A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)
Answers
Answer:
D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)
E: CH3-CH2-(CH3)-CH2-CH3
(Please see the figures enclosed )
Explanation:
D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically inactive. The reason is that two enantiomers are present in an equal amount).
E is optically inactive, so its structure has to be symmetric.
Related Questions
Water was poured over a large oil fire to extinguish it. What would happen and why?
Answers
Answer:
I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water
Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.
What happens when you pour water on an oil fire?A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.
Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.
The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.
Thus, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.
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The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 6.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?
Answers
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answers
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Clues about the history of the earth have been obtained from the study of
fossil fuels
rain forest materials
soll samples
O synthetic plastics
Answers
Answer: Fossil fuels
Explanation:
Fossil fuels such as petroleum, oil, and natural gas, are non-renewable energy resources which are formed from the remains of prehistoric ancient plants and animals beneath layers of rock of the earth surface.
By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.
The type of nuclear decay an unstable nucleus will undergo depends on its ratio of neutrons to protons. The radioisotope cobalt-65 has a ratio of neutrons to protons of 1.41, which is too high for a nucleus of this size. What nuclear changes could reduce this ratio
Answers
Answer:
Explanation:
In cobalt - 65 ,
no of protons is 27 ( p )
no of neutron = 65 - 27 ( n )
= 38
n / p ratio
= 38 / 27
= 1.41
If case of emission of alpha particle
no of proton p = 27 - 2 = 25
no of neutrons = 38 - 2 = 36
n / p ratio = 36 / 25
= 1.44
So it increases
In case of emission of beta particle
No of neutron n = 38 - 1 = 37
No of proton = 27 + 1 = 28
n / p ratio = 37 / 28
= 1.32
Hence ratio decreases.
Hence beta ray decay will result in decrease in n / p ratio.
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.904 g of precipitate, what is the molarity of lead(II) ion in the original solution
Answers
Answer:
[tex]M=0.0637M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]
Finally, the resulting molarity in 30.8 mL (0.0308 L):
[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]
Regards.
The boiling of water is a:_______.
a. chemical change because a gas (steam) is given off.
b. chemical change because heat is needed for the process to occur.
c. physical change because the water merely disappears chemical and physical damage.
d. physical change because the gaseous water is chemically the same as the liquid.
Answers
Answer:
D
Explanation:
trust me its correct i think
ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]
Answers
Answer:
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Explanation:
Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]
Photon momentum = h/wavelength
[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]
Mass of H2 molecule m = molar mass/Avogadros number
[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]
Since momentum is conserved:
Photon momentum = H2 molecule momentum = mass x velocity of H2
[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]
velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]
Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2
[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
Answers
11. Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. (a) What is the empirical formula of caproic acid
Answers
Answer:
C3H6O
Explanation:
Step 1:
Data obtained from the question include the following:
Mass of the compound = 0.225g
Mass of CO2 = 0.512g
Mass of H2O = 0.209g
Step 2:
Determination of the masses of carbon, hydrogen and oxygen present in the compound.
This is illustrated below:
For Carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C in CO2 = 12/44 x 0.512 = 0.1396g
For Hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H in H2O = 2/18 x 0.209 = 0.0232g
For Oxygen, O:
Mass of O = 0.225 – (0.1396 + 0.0232)
Mass of O = 0.0622g
Step 3:
Determination of the empirical formula for caprioc acid.
This can be obtain as follow:
C = 0.1396g
H = 0.0232g
O = 0.0622g
Divide by their molar mass
C = 0.1396/12 = 0.0116
H = 0.0232/1 = 0.0232
O = 0.0622/16 = 0.0039
Divide by the smallest
C = 0.0116/0.0039 = 3
H = 0.0232/0.0039 = 6
O = 0.0039/0.0039 = 1
Therefore, the empirical formula for caprioc acid is C3H6O
What happens in a double replacement reaction
Answers
Answer: D
Explanation: The elements in two compunds switch places
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answers
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
what is the sign of Mercury
Answers
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answers
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
When 200g of AgNO3 solution mixes with 150 g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34oC. Assume 350 g of solution and a specific heat capacity of 4.184 J/g•oC. Calculate H for the following: Ag+(aq) + I- (aq) → AgI(s)
Answers
Answer:
[tex]\Delta H=1962.3J[/tex]
Explanation:
Hello,
In this case, we can compute the change in the solution enthalpy by using the following formula:
[tex]\Delta H=mC\Delta T[/tex]
Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:
[tex]\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J[/tex]
It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.
Best regards.
At 25 oC, the rate constant for the first-order decomposition of a pesticide solution is 6.40 x 10-3 min-1. If the starting concentration of pesticide is 0.0314 M, what concentration will remain after 62.0 minutes at 25 oC? 3.12 x 10-2 M 47.4 M 2.11 x 10-2 M 4.67 x 10-2 M 8.72 M
Answers
Answer:
[tex]2.11\ * 10^{-2}[/tex] is the correct answer to the given question.
Explanation:
Given k=6.40 x 10-3 min-1.
According to the first order reaction .
The concentration of time can be written as
[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]
Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.
So [tex][\ A\ ]_{0}= 0.0314 M[/tex]
Putting this value into the above equation.
[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]
=0.211 M
This can be written as
[tex]=\ 2.11 *\ 10^{-2}[/tex]
A sample of helium has a volume of 325 mL and a pressure of 655 mmHg. What will be the pressure, in mmHg, if the sample of helium is compressed to 125 mL (T, n constant)? (Show calculations.)
Answers
Answer:
1703 mmHg
Explanation:
Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:
(13/5)(655 mmHg) = 1703 mmHg
In TLC chromatography of plant pigments, why do different pigments travel up the plate at different rates
Answers
Could someone please help me with this chemistry question I will mark the correct answer as brainliest
Answers
I’m pretty sure hope you get it right!
:)
Give the IUPAC name for the following structure
Answers
Answer:
6-metyl-2-heptyne
Explanation:
C-C-C-C-C-C-C hept
2
C-C≡C-C-C-C-C 2-heptyne
C
| 6
C-C≡C-C-C-C-C
6-metyl-2-heptyne
The IUPAC name for the above structure is 6 methyl, hept-2-yne.
What is IUPAC?IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.
The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.
According to this question, a structure is given. The following applies;
The compound has a triple bond located on the second carbon, hence, belongs to alkyne group. It has seven carbon atoms, hence, is heptyne. The methyl group is on the sixth carbon.Learn more about IUPAC at: https://brainly.com/question/33646537
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If a solution containing 23.81 g of lead(II) acetate is allowed to react completely with a solution containing 7.410 g of sodium sulfate, how many grams of solid precipitate will be formed g
Answers
Answer:
The correct answer is 15.80 grams.
Explanation:
The reaction taking place in the given question,
Pb(CH₃COO)₂ + Na₂SO₄ ⇒ PbSO₄ + 2NaCH₃COO
The number of moles can be calculated by using the formula,
n = weight / molecular mass
Based on the given question, the weight of lead (II) acetate is 23.81 grams and the weight of sodium sulfate is 7.410 grams.
The number of moles of Pb(CH₃COO)₂ is,
n = 23.81 g / 325.29 g/mol = 0.0732 moles
The number of moles of Na₂SO₄ is,
n = 7.410 g / 142.04 g/mol = 0.0521 moles
As one mole of lead (II) acetate needs one mole of sodium sulfate. Therefore, 0.0732 moles of lead (II) acetate needs 0.0732 moles of sodium sulfate.
However, as sodium sulfate is less, that is, 0.0521, therefore, Na₂SO₄ is a limiting reactant.
One mole of sodium sulfate produces one mole of PbSO₄. So, 0.0521 moles of Na₂SO₄ produces 0.0521 moles of PbSO₄.
Now the mass of PbSO₄ is,
mass = moles × molecular mass
mass = 0.0521 × 303.26 g/mol
mass = 15.80 grams.
2. In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a polar solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography
Answers
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answers
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
In the compound Fe2O3, iron's oxidation number is +3, and oxygen's oxidation
number is
Answer here
Answers
Answer: The oxygen's oxidation number is -2.
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as cation and oxygen is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]
The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answers
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
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?
Answers
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.
9. Predict the major products formed when: (a) Toluene is sulfonated. (c) Nitrobenzene is brominated. (b) Benzoic acid is nitrated. (d) Isopropylbenzene reacts with acetyl chloride and AlCl3. If the major products would be a mixture of ortho and para isomers, you should so state.
Answers
Answer:
a) ortho-para isomers predominates
b) 3-nitrobenzoic acid ( meta isomer predominates)
c) 3-bromo nitrobenzene ( meta isomer predominates)
d) the ortho- para isomers predominates
Explanation:
a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.
b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.
c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.
d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .
Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~50g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.
Answers
Answer:
The correct answer is 66.35 kilograms.
Explanation:
Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.
Hence, the total ATP produced in the process is,
ATP = 4000 kJ / 30.6 kJ/mol
= 130.7189 mol.
Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.
The mass of ATP can be calculated by using the formula,
moles = mass/molecular mass
The molecular mass of ATP is 507.18 g per mol
Now by putting the values we get,
mass of ATP = 130.7189 mol * 507.18 g/mol
= 66298.011 g or 66.298 kg
It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.
= Total production of ATP + Total ATP available
= 66.298 kg + 0.05 kg
= 66.348 kg
Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.
Calculate the pH of this solution 0.0043 M of H2SO4=
Answers
Answer:
pH = - log [concentration]
pH = - log (0.0043M)
pH = 2.37
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.850 A that flows for 60.0 min.
Answers
Answer:
Mass of Ga = 0.73694 gram
Explanation:
Given:
Current = 0.850 A
Time = 60 minutes
Find:
Amount of gas deposit.
Computation:
Total charge = Current × Time in second
Total charge = 0.850 × 60 × 60
Total charge = 3,060 C
Mole of electron = Total charge / Faraday constant [Faraday constant = 96,485.3329]
Mole of electron = 3,060 / 96,485.3329
Mole of electron = 0.0317146
Moles of Ga = 1/3 [Mole of electron]
Moles of Ga = 1/3 [0.0317146]
Moles of Ga = 0.01057
Mass of Ga = molar mass × Moles of Ga
Mass of Ga = 69.72 × 0.01057
Mass of Ga = 0.73694 gram