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
[tex]\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}[/tex]
V₂ = 7.87 litres
22.4 litres of any gas is equivalent to 1 mole
7.87 litres of air will be equivalent to
7.87 / 22.4 moles
= .35 moles .
Match the symbol with the appropriate definition.
a. ΔG
b. ΔS
c. Keq
d. ΔH
e. ΔG°
f. J
1. Equation constant
2. Change in size
3. Change in entropy
4. Change in gas pressure
5. Change in stored energy
6. Equilibrium constant
7. Change in enthalpy
8. Joule
9. Standard free energy charge
10. Change in energy
Answer:
a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change.
b. ΔS : 3. Change in entropy.
c. Keq : 6. Equilibrium constant.
d. ΔH : 7. Change in enthalpy.
e. ΔG° : 9. Standard free energy change.
f. J : 8. Joule.
Explanation:
Hello,
In this case, we can match the symbol with the proper definition as shown below:
a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change and it uses G since it is better referred to the Gibbs free energy.
b. ΔS : 3. Change in entropy.
c. Keq : 6. Equilibrium constant.
d. ΔH : 7. Change in enthalpy.
e. ΔG° : 9. Standard free energy change.
f. J : 8. Joule.
Best regards.
Rank the solutions in order of decreasing [H3O+]. Rank solutions from largest to smallest hydronium ion concentration.
a. 0.10 M HNO3HNO3
b. 0.10 M HClO2HClO2
c. 0.10 M HCNHCN
d. 0.10 M HC2H3O2
The table below shows the electronegativity values of various elements on the periodic table. Electronegativities A partial periodic table. Which pair of atoms would form a covalent bond ? calcium (Ca) and bromine (Br) rubidium (Rb) and sulfur (S) cesium (Cs) and nitrogen (N) oxygen (O) and chlorine (Cl)
Answer:
Oxygen and Chlorine
Explanation:
Covalent bonds involve the sharing of electrons between nonmetals.
Answer:
oxygen (O) and chlorine (Cl)
Explanation:
cuz i said so
Calculate the ratio of moles of H2O to moles of anhydrous CuSO4 in CuSO4.
1H2O, CuSO4, 3H2O, CuSO4, 5H2O, CuSO4, 7H2O, and CuSO4, 9H2O.
(Mol wt. of H2O- 18.02g; Mol wt. of CuSO4 - 159.61g).
Answer:
CuSO₄.H₂O = 1
CuSO₄.3H₂O = 3
CuSO₄.5H₂O = 5
CuSO₄7.H₂O = 7
CuSO₄.9H₂O = 9
Explanation:
Some salts as CuSO₄ are presented in the hydratated form to give some stability in their uses.
Ratio of moles represents moles of H₂O / moles of CuSO₄.
In CuSO₄.H₂O you have 1 mole of water per mole of CuSO₄, Ratio is 1/1 = 1.
For CuSO₄.3H₂O are 3 moles of water per mole of CuSO₄. Ratio is 3/1 = 3
For CuSO₄.5H₂O are 5 moles of water per mole of CuSO₄. Ratio is 5/1 = 5
For CuSO₄.7H₂O are 7 moles of water per mole of CuSO₄. Ratio is 7/1 = 7
For CuSO₄.9H₂O are 9 moles of water per mole of CuSO₄. Ratio is 9/1 = 9
To format the electron configuration correctly make sure to superscript where needed, do not add spaces to your answer and italics are not possible in this module. Write the ground-state electron configuration for calcium, Ca. You may write either the full or condensed electron configuration.
Answer:
[tex]1s^{2}2s^{2} 2p^{6}3sx^{2}3p^{6} 4s^{2}[/tex] [full configuration]
Explanation:
Calcium has 20 electrons in its nucleus, and by virtue of its position on the periodic table (fourth row, second column), it is an s-block element with its entire lower orbital filled. it has two valence electrons.
Therefore, its configuration is:
[tex]1s^{2}2s^{2} 2p^{6}3sx^{2}3p^{6} 4s^{2}[/tex]
To condense this, you can utilize Argon's stable configuration before [tex]4s^{2}[/tex] to show that all orbitals before 4s are filled.
I hope this was helpful.
What is the magnetic quantum number value for an element with n = 1?
Answer:
ml= 0
Explanation:
The magnetic quantum number usually gives information regarding the directionality of an atomic orbital. The values of the magnetic quantum number ranges between integer values that lie between +l to -l.
However, the n=1 level contains only the 1s orbital which is spherically symmetrical. The s orbital having l=0 also has magnetic quantum number (ml) =0, hence the answer.
Treatment of 1 mole of dimethyl sulfate with 2 moles of sodium acetylide results in the formation of propyne as the major product.
A) Draw a reasonable mechanism accounting for the formation of the byproduct 2-butyne.
B) 2-Butyne is observed as a minor product of this reaction. Draw a mechanism accounting for the formation of this minor product and explain how your proposed mechanism is consistent with the observation that acetylene is present among the reaction products.
C) Predict the major and minor products that are expected if diethyl sulfate is used in place of dimethyl sulfate.
Answer:
(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation
(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.
(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.
Explanation:
attached below is diagram
Concentrated hydrochloric acid, HCl, comes with an approximate molar concentration of 12.1 M. If you are instructed to prepare 350.0 mL of a 0.975 M HCl solution, how many milliliters of the stock (concentrated) HCl solution will you use
Answer:
28.20 mL of the stock solution.
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 12.1 M
Volume of diluted solution (V2) = 350.0 mL
Molarity of diluted solution (M2) = 0.975 M
Volume of stock solution needed (V1) =..?
The volume of stock solution needed can be obtained by using the dilution formula as shown below:
M1V1 = M2V2
12.1 x V1 = 0.975 x 350
Divide both side by 12.1
V1 = (0.975 x 350)/12.1
V1 = 28.20 mL.
Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.
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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Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH
Answer:
pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa
pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa
Explanation:
pKa acetic acid, CH3COOH = 4.7
It is possible to determine pH of a buffer using H-H equation:
pH = pka + log [A⁻] / [HA]
For the acetic buffer,
pH = 4.7 + log [CH3COONa] / [CH3COOH]
As you want a pH 4.7 buffer:
4.7 = 4.7 + log [CH3COONa] / [CH3COOH]
1 = [CH3COONa] / [CH3COOH]
That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:
5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONaFor pH 5.7:
5.7 = 4.7 + log [CH3COONa] / [CH3COOH]
1 = log [CH3COONa] / [CH3COOH]
10 = [CH3COONa] / [CH3COOH] (1)
That means you need 10 times [CH3COONa] over [CH3COOH]
And as you know:
10mL= [CH3COONa] + [CH3COOH] (2)
Replacing (1) in (2):
10 = 10mL + [CH3COOH] / [CH3COOH]
10[CH3COOH] = 10mL + [CH3COOH]
11[CH3COOH] = 10mL
[CH3COOH] = 0.91mL
And [CH3COONa] = 10mL - 0.91mL =
[CH3COONa] = 9.09mL
That is:
0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONaThe volumes according to the pH are as follows:
(i) 5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa for pH 4.7
(ii) 0.91mL of 0.10 M CH₃COOH and 9.09mL 0.10 M CH₃COONa pH 5.7
Calculating the volume of chemicals needed:Given that pKa of acetic acid, CH₃COOH = 4.7
The pH of a buffer using the H-H equation is given by:
pH = pKa + log [A⁻] / [HA]
For the acetic buffer,
pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]
4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]
0 = log [CH₃COONa] / [ CH₃COOH]
takin antilog on both sides of the equation we get:
1 = [CH₃COONa] / [CH₃COOH]
It implies that the same amount of both species is needed to make the pH 4.7 buffer.
So,
5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa makes a buffer of pH 4.7
Similarly:
5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]
1 = log [CH₃COONa] / [CH₃COOH]
takin antilog on both sides of the equation we get:
10 = [CH₃COONa] / [CH₃COOH]
10[CH₃COOH] = [CH₃COONa]
It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]
We have to prepare 10 mL of buffer, so:
10mL= [CH₃COONa] + [CH₃COOH]
10mL = 11[CH₃COOH]
[CH₃COOH] = 0.91mL
So, [CH₃COONa] = 10mL - 0.91mL
[CH₃COONa] = 9.09mL
Therefore,
0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7
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5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g) 3H 2(g) + N 2(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K c for the reaction.
Explanation:
system at equilibrium, will the reaction shift towards reactants ~
--?'
2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an
exothermic reaction. Will heating the equilibrium system increase o~e amount of
ammonia produced? . .co:(
3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will
the reaction shift? ':'\
.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,
4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)
Temperature (K) Kc
300 1.5x104
600 55 k ' pr, cl l<..J~
e- ~ r fee, ct o. ~ 1<
900 3.4 X 10-3
Is the reaction endothermic or exothermic (explain your answer)?
t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\
exothe-rnh't.-- ,.. ..,. (/.., ,~.
5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.
What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)
N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..
~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb
J. [,v 1+3] ~
I
4,:i.~ = 0,05
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:
[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]
At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:
[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]
We can calculate the concentrations of all the species at equilibrium using an ICE chart.
2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)
I 1.00 0 0
C -2x +3x +x
E 1.00-2x 3x x
Since the concentration of ammonia at equilibrium is 0.200 M,
[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]
The concentrations of all the species at equilibrium are:
[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]
The concentration equilibrium constant (Kc) is:
[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
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The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Answer:
Approximately 100 °C.
Explanation:
Hello,
In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]
We can solve for the temperature as follows:
[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]
Thus, with the proper units, we obtain:
[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]
Hence, answer is approximately 100 °C.
Best regards.
Consider a triangle ABC like the one below. Suppose that C=83°, a = 43, and b = 44. Solve the triangle.
Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
If there is more than one solution, use the button labeled "or".
Answer:
Explanation:
In a triangle
a / sin A = b / sinB = c / sinC
Putting the values
43 / sin A = 44 / sinB
sinA / sinB = 43 / 44 = 1 / 1.023
A + B = 180 - 83 = 97
sinA / sin ( 97 - A ) = 1 / 1.023
sin 97 cos A - cos 97 sin A = 1.023 sin A
= .9925 cos A + .122 sin A = 1.023 sin A
.9925 cos A = .901 sin A
squaring
.985 cos²A = .8118 sin²A
.985 - .985 sin²A = .8118 sin²A
.985 = 1.7968 sin²A
sinA = .74
A = 47.73
B = 49.27
c / sin C = b / sin B
c = b sinC / sinB
= 44 x sin 83 / sin 49.27
= 44 x .9925 / .7578
= 57.62
What is titration? Question 1 options: The process of quickly adding one solution to another until a solid is formed. The process of slowly adding one solution to another until the reaction between the two is complete. The process of mixing equal volumes of two solutions to observe the reaction between the two. The process of combining two solids until the reaction between the two is complete.
Answer:
The process of slowly adding one solution to another until the reaction between the two is complete.
Explanation:
When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.
Hope that helps.