Answer:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
[tex]\frac{600 torr(10L)}{500K} =\frac{200 torr(8L)}{xK}[/tex]
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
[tex]xK = \frac{500K(200 torr)(8L)}{600 torr(10L)}[/tex]
x = 400/3°K or 133.333°K
8. A 25.0 mL sample of an H2SO4 solution is titrated with a 0.186 M NaOH solution. The equivalence point is reached with 12.9 mL of base. The concentration of H2SO4 is ________ M. (Hint: write a balanced chemical equation first!)
Answer:
0.0480 M
Explanation:
The reaction is ...
H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O
That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...
moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M
2) Which type movement do pivot joints allow?
Select the oxidation reduction reactions??
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
Consider the addition of an electron to the following atoms from the fourth period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations.
Atom or ion Electron configuration
Br 1s22s22p63s23p64s23d104p5
Ge 1s22s22p63s23p64s23d104p2
Kr 1s22s22p63s23p64s23d104p6
Answer:
Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell
Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell
Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell
Explanation:
The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.
The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.
Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).
Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.
Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.
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The activation energy for the decomposition of HI is 183 kJ/mol. At 573 K, the rate constant was measured to be 2.91 x 10^{-6} M/s. At what temperature in Kelvin does the reaction have a rate constant of 0.0760 M/s
Answer:
[tex]T_2=453.05K[/tex]
Explanation:
Hello,
In this case, the temperature-variable Arrhenius equation is written as:
[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]
Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:
[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]
Regards.
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
Write a net ionic equation to show that benzoic acid, C6H5COOH, behaves as a Brønsted-Lowry acid in water.
Answer:
H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The complete ionic equation includes all the ions and molecular species.
C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)
Balance the chemical equation
Fe2O3 (s) + CO (g) 2 Fe(s) + CO2 (g)
Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
[tex]Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]
Explanation:
[tex]Fe_2O_3+CO\Rightarrow \:2Fe+CO_2\\\\Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]
Best Regards!
When solutions of hydrochloric acid and sodium hydroxide are mixed, a chemical reaction occurs forming aqueous sodium chloride and water. What would you expect to observe if you ran the reaction in the laboratory
Answer:
a change in temperature would be observed(ΔH is -ve)
Explanation:
Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water
HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)
There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.
Since neutralization is exothermic(heat is evolved), therefore ΔH is negative
100. mLmL of 0.200 MMHClHCl is titrated with 0.250 MMNaOHNaOH. Part A What is the pH of the solution at the equivalence point? Express the pH numerically.
Answer:
pH = 7.0
Explanation:
When HCl reacts with NaOH, H₂O and NaCl are produced, thus:
HCl + NaOH → H₂O + NaCl
At equivalence point, all HCl reacts with NaOH. The only you will have is water.
Equilbrium of water is:
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
K = 1x10⁻¹⁴ = [H⁺] [OH⁻]
As H⁺ = OH⁻ because both are produced from the same water-
1x10⁻¹⁴ = [H⁺]²
1x10⁻⁷M = [H⁺]
As pH = -log= [H⁺]
pH = 7.0-The pH at equivalence point in the titration of a strong acid with a strong base is always 7.0-
The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹A student mixed 115 g of sugar, 350 g of water and 5 g of spices. What will be the mass of the solution?
these are the options
470g
465g
350g
120g
Answer:
[tex]m_{solution}=470g[/tex]
Explanation:
Hello,
In this case, a solution is formed when a solute is completely dissolved in a solvent, thus, for this situation, the sugar is the solute and the water the solvent but in addition to them we find spices which are also considered in the total mass of the solution. In such a way, for computing the total mass we must add the mass of three constituents (115 g sugar, 350 g water and 5 g spices) as shown below:
[tex]m_{solution}=115g+350g+5g\\\\m_{solution}=470g[/tex]
Best regards.
For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, [SO3] = 0.254 M, [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will it go?
Answer:
To the left.
Explanation:
Step 1: Write the balanced reaction at equilibrium
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
Step 2: Calculate the reaction quotient (Qc)
Qc = [SO₃]² / [SO₂]² × [O₂]
Qc = 0.254² / 0.500² × 0.00855
Qc = 30.2
Step 3: Determine in which direction will proceed the system
Since Qc > Kc, the system will shift to the left to attain the equilibrium.
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
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which process is used to produce gases from solutions of salts dissolved in water or another liquid?
A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding
Answer:
A.Electrolysis
Explanation:
A.Electrolysis
For example, electrolysis of solution of NaCl in water gives H2 and O2.
A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate minerals. If it undergoes lots of chemical weathering, which sand grains will be quickly chemically weathered away?
a. Calcite
b. ferromagnesian silicate minerals
c. non-ferromagnesian silicate minerals
The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.
What are ferromagnesian silicate minerals?It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.
Hence, the correct option is b.
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Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be produced? Fe + CuSO4 -> Cu + FeSO4
Answer:
0.83 g
Explanation:
Step 1: Write the balanced equation
Fe + CuSO₄ ⇒ Cu + FeSO₄
Step 2: Calculate the moles corresponding to 0.75 g of Fe
The molar mass of Fe is 55.85 g/mol.
[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]
Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe
The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.
Step 4: Calculate the mass corresponding to 0.013 moles of Cu
The molar mass of Cu is 63.55 g/mol.
[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]
Answer:
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
Explanation:
You know the following balanced reaction:
Fe + CuSO₄ ⇒ Cu + FeSO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:
Fe: 1 moleCuSO₄: 1 moleCu: 1 moleFeSO₄: 1 moleBeing:
Fe: 55.85 g/moleCu: 63.54 g/moleS: 32 g/moleO: 16 g/molethe molar mass of the compounds participating in the reaction is:
Fe: 55.85 g/moleCuSO₄: 63.54 g/mole + 32 g/mole+ 4* 16 g/mole= 159.54 g/moleCu: 63.54 g/moleFeSO₄: 55.85 g/mole + 32 g/mole+ 4* 16 g/mole= 151.85 g/moleThen, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:
Fe: 1 mole*55.85 g/mole= 55.85 gCuSO₄: 1 mole* 159.54 g/mole= 159.54 gCu: 1mole* 63.54 g/mole= 63.54 gFeSO₄: 1 mole* 151.85 g/mole= 151.85 gThen you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?
[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]
mass of Cu= 0.85 grams
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. What is the final level of the electron?(c=3.00×10^8m/s, h=6.63×10^-34 J·s, RH=2.179×106-18J)a. 5
b. 6
c. 8
d. 9
e. 1
Explanation:
It is given that,
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,
[tex]n_i=8[/tex]
[tex]\lambda=3745\ nm[/tex]
The amount of energy change during the transition is given by :
[tex]\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]
And
[tex]\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]
Plugging all the values we get :
[tex]\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5[/tex]
So, the final level of the electron is 5.
Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
CaF2 + CO3- ----> CaCO3 + 2 F-
Explanation:
The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.
Calcium carbonate is a quaternary salt
A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Answer:
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g = [tex]\frac{0.0941}{74.44}[/tex] = 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
c) What is the pH of the buffer system in part a when 0.030 moles of strong acid are added (without a change in volume)
Answer:
remain the same
Explanation:
The pH of the buffer system remain the same when 0.030 moles of strong acid are added because buffer system has the property to resist any change in the pH when acid or base is added to the solution. In buffer system, one molecule is responsible for neutralizing the pH of the solution by giving H+ or OH-.This molecule is known as buffer agent. If more base is added, the molecule provide H+ and when more acid is added to the solution, then the molecule add OH- to the solution.
A correct name for the following compound is:_________.a) 4-bromo-3,8-dimethylbicyclo[5.2.2]nonane b) 3,8-dimethyl-4-bromo-bicyclo[5.2.O)nonane c) 4-bromo-3,8-dimethylbicyclo[5.2.1]decane d) 7-bromo-2,6-dimethylbicyclo[5.2.0]nonane e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane
Answer:
e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane
Explanation:
The missing image of the the compound we are to name is attached below.
Before we can name an organic compound; It is crucial we know the guiding rules in naming them.
1. Select the longest continuous carbon chain as the root hydrocarbon and name according to the number of carbon atoms it contains, adding appropriate suffix to indicate the principal substituent group.
2. Number the carbon atoms in the root hydrocarbon from the end which gives the lowest number to the substituents.
3. If the same substituent is present two or more times in a molecule; the number of this substituent is indicated by the prefix di -(2), tri - (3) , tetra - (4) etc attached to the substituent name.
4. If there is more than one type of substituent in the molecule ; the substituents are named according to the alphabetical order but where there are mixed substituents ; the inorganic are named first.
5. In selecting and numbering the longest continuous chain, the functional groups are given preference over substituents., i.e the functional group is given the smallest possible number.
In the light of the above guiding rules; we were able to name the given compound because the compound contains nine carbons in the ring form which result to root name nonane. The two methyl are on the third and eight carbon; bromine is on the fourth carbon ; there are two cyclic ring present in the compound where we have 5 carbons in one structure, another 2 carbons in the second structure and zero carbon in the bridge structure which eventually result to the correct name:
4-bromo-3,8-dimethylbicyclo[5.2.0]nonane
Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
the iupac name of the compound
Answer:
3-Pentyn-1-ol
Explanation:
tripple bond is at 3 postion from alochol
carbon are 5 atoms so pent
yn becauese its alkyne
Which land feature supports the theory of continental drift?
A.canyons B.volcanoes C.coal fields D.oceans
Answer:
Coals
Explanation:
The land feature that supports the theory of continental drift is ; ( C ) coal fields
Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.
Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.
Hence we can conclude that the land feature that supports the theory of continental drift is coal fields
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A(n) _____ reaction occurs when an acid and a base are present in the same solution.
Answer:
The answer is Neutralization reaction
It occurs when an acid and a base are present in the same solution and react to form salt and water only
Hope this helps you
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reactionATP(aq)+ H2O(l) → ADP(aq)+ HPO4^-2 (aq)for which ΔGrxn = -30.5 kj/mol at 37.0C and pH 7.0. Required:a. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and HPO4^-2= 5.0mMb. Is the hydrolysis of ATP spontaneous under these conditions?
Answer:
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous
Explanation:
From the question; The equation for this reaction can be represented as :
[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]
where:
[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol
= -30.5 kJ/mol × 1000 J/ 1 kJ
= -30.5 × 10 ⁻³ J/mol
Temperature T = 37 ° C
= (37+273)
= 310 K
pH = 7.0
[ATP] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
[ADP] = 0.30 mM
= 0.30 mM × 1M/1000mM
= 0.0003 M
[tex][HPO_4^{2-}}][/tex] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis of ATP is spontaneous under these conditions.
Now;
From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:
[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]
[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]
[tex]K_{eq} = 3*10^{-4}[/tex]
The Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]
Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous
The authors state in the general procedures that the reaction was monitored by TLC. How would this be done? What would you spot in each lane? How would you know the reaction was done?
Answer:
Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.
The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.
To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.
What is the name of CaCl2 7H2o
Answer:
calcium chloride dihydrate
The reduction of iron(III) oxide to iron metal is an endothermic process: Fe2O3(s) + 2 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = +26.3 kJ How many kilojoules of energy are required to produce 1.00 kilogram of iron metal?
Answer: Thus 234 kJ of energy are required to produce 1.00 kilogram of iron metal
Explanation:
To calculate the number of moles , we use the equation:
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
Putting values , we get:
[tex]\text{Moles of iron}=\frac{1000g}{56g/mol}=17.8moles[/tex] (1.00kg=1000g)
The balanced chemical reaction is:
[tex]Fe_2O_3(s)+2CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex] [tex]\Delta H=+26.3kJ[/tex]
Given :
Energy released when 2 moles of [tex]Fe[/tex] is produced = 26.3 kJ
Thus Energy released when 17.8 moles of [tex]Fe[/tex] is produced =
= [tex]\frac{26.3kJ}{2}\times 17.8=234kJ[/tex]
Thus 234 kJ of energy are required to produce 1.00 kilogram of iron metal