Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
Use 1-Butanol as the only organic compound, design a method to synthesize 5-Nonanone. You may use any other inorganic reagents. Any organic reagents have to be made from 1-butanol.
Answer:
See attached picture.
Explanation:
Hello,
In this case, starting by 1-butanol, we can make it react with hydrogen bromide (1st step) in order to yield 1-bromobutane. Next, the formed alkyl halide is treated magnesium in the presence of an ether in order to yield butyl magnesium bromide which is a Grignard reagent (2nd step). Finally, by adding carbon dioxide, water and extra hydrogen chloride, a carbonyl group can be formed between two butyl radicals in order to form the 5-nonanone (3rd step) as shown on the attached picture.
Best regards.
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
Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Answer:
The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.
Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
Considering the definition of ionization energy,
Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.
You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.
In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.
In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.
Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
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https://brainly.com/question/24409114https://brainly.com/question/14158485?referrer=searchResultshttps://brainly.com/question/14454446?referrer=searchResultsTrue 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.
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.
29. Which alcohol combines with carboxylic acid to produce the ester called ethyl butanoate?
A) butan-2-ol
B) propan-1-ol
C) butan-1-ol
D) ethanol
Answer:
The answer is option D.
ethanol
Hope this helps you
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61MWhat element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)
Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
A. Silicon (Si)
Explanation:
Silicon (Si) is primarily used as a semiconductor material to make electronic chips.
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.
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.
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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Suppose there is 1.00 L of an aqueous buffer containing 60.0 mmol of formic acid (pKa=3.74) and 40.0 mmol of formate. Calculate the pH of this buffer.
Answer:
pH = 3.56
Explanation:
The pH of the buffer producing from the mixture of formic acid and formate ion can be found using H-H equation:
pH = pka + log [A⁻] / [HA]
pH = 3.74 + log [Formate] / [formic acid]
Where [] represents molar concentrations -or moles- of formate and formic acid in the solution.
Replacing knowing moles of formic acid = 0.0600 and moles formate = 0.0400:
pH = 3.74 + log [Formate] / [formic acid]
pH = 3.74 + log [0.0400] / [0.0600]
pH = 3.56
What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.
Answer: B. The period number tells which is the highest energy level occupied by the electrons
Explanation:
The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.
hence, the correct option is B. The period number tells which is the highest energy level occupied by the electrons.
The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.
What are energy levels ?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.
When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.
Therefore, option B , second option is correct.
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A________ chemical bond is a strong attraction between two or more atoms.
Answer:
Covalent bond is a chemical bond is a strong attraction between two or more atoms.
Answer:
Ionic
Explanation:
An ionic chemical bond is a strong attraction between two or more atoms. Two atoms are more strongly attracted by ionic bonds.
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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In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process.
Answer:
37%
Explanation:
From the question, the equation goes does.
C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.
Mair=Mair/Rin
( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.
33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1
= 176/176+8
X= 0.37
0.37 × 100
X= 37%
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
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.
Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water and carbon dioxide chemically react to form the simple sugar glucose and oxygen gas . What mass of simple sugar glucose is produced by the reaction of 4.9 of carbon dioxide?
Answer:
3.3 g of glucose, C6H12O6.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
6CO2 + 6H2O —> C6H12O6 + 6O2
Next, we shall determine the mass of CO2 that reacted and the mass of C6H12O6 produced from the balanced equation.
This is illustrated below:
Molar mass of CO2 = 12 + (2x16) = 44 g/mol
Mass of CO2 from the balanced equation = 6 x 44 = 264 g
Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol
Mass of C6H12O6 from the balanced equation = 1 x 180 = 180 g
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Finally, we shall determine the mass of C6H12O6 produced by reacting 4.9 g of CO2 as follow:
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Therefore, 4.9 g of CO2 will react to produce = (4.9 x 180)/264 = 3.3 g of C6H12O6.
Therefore, 3.3 g of glucose, C6H12O6 were obtained from the reaction.
A 45.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH.? Determine the pH of the solution after adding 35.0 mL of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + NaOH (aq) D NaC2H3O2(aq) + H2O (l) (Hint: Calculate new concentration and ICE table)
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
.02M
CH₃COOH = CH₃COO⁻ + H⁺
C xC xC
Ka = xC . xC / C = x² C
1.8 x 10⁻⁵ = x² . .02
x² = 9 x 10⁻⁴
x = 3 x 10⁻²
= .03
concentration of H⁺ = xC = .03 . .02
= 6 x 10⁻⁴ M , volume = 45 x 10⁻³ L
moles of H⁺ = 6 X 10⁻⁴ x 45 x 10⁻³
= 270 x 10⁻⁷ moles
= 2.7 x 10⁻⁵ moles
concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L
moles of Na OH = 2 X 10⁻² x 35 x 10⁻³
= 70 x 10⁻⁵ moles
=
NaOH is a strong base so it will dissociate fully .
there will be neutralisation reaction between the two .
Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles
= 67.3 x 10⁻⁵ moles of NaOH
Total volume = 45 + 35 = 80 x 10⁻³
concentration of NaOH after neutralisation.= 67.3 x 10⁻⁵ / 80 x 10⁻³ moles / L
= 8.4125 x 10⁻³ moles / L
OH⁻ = 8.4125 x 10⁻³
H⁺ = 10⁻¹⁴ / 8.4125 x 10⁻³
= 1.1887 x 10⁻¹²
pH = - log ( 1.1887 x 10⁻¹² )
= 12 - log 1.1887
= 12 - .075
= 11.925 .
Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. Use the theoretical yield of liquid iron and the mass or iron ingots to calculate the percent yield of the reaction.
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 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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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
HELPPP 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
For the imine synthesis reaction, the two reactants react in melting state. How is that possible since the melting points for both ortho-vanilin and para-toluidine are above room temperature
Answer:
I have no clue at all im in 11 and dont know anything lol byeeee
Explanation:um um i am lost
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
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.
Which types of electron orbitals will have higher energy than a 4d orbital?
A. 4p
B. 3s
C. 4f
D. 5s
Answer:
D: 5s
Explanation:
hope this helps :)
How would you monitor the progress of a neutralization reaction? Question 2 options: We will use a funnel to separate the solid as it forms We will use a balance to see the changes in mass We will use a thermometer to check the changes in temperature We will use an acid-base indicator to see changes in color depending on the pH
Answer:
We will use an acid-base indicator to see changes in colour depending on the pH
Explanation:
The pH changes during a titration, so you could use an acid-base indicator to follow the changes in pH.
A is wrong. An acid-base titration does not usually form a solid, and it would be impractical to isolate a solid with a funnel.
B is wrong. There are no changes in mass.
C is wrong. Any changes in temperature would be too small to measure precisely with an ordinary thermometer.
The best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.
The chemical reaction that occurs when you mix an acid and a base together is referred to as neutralization reaction.
In a neutralization reaction, what is formed is salt and water.
Acid-base titration is a neutralization method.
During acid-base titration, the neutralization reaction that occurs is usually monitored by observing the pH changes that occurs.
Change in pH is an indicator that there is progress in the neutralization reaction.
An acid-base indicator, can be used to detect the changes that occur via the pH changes in relation to the color change.
Therefore, the best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.
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Compare strontium with rubidium in terms of the following properties:
a. Atomic radius, number of valence electrons, ionization energy.
b. Strontium is smaller than rubidium.
c. Rubidium is smaller than strontium.
d. Strontium has more valence electrons.
e. Rubidium has more valence electrons.
f. Strontium has a larger ionization energy.
g. Rubidium has a larger ionization energy.
Answer:
Strontium is smaller
Strontium has the higher ionization energy
Strontium has more valence electrons
Explanation:
It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table
While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)
Since they are in the same period, periodic trends would be useful in evaluating their properties
In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size
Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius
Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has
In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case