Answer: seen below
Explanation:
HCH3CO2 + NaOH --------------> CH3CO2- + H2O
Acid specie- HCH3CO2
base- NaOH
Neutral- Na+
CH3COOH + KOH ----------> CH3COOK + H20
Acid- CH3COOH
Base- KOH
Neutral- K+
The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol . Calculate ΔG for this reaction at 298 K when [dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00100 M .
Answer:
ΔG = -3879.6 J/mol = -3.88 kJ/mol
Explanation:
Step 1: Data given
The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol
Temperature = 298 K
[dihydroxyacetone phosphate]=0.100 M
[glyceraldehyde-3-phosphate]=0.00100 M .
Step 2: Calculate ΔG for this reaction
ΔG = ΔG° + RT ln ([glyceraldehyde-3-phosphate]/ [dihydroxyacetone phosphate])
⇒with ΔG° = 7.53 kJ/mol = 7
⇒with R = 8.314 J/mol*K
⇒with T = 298 K
⇒ with [glyceraldehyde-3-phosphate]=0.00100 M
⇒ with [dihydroxyacetone phosphate]=0.100 M
ΔG = 7530 J/mol + 8.314 * 298 * ln(0.001/0.1)
ΔG = -3879.6 J/mol = -3.88 kJ/mol
The Gibbs free energy change ΔG is approximately -3882 J/mol (or -3.88 kJ/mol).
To calculate the change in Gibbs free energy (ΔG) for the given reaction at 298 K, we use the relationship:
ΔG = ΔG°' + RT ln(Q)
where:
ΔG°' is the standard Gibbs free energy change, which is 7.53 kJ/mol (or 7530 J/mol since 1 kJ = 1000 J).R is the gas constant, 8.314 J/(mol·K).T is the temperature, 298 K.Q is the reaction quotient, given by the ratio of the concentrations of products to reactants.For the reaction, Q is calculated as:
Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]Given:
[dihydroxyacetone phosphate] = 0.100 M[glyceraldehyde-3-phosphate] = 0.00100 MTherefore:
Q = 0.00100 / 0.100 = 0.01Now, substituting the values into the equation:
ΔG = 7530 J/mol + (8.314 J/(mol·K) * 298 K * ln(0.01))Calculating the term involving the natural logarithm:
ln(0.01) ≈ -4.605Thus, the calculation is:
ΔG = 7530 J/mol + (8.314 J/(mol·K) * 298 K * -4.605)ΔG ≈ 7530 J/mol - 11412 J/molΔG ≈ -3882 J/molTherefore, the Gibbs free energy change ΔG for the reaction under the given conditions is approximately -3882 J/mol (or -3.88 kJ/mol).
In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the object is 40 g, the specific heat capacity of the chemical would be
Group of answer choices
25 J /g* °C
75 J /g* °C
100 J /g* °C
50 J /g* °C
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Compound B, C6H12O2, was found to be optically active, and it was oxidized to an optically active carboxylic acid A, by Ag (aka, Tollens Reagent). Oxidation of B by PCC gave an optically inactive compound X that reacted with Zn amalgam/HCl to give 3-methylpentane. With Na2Cr2O7/H2SO4, compound B was oxidized to an optically inactive dicarboxylic acid C, C6H10O4. Provide the structures of A, B, and C (ignore specific configuration of any stereocenters).
Answer:
Check the explanation
Explanation:
Acidipic Acid (which is an essential dicarboxylic acid for manufacturing purposes with about 2. 5 billion kilograms produced per year. It is mainly used for the production of nylon and its related materials.)
Going by the question, since the [tex]H_{2} CrO_{4}[/tex] is comparatively mild oxidizing agent than the [tex]CrO_{3}[/tex], it only oxidizes the carbon group
Kindly check the attached image below for the full explanation to the question above.
g Choose the best fit. Only choose one option once. A(n) ________ is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation. Answer 1 The ________ is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism. Answer 2 The ________ of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the ________ gives the balanced chemical reaction.
Answer:
A)An elementary step reaction
B) Molecularity
C) Mechanism of reaction
D.)Balance chemical equation
Explanation:
Molecularity is the number of molecules that react in an elementary reaction which is equal to the sum of stoichiometric coefficients of all reactants that participate in the elementary reaction.The reaction can be unimolecular or bimolecular depending on the number of molecules.
Mechanism of reaction reffered to the step by step sequence of elementary reactions through which the overall chemical reaction change occurs. It gives detail of the reaction in each stage of an overall chemical reaction.
An elementary step reaction reffered to is one step of reaction among the series of simple reactions that show how reaction is progressing at the molecular level.
In the redox persulfate-iodide experiment, a 100 mL reaction mixture is prepared for one of the runs as follows. 0.200 M KI 20 mL 0.200 M KNO3 15 mL 0.25% starch 5 mL 0.01 M Na2S2O3 10 mL 0.1 M EDTA 1 drop water to make 65 mL 0.200 M (NH4)2S2O8 35 mL a. How many moles of S2O32- would react when the solution turns dark blue? b. How many moles of S2O82- would react when the solution turns dark blue? c. What would be the initial rate of reaction if this reaction mixture took 200 seconds to turn dark blue?
Answer:
a) The number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b) The number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c) rate = 2.5 x 10-6M/L/s
Explanation:
a)
The reactions taking place in this experiment are represented by the ionic equations,
S₂O₈ 2- + 2I- ----------> 2SO₄2- + I₂ --------------------(1)
2 S₂O₃2- + I₂ -----------> S₄O₆ +2 I- -------------------(2)
The persulphate ions react with the iodide ions to produce free iodine which is in turn reduced by the thiosulphate ions to produce iodide ions again. This reaction proceeds till all the thiosulphate ions are used up. Therefore the rate of the reaction will be the rate at which iodine is formed and used up.
When there is free iodine the reaction mixture, the solution gives a dark blue coloration. This happens when all the thiosulphate ions are used up.
The volume of sodium thiosulphate ( Na₂S₂O3) solution added to the reaction vessel = 10ml
Molarity of sodium thiosulphate ( Na₂S₂O3) solution = 0.01M
Number of moles of ( Na2S2O3) = 0.01ml x 10M /1000ml = 0.0001M (molarity x volume in L)
The number of moles of thiosulphate reacted will be equal to the number of moles taken since the reaction proceeds till all the thiosulphate is consumed.
Hence, the number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b)
To calculate S₂O₈ 2-
The permanent blue color is produced once all the thiosulphate ions are used up and persulphate reacts with iodide ions to produce iodine, so, the number of moles of persulphate ions will be equal to the number of moles of iodine formed
According to the stoichiometry of equation 1.
1 mole of S₂O₈ 2-produces 1 mole of iodine.
According to the stoichiometry of equation 2,
1mole of iodine produced consumes 2 moles of S₂O₃2-
The number of moles of S₂O₃2- taken = 0.0001M.
2 moles of S₂O₃2- is equivalent to 1 mole I₂
therefore
0.0001 mole of S₂O₃2- = 0.0001/2 = 0.00005M of I₂
Since stoichimetrically,
1 mole of S₂O₈2- is equivalent to 1 mole I2, the number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c)
The initial reaction rate is given by
rate =change in concentration of persulphate ion [S₂O₈2-] / time
rate = change in Concentraion of I₂ / t
since initial concentration of I₂ = 0.
rate = Concentraion of I₂/ t
The concentration of I₂ = number of moles of iodine / total volume of solution in L
= 0.00005M/ 0.1L = 0.0005M/L (Volume of the solution = 100ml = 0.1L)
rate = Concentraion of I₂ / t
= 0.0005 /200s
rate = 2.5 x 10-6M/L/s
What is the solubility in moles/liter for iron(III) hydroxide at 25 oC given a Ksp value of 2.0 x 10-39. Write using scientific notation and use 1 or 2 decimal places (even though this is strictly incorrect!)
Answer:
9.28 × 10⁻¹¹ mol/L
Explanation:
Let's consider the solution of iron(III) hydroxide.
Fe(OH)₃(s) ⇄ Fe³⁺(aq) + 3 OH⁻(aq)
We can relate the solubility (S) of the hydroxide with the solubility product (Ksp) using an ICE chart.
Fe(OH)₃(s) ⇄ Fe³⁺(aq) + 3 OH⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product is:
Ksp = [Fe³⁺] × [OH⁻]³ = S × (3S)³ = 27 S⁴
[tex]S=\sqrt[4]{\frac{Ksp}{27} } = \sqrt[4]{\frac{2.0 \times 10^{-39} }{27} } = 9.28 \times 10^{-11} mol/L[/tex]
Answer:
S = 9.28 E-11 M
Explanation:
Fe(OH)3 ↔ Fe3+ + 3OH-S S 3S
∴ Ksp Fe(OH)3 = 2.0 E-39
⇒ Ksp = [Fe3+]*[OH-]³ = (S)*(3S)³ = 27(S)∧4
⇒ 27(S)∧4 = 2.0 E-39
⇒ (S)∧4 = 7.407 E-41
⇒ S = (7.407 E-41)∧(1/4)
⇒ S = 9.28 E-11 M
Describe relationship between frequency, wavelength, and energy as you move up and down the spectrum.
As you move up the electromagnetic spectrum, wavelength decreases, frequency increases, and energy increases. Conversely, moving down the spectrum results in increased wavelength, decreased frequency, and decreased energy.
The relationship between frequency, wavelength, and energy in the electromagnetic spectrum is fundamental to understanding various forms of electromagnetic radiation.
As you move up the spectrum (toward X-rays and gamma rays), the wavelength decreases, the frequency increases, and thus the energy of the electromagnetic waves increases. Conversely, as you move down the spectrum (toward radio waves), the wavelength increases, the frequency decreases, and the energy decreases.
This relationship can be summarized with the following points:
Wavelength and frequency are inversely proportional, meaning that as one increases, the other decreases.Energy is directly proportional to frequency — higher frequency means higher energy.Energy is inversely proportional to wavelength — longer wavelengths correspond to lower energy.These relationships are crucial because they help us understand why different types of electromagnetic radiation interact with matter in various ways and are used in different technologies.
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the
next few minutes?
A)Molecules in both the metal and the surrounding air will start moving at lower speeds.
B)Molecules in both the metal and the surrounding air will start moving at higher speeds.
C)The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up
D)The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down
D) The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
This means the air around will heat up, and the plate will cool down. They are trying to reach a thermal equilibrium.
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down is the event which would most likely to take place over next few minutes.
Hence, option (D) is correct answer.
What is Exothermic Reaction ?Exothermic reaction is a reaction that is chemical in nature and in which the energy is released in form of heat, electricity or light. Energy is released when the new bonds are formed and this bond making process is an exothermic process.
What is Exothermic Reaction ?Endothermic reaction is a chemical reaction that absorbs heat from the surroundings. Energy is absorbed when the bonds breaks and this bond breaking process is endothermic process.
Here in this process the heat is evolved so it is exothermic process.
Thus, from the above conclusion we can say that The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down is the event which would most likely to take place over next few minutes.
Hence, option (D) is correct answer.
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18-19. If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absor- bance in a 1-cm cell. Will the absorbance of the reagent-blank solu- tion also be increased by a factor of 10?\
Answer:
As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Explanation:
Absorbance of light by a reagent of concentration c, is given as
A = εcl
A = Absorbance
ε = molar absorptivity
c = concentration of reagent.
l = length of light path or length of the solution the light passes through.
So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.
But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.
Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Hope this Helps!!!
Answer:
As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Explanation:
Absorbance of light by a reagent of concentration c, is given as
A = εcl
A = Absorbance
ε = molar absorptivity
c = concentration of reagent.
l = length of light path or length of the solution the light passes through.
So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.
But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.
Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
The reaction C4H10 ---> C2H6 + C2H4 has activation energy (Ea) of 450 kJ/mol, and the Ea of the reverse reaction is 250 kJ/mol. Estimate ΔH, in kJ/mol, for the reaction as written above.
The enthalpy change (ΔH) of the reaction C4H10 ---> C2H6 + C2H4 can be calculated as the difference between the activation energies of the forward and reverse reactions, giving a result of 200 kJ/mol. Activation energy, the minimum energy that reactants need to react, influences the reaction rate.
Explanation:The question pertains to the activation energy and enthalpy change in the chemical reaction C4H10 ---> C2H6 + C2H4. The activation energy (Ea) is the minimum energy that reactants need to undergo a reaction, and it varies depending on the direction of the reaction. Using the given activation energies, the enthalpy change (ΔH) can be estimated as ΔH = Ea(forward) - Ea(reverse). Substituting the given values, ΔH = 450 kJ/mol - 250 kJ/mol = 200 kJ/mol.
Activation energy is highly significant to the rate of a chemical reaction. If the activation energy is larger than the average kinetic energy of the reactants, the reaction will occur slowly as only a few high-energy molecules can react. Conversely, if the activation energy is smaller, more molecules can react and the reaction rate is higher.
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Yousef measured the height of each seedling on day 1 and day 7. These are his results.
Group
Seedling
Increase in
height/mm
A no water
6.5
0.5
0.5
0.5
5.5
B 2 cm water
Height of seedling/mm
Day 1
Day 7
- 6.0
5.5
6.0
6.0
5.5
7.5
6.0
8.0
6.0
8.5
6.0
9.5
5.5
9.5
6.0
10.0
2.0
00 UN
2.0
2.5
C5cm water
a
b
C
Calculate the increase in height for each of the seedlings 7, 8 and 9.
Calculate the mean increase in height for each group of seedlings.
On graph paper, draw a bar chart to show Yousef's results. Put volume of
water on the x-axis, and mean increase in height on the s-axis.
Write a conclusion that Yousef could make from his results.
d
1 Plants
The main difference causing variation in mean stem length between the plants from the two dishes is the presence or absence of light, which affects photosynthesis and plant growth.
Explanation:The most probable cause for the difference in mean stem length between plants in dish A (no light) and dish B (light cycle) is the effect of light on plant growth. Plants in dish A, with no exposure to light, likely did not undergo photosynthesis effectively, resulting in shorter stems. On the other hand, the plants in dish B received a 14-hour light cycle, allowing them to photosynthesize and grow taller.
In terms of experimental variables, the key difference between the groups is the presence of light, a critical factor for photosynthesis, and consequently, plant growth. This is supported by the fact that other conditions were kept consistent for both dishes.
Mg(OH)2 + 2 HCl = MgCl2 + 2 H20
How many molecules of water are produced?
Answer:
2 molecules
Explanation:
From the stoichiometric equation, only 2-molecules of water is produced
Calculate the [H+] in 1.0 M solution of Na2CO3 (for H2CO3, Ka1 = 4.3 × 10–7; Ka2 = 5.6 × 10–11). 7.5 × 10–6 M 1.3 × 10–2 M 7.5 × 10–13 M 6.6 × 10–4 M None of these choices are correct.
Answer:
7.5x10⁻¹³M = [H⁺]
Explanation:
When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:
Na₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁺(aq) + OH⁻(aq) + 2Na⁺
Where Kb is defined from equilibrium concentrations of reactants, thus:
Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)
It is possible to obtain Kb value from Ka2 and Kw thus:
Kb = Kw / Ka2
Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹
Kb = 1.8x10⁻⁴
Replacing in (1):
1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]
The equilibrium concentrations are:
[Na₂CO₃] = 1.0M - X
[HCO₃⁺] = X
[OH⁻] = X
Thus:
1.8x10⁻⁴ = [X][X] / [1-X]
1.8x10⁻⁴ - 1.8x10⁻⁴X = X²
X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0
Solving for X:
X = -0.0135 → False answer, there is no negative concentrations
X = 0.0133
As [OH⁻] = X;
[OH⁻] = 0.0133
From Kw:
Kw = [OH⁻] [H⁺]
1x10⁻¹⁴ = 0.0133[H⁺]
7.5x10⁻¹³M = [H⁺]
The concentration of [H+] in a 1.0M solution of Na2CO3 is approximately 1.2 x 10-4M, closest to 6.6 × 10^-4 M
Explanation:To solve this problem, we will use the dissociation constants (Ka1 and Ka2) provided for the diacid H2CO3, as well as the fact that Na2CO3 will dissociate completely in water to form 2Na+ and CO32-. The CO32- anion can then react with water (H2O) to produce HCO3- and OH-, the latter of which will increase the pH of the solution. However, the HCO3- anion is amphiprotic and can further react with water to produce H2CO3 and OH-, again increasing the pH.
The calculations necessary to solve this question require solving the equilibrium problems for two equations: HCO3- + H2O <=> H2CO3 + OH- and H2CO3 + H2O <=> H3O+ + HCO3-. The concentrations at equilibrium are given as [H2CO3] = 0.033M, [HCO3-] = 1.2 × 10-4 M, [CO32-] = 4.7 x 10-11M, [H3O+] = 1.2 × 10-4M. Hence, [H+] = [H3O+] = 1.2 x 10-4 M. Given the multiple choice options, the closest is 6.6 × 10-4M.
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The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answer choices ortho and para at a 50:50 ratio mostly ortho mostly para meta
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
A sample of an ideal gas at 1.00 atm and a volume of 1.50 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 25.0 atm, what was the volume of the sample? Assume that the temperature was held constant.
Answer:
Explanation:
Using Boyle's law
P₁ V₁ = P₂V₂ and temperature is constant
where P₁(pressure) = 1.00atm, P₂ = 25 atm, V₁( volume) = 1.50L V₂ =
V₂ = ( P₁ V₁ ) / P₂ = ( 1 atm × 1.50L ) / 25 atm = 0.06 L
Chloromethane, , arises from microbial fermentation and is found throughout the environment. It is also produced industrially, is used in the manufacture of various chemicals, and has been used as a topical anesthetic. How much energy is required to convert 74.0 g of liquid to a vapor at its boiling point, –24.09 °C? (The heat of vaporization of is 21.40 kJ/mol.) Energy required = kJ
Answer:
31.36 KJ/mol
Explanation:
mass of the liquid = 74.0 g
heat of vaporization = 21.40 kJ/mol
constant boiling point temperature = 24.09 °C
heat required to vaporize 74.0g = heat of vaporization × mole
mole of Chloromethane = 74.0 g / 50.49 (g/mol) = 1.466 mol
energy required = 1.466 mol × 21.40 KJ/mol = 31.36 KJ/mol
Which of the following are the bark, roots, seeds, buds or berries of an aromatic plant?
-Spices
-Herbs
-Vegetables
-Fruits
Answer:
Spices
Explanation:
Herbs are the leaves.
Vegetables are the seeds.
Fruits are the seeds.
So spices are the only option left.
The correct term for the aromatic parts of plants such as bark, roots, seeds, buds, or berries is Option a i.e, spices. Spices are used for their flavor and aroma and come from various dried parts of plants. Examples include black pepper from the fruit of Piper nigrum and cinnamon from tree bark.
When referring to the aromatic parts of plants such as bark, roots, seeds, buds, or berries, the correct term is spices. Spices are aromatic substances derived from various dried parts of plants including roots, shoots, fruits, bark, and leaves. They are often used in cooking to add flavor and aroma and are sold in forms such as whole spices, ground spices, or seasoning blends. The correct option is a i.e, Spices.
An example is black pepper, which comes from the fruit of the Piper nigrum plant, and cinnamon, which is derived from the bark of a tree in the Laurales family.
How many molecules are in 2.00 mol of water?
There are 1.2044 × 10²⁴ molecules in 2 moles of water.
Explanation:
Given: 2 moles of water.
To convert moles of water into molecules by multiplying the number of moles by Avogadro's number (6.022 × 10²³ molecules).
We have to find the number of molecules as,
Number of molecules = Moles × Avogadro's number
Plugin the values as,
Number of molecules = 2 × 6.022 × 10²³ molecules
= 1.2044 × 10²⁴ molecules.
So there are 1.2044 × 10²⁴ molecules in 2 moles of water.
what is the main difference between a physical reaction and a
chemical reaction?
Answer:
physical reactons dont change the substance
Explanation:
physical changes are things like shape and color it doesnt change the base material but a chemical change changes the base aterial baking soda and vinegar create co2 but cutting wood still results in wood hope this helps god bless
Triphenylmethanol is insoluble in water, but when it is treated with concentrated H2SO4, a bright yellow solution results. As this yellow solution is diluted with water, its color disappears, and a precipitate of triphenylmethanol reappears. Suggest a structure for the bright yellow species.
Answer:
The product is triphenylmethane dye
Explanation:
The H2SO4 removes the OH
Leaving triphenylmethane dye
Find attached the suggested structure
Why does an incomplete combustion reaction occur?
The formation of tert-butanol is described by the following chemical equation:
(CH3)3CBr (aq) + OH- (aq) → Br- (aq) + (CH3)3COH (ǎq )
Suppose a two-step mechanism is proposed for this reaction, beginning with this elementary reaction
(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)
Suppose also that the second step of the mechanism should be bimolecular. Suggest a reasonable second step. That is, write the balanced chemical equation of a bimolecular elementary reaction that would complete the proposed mechanism
Answer:
Second step
(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)
Explanation:
This reaction involves;
First the ionization of the tertiary halide to firm a carbocation
Secondly the attack of the hydroxide ion on the carbocation to form tert-butanol
First step;
(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)
Second step
(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)
This second step completes the reaction mechanism.
Final answer:
The second step of the tert-butanol formation mechanism involves the tert-butyl cation reacting with a hydroxide ion to form tert-butanol, exemplifying a bimolecular reaction in a SN1 mechanism.
Explanation:
The question pertains to the formation of tert-butanol from tert-butyl bromide in a two-step mechanism. Given that the first step involves the formation of a tert-butyl cation and bromide ion, a reasonable second step in this bimolecular reaction would involve the tert-butyl cation reacting with hydroxide ion (OH-). The second and final step in the mechanism, thus, can be written as:
(CH3)3C+ (aq) + OH- (aq) → (CH3)3COH (aq)
This step involves the nucleophilic attack of the hydroxide ion on the carbocation, leading to the formation of tert-butanol. It exemplifies the bimolecular nature (involving two reactant species) of the second step, consistent with a SN1 mechanism where the first step is the rate-determining step.
How much heat in kJ is produced by the oxidation of 18.6 g of Mn?
3 Mn(s) + 2 O2(g) => Mn3O4(s) ΔHo = -1388 kJ
The oxidation of 18.6 g of manganese releases approximately -155.7 kJ of heat when reacting to form Mn₃O₄(s), using stoichiometry and the molar mass of manganese.
The student is asking about the heat produced by the oxidation of manganese (Mn) to form Mn₃O₄. The provided chemical equation indicates that the enthalpy change (ΔH⁰) for the reaction is -1388 kJ for the oxidation of 3 moles of manganese.
To find the heat released by the oxidation of 18.6 g of Mn, we must first calculate the number of moles of Mn in 18.6 g. Manganese has a molar mass of approximately 54.94 g/mol. Therefore:
Moles of Mn = 18.6 g ÷ 54.94 g/mol = 0.3386 mol Mn
Now, we will use the stoichiometry of the reaction to find the enthalpy change for 0.3386 mol Mn given that -1388 kJ is the enthalpy change for 3 moles of Mn:
Heat produced = 0.3386 mol Mn × (-1388 kJ ÷ 3 mol Mn) = -155.7 kJ (rounded to one decimal place)
The oxidation of 18.6 g of manganese produces approximately -155.7 kJ of heat.
A comic book villain is holding you at gun point and is making you drink a sample of acid he gives you a beaker with 100 ml of a strong acid with ph=5 he also gives you a beaker of a strong base with a ph=10 you can add as much of the strong base to the strong acid as you want, and you must then drink the solution. You'd be best off trying to make the solution neutral before drinking it. How much of the base should you add?
a. 1 ml
b. 10 ml
c. 100 ml
d. 1000 ml
Answer:
b. 10 mL
Explanation:
First we calculate the amount of H⁺ moles in the acid:
pH = -log [H⁺][H⁺] = [tex]10^{-pH}[/tex][H⁺] = 10⁻⁵ = 1x10⁻⁵M100 mL ⇒ 100 / 1000 = 0.100 L
1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺In order to have a neutral solution we would need the same amount of OH⁻ moles.
We can use the pOH value of the strong base:
pOH = 14 - pHpOH = 14 - 10 = 4Then we calculate the molar concentration of the OH⁻ species in the basic solution:
pOH = -log [OH⁻][OH⁻] = [tex]10^{-pOH}[/tex] = 1x10⁻⁴ MIf we use 10 mL of the basic solution the number of OH⁻ would be:
10 mL ⇒ 10 / 1000 = 0.010 L
1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻It would be equal to the moles of H⁺ so the answer is b.
Answer:
We have to add 10 mL of base ( option B is correct)
Explanation:
Step 1: Data given
Volume of the strong acid = 100 mL = 0.100 L
pH = 5
pH of the strong base = 10
Step 2: Calculate molarity of the strong acid
pH =[H+] = 5
[H+]= 10^-5 M
Step 3: Calculate moles of the strong acid
Moles = molarity * volume
Moles = 10^-5 M * 0.100 L
Moles = 10^-6 moles
Step 4: Calculate pOH
pOH = 14 - 10 = 4
Step 5: Calculate [OH-]
[OH-] = 10^-4 M
Step 6: Calculate volume need
We need 10^-6 moles of base
Volume = moles / molarity
Volume = 10^-6 moles / 10^-4 M
Volume = 0.01 L
Volume = 10 mL
We have to add 10 mL of base ( option B is correct)
How much concentrated 5.24M sulfuric acid is need to prepare 4.4 mL of a 5.29M solution
Answer:
V1= 4.44ml
Explanation:
C1= 5.24M V2=?, C2= 5.29, V2= 4.4ml
Applying dilution formula
C1V1=C2V2
5.24×V1 = 5.29×4.4
V1 =4.44ml
Determine the number of Ni2+ ions involved in metalization (Ni plating) of an electrode for 30 sec in an electrolysis cell operated at 0.020A at 78 % current efficiency? The Faraday constant, F = 96485 C/mol and Avogadro # NA = 6.023×1023. What will be the number of Ag+ ions if Ag is plated under the same experimental conditions instead of Ni2+?
Answer: The number of Ni2+ ions is 1.46 × 10^18 ions
The number of Ag+ ions is 2.92 × 10^18 ions
Explanation: Please see the attachments below
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2SO3(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: SO42−(aq)+Sn2+(aq)+ –––→H2SO3(aq)+Sn4+(aq)+ ––– Part B What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.
Answer:
The final balanced equation is :
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
Explanation:
[tex]SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq) [/tex]
Balancing in acidic medium:
First we will determine the oxidation and reduction reaction from the givne reaction :
Oxidation:
[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)[/tex]
Balance the charge by adding 2 electrons on product side:
[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-[/tex]....[1]
Reduction :
[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq) [/tex]
Balance O by adding water on required side:
[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]
Now, balance H by adding [tex]H^+[/tex] on the required side:
[tex]SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]
At last balance the charge by adding electrons on the side where positive charge is more:
[tex]SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]..[2]
Adding [1] and [2]:
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
The final balanced equation is :
[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]
The coefficients of the reactants and products in the balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution are 1, 1, 10, 1, 1, 5.
Explanation:The balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution can be written as:
SO₄²⁻(aq) + Sn₂⁺(aq) + 10H⁺(aq) → H₂SO₃(aq) + Sn₄+(aq) + 5H₂O(l)
The coefficients of the reactants and products in the balanced equation are 1, 1, 10, 1, 1, 5 respectively.
Learn more about Balanced equation here:https://brainly.com/question/7181548
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) Consider the starting materials and reagents. What do you expect to happen at the beginning of this reaction (Boxes 1-3 on the following page)? Explain your answer. Be sure to identify the species that act as electrophiles, nucleophiles, acids, or bases in Boxes 1-3 as part of your answer. (100 words max)
Answer:
Check the explanation
Explanation:
functional group found in the major organic product = alpha -beta unsaturated ketone
Reaction used to form this functional group = Michael condensation reaction
Also other reactions are - Aldol condensation , Robinson annulation reaction.
Kindly check the attached image below to see the step by step solution to the question above.
The ksp of lead(ii) carbonate, pbco3, is 7.40 × 10-14. calculate the solubility of this compound in g/l.
The solubility of lead(II) carbonate (PbCO₃) is approximately 1.04 × 10⁻⁸ g/L.
To calculate the solubility of lead(II) carbonate (PbCO₃) in g/L, we first need to understand the concept of Ksp (solubility product constant). The Ksp represents the equilibrium constant for the dissolution of an ionic compound in water.
The balanced equation for the dissociation of PbCO₃ is:
PbCO₃(s) ⇌ Pb₂ + (aq) + CO₃²-(aq)
Using the Ksp expression, we can write:
Ksp = [Pb₂+][CO₃²-]
Given that the Ksp of PbCO₃ is:
PbCO₃(s) ⇌ Pb₂ + (aq) + CO₃²-(aq)
Using the Ksp expression, we can write:
Ksp = [Pb₂+] [CO₃²-]
Given that the Ksp of PbCO₃ is 7.40 × 10⁻¹⁴, we can assume that the concentration of Pb₂+ and CO₃²- ions at equilibrium is the same. Let x be the solubility of PbCO₃ in mol/L.
So, Ksp = x * x = x²
Solving for x:
x = √(Ksp)
x = √(7.40 × 10⁻¹⁴) ≈ 8.60 × 10⁻⁸ mol/L
Now, to convert the solubility from mol/L to g/L, we use the molar mass of PbCO₃ (267.2 g/mol):
Solubility = (8.60 × 10⁻⁸ mol/L) * (267.2 g/mol) ≈ 1.04 × 10⁻⁸ g/L
Therefore, the solubility of lead(II) carbonate in g/L is approximately 1.04 × 10⁻⁸ g/L.
Complete Question:
The Ksp of lead(II) carbonate, PbCO₃, is 7.40 × 10⁻¹⁴. Calculate the solubility of this compound in g/l.
The answer is:[tex]2.30 \times 10^{-4} \text{ g/L}.[/tex] is solubility of this compound in g/l.
The solubility of lead(II) carbonate, PbCO₃, can be calculated using its solubility product constant (Ksp). The Ksp expression for PbCO₃ is given by:
[tex]\[ K_{sp} = [Pb^{2+}][CO_3^{2-}] \][/tex]
where the concentrations of Pb⁺² and CO₃²⁻- ions are equal at equilibrium because PbCO₃ dissociates into one Pb⁺² ion and one CO₃²⁻- ion. Let's denote the solubility of PbCO₃ as S, which is the concentration of Pb⁺² ions (and also CO₃²⁻ ions) at equilibrium. Therefore, we can write:
[tex]\[ K_{sp} = S \times S = S^2 \][/tex]
Given that Ksp = 7.40 — 10⁻¹⁴, we have:
[tex]\[ S^2 = 7.40 \times 10^{-14} \][/tex]
Solving for S:
[tex]\[ S = \sqrt{7.40 \times 10^{-14}} \] \[ S = 8.60 \times 10^{-7} \text{ M} \][/tex]
Now, to convert this molar solubility (S) to solubility in g/L, we need to know the molar mass of PbCO₃. The molar mass of PbCO₃ is calculated as follows:
Molar mass of Pb = 207.2 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
[tex]\[ \text{Molar mass of PbCO}_3 = 207.2 + 12.01 + 3 \times 16.00 \] \[ \text{Molar mass of PbCO}_3 = 207.2 + 12.01 + 48.00 \] \[ \text{Molar mass of PbCO}_3 = 267.21 \text{ g/mol} \][/tex]
Now, we can calculate the solubility in g/L:
[tex]\[ \text{Solubility in g/L} = S \times \text{Molar mass of PbCO}_3 \] \[ \text{Solubility in g/L} = 8.60 \times 10^{-7} \text{ M} \times 267.21 \text{ g/mol} \] \[ \text{Solubility in g/L} = 2.30 \times 10^{-4} \text{ g/L} \][/tex]
Therefore, the solubility of lead(II) carbonate in g/L is:
[tex]\[ \boxed{2.30 \times 10^{-4} \text{ g/L}} \][/tex]
Draw the arrow-pushing mechanism of a generic esterification reaction: b) From the spectral data (NMR, IR, MS) you were given, identify the structure of your ester product. c) Based on your answer to part b), what is the structure of your starting alcohol? 2) Critical analysis (6 points): a) Fully assign the 1H NMR spectrum of your product (i.e. determine which peaks in the 1H NMR correspond to which hydrogens in the product). You will not receive full marks for determination of the unknown unless you assign the 1H NMR spectrum completely. b) You will notice that we gave you the m/z value of the M+ ion of your product on the EI (electron impact) mass spectrum, but the actual peak on the spectrum is very small (or even non-existent). Read Solomons chapter 9 and explain why the M+ peak is so small.
Answer:
See explaination for detailed answer
Explanation:
In the IR spectrum, the broad peak at 3322 cm-1 corresponds to OH stretching while the peaks at 2929-2961 cm-1 correspond to C-H stretching. Thus the presence of alcohol is evident in the IR spectrum.
The 13C NMR suggests the presence of seven C-atoms in this ester. The peak corresponding to carbonyl carbon appear most downfield at ~172 ppm. The other six peaks are in the aliphatic region suggesting an aliphatic ester.
In the 1H NMR, we see a singlet at 2.0 ppm with the integral value of 3. This singlet is characteristic to the protons of the acetate (CH3CO) group as seen in ethyl acetate. This suggests that the acetic acid was employed in this esterification reaction. Using this information along with what we know from 13C NMR we can be certain that the given alcohol contains 5 carbons (total 7 carbon – 2 carbon from acetate group). Therefore the starting material must be pentyl alcohol.
The 1H NMR peaks for the pentyl group are
The most downfield triplet at 4.1 ppm corresponding to OCH2. This is due to deshielding of the CH2 by the electronegative O-atom
The most upfield triplet at 0.9 ppm corresponding to CH3.
A multiplet at 1.3 ppm corresponding to CH2 CH2 which is attached toCH3 moiety
A pentate at 1.6 ppm corresponding to CH2 which is attached toOCH3 moiety
Therefore, the given alcohol is n-pentyl alcohol and the ester is pentyl acetate (Molar mass 130.0994)
See attachment for diagram