Answer:
Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}
or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4
Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.
Explanation:
The Mannich reaction is one of the few three-component reactions in organic chemistry. In this reaction, a ketone, an aldehyde and an amine react together under acid catalyzed conditions to form the final product. The mechanism involves the following steps: 1. Following initial protonation of the carbonyl oxygen, nucleophilic attack by the amine forms a protonated carbinolamine 1; 2. Proton transfer and elimination of water forms iminium ion 2; 3. The enol form of the ketone attacks the iminium ion to form adduct 3; 4. Deprotonation of adduct 3 leads to the final product. Write out the mechanism on a separate sheet of paper and then draw the structure of iminium ion 2.
Answer:
Step 1, formation of the iminium ion
2) enol reaction with iminium ion to form beta-amino enone
Explanation:
Please find attached the detailed reaction mechanism and the structure of iminium ion separately
The Mannich reaction is a three-component reaction that forms a ß-amino ketone or aldehyde. It involves protonation, nucleophilic attack, formation of an iminium ion, enol attack, and deprotonation.
The Mannich reaction is a three-component organic reaction involving a ketone, an aldehyde, and an amine under acid-catalyzed conditions to form a ß-amino ketone or ß-amino aldehyde. The mechanism of the reaction can be broken down into several steps:
Protonation of the carbonyl oxygen: Initially, the carbonyl oxygen of the ketone or aldehyde is protonated by an acid catalyst.Nucleophilic attack by the amine: The nucleophilic amine attacks the protonated carbonyl, forming a protonated carbinolamine.Formation of iminium ion: Following a proton transfer, the carbinolamine dehydrates to form an iminium ion.Enol attack: The enol form of the ketone then attacks the iminium ion, leading to the formation of adduct 3.Deprotonation: Finally, adduct 3 undergoes deprotonation to yield the final ß-amino product.Given these steps, the structure of iminium ion 2 involves a nitrogen atom double-bonded to a carbon atom derived from the original carbonyl.
What is the mass of 0.75 moles of (NH4)3PO4?
The mass of (NH4)3PO4 is 111.75grams.
HOW TO CALCULATE THE MASS:
The mass of a substance can be calculated by multiplying the number of moles by its molecular mass. That is;mass (g) = moles (mol) × molar mass (g/mol)
Molar mass of (NH4)3PO4 is calculated as follows:{14 + 1(4)}3 + 31 + 16(4)
= (14+4)3 + 31 + 64
= 54 + 31 + 64
= 149g/mol
mass in grams = 149 × 0.75
mass in grams = 111.75g
Therefore, the mass of (NH4)3PO4 is 111.75grams.
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Calcium hydroxide, Ca(OH)2, is used as a calcium nutritional supplement in some foods and beverages, such as orange juice. What is the pH of a solution of 0.0012M calcium hydroxide at 25.0∘C?
Answer:
pH = 11.38
Explanation:
The pH of a solution is given as:
pH = pKw − pOH = 14.00 + log[OH-]
Since the concentration of Ca(OH)2 is 0.0012 M, the [OH−] is twice that, or 0.0024 M, resulting in pOH = 2.62 and pH = 11.38.
Final answer:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C is 11.38.
Explanation:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C can be calculated using the concentration of hydroxide ions ([OH-]).
Calcium hydroxide is a strong base that dissociates completely in water to form two hydroxide ions for every formula unit dissolved.
The concentration of the solute is 0.0012M, but because Ca(OH)2 is a strong base, the actual concentration of hydroxide ions ([OH-]) is two times this, or 2 × 0.0012M = 0.0024M.
Since [OH-] = 0.0024M, the pOH can be calculated as pOH = -log([OH-]) = -log(0.0024) = 2.62. The pH can then be calculated using the expression pH = 14 - pOH = 14 - 2.62 = 11.38.
The reaction: 2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Answer:
Explanation:
question solved.
The rate law is defined as the rate of a chemical reaction, which is dependent on the concentration of the reactants.
It can be calculated by the formula:
[tex]\text v_o &= \text k [\text A]^x [\text B]^y[/tex]
where,
Vo = ratek = rate constantA = concentration of species Ax = order of reaction with respect to AB = concentration of species By = order of reaction with respect to BIn the given reaction:
[tex]\begin{aligned}2\;\text{ClO}_2\left(aq \right )+2\text{OH}^-\left(aq \right )\rightarrow\text{ClO}_3^-\left(aq \right )+\text{ClO}_2^-+\text{H}_2\text{O} \end{aligned}[/tex]
Using the rate law formula, we get:
Rate = K [Cl]ⁿ [OH⁻]ⁿ
0.0248 = K [0.060]ⁿ [0.030 ]ⁿ ........(1)
0.00276 = K [0.020]ⁿ [0.030 ]ⁿ .........(2)
0.00828 = K [0.020]ⁿ [0.090]ⁿ ..........(3)
Dividing (1) equation by (2), we get:
[tex]\begin {aligned} \dfrac {0.0248}{0.00276}&= \dfrac{\text K (0.06)^n (0.03)^n} {\text K (0.02)^n (0.03)^n}\\\\9 &= \dfrac{0.06}{0.02}^n\\\\3 ^2 &= 3 ^ m\\\text m &=2\\\end {aligned}[/tex]
Dividing the equation (2) by (3), we get:
[tex]\begin {aligned} \dfrac {0.00276}{0.00828 }&= \dfrac{\text K (0.02)^n (0.03)^n} {\text K (0.02)^n (0.09)^n}\\\\\dfrac{1}{3} &= \dfrac{0.03}{0.09}^n\\\\ \dfrac{1}{3} ^1 &= \dfrac{1}{3} ^n\\\\\text n &=2\\\end {aligned}[/tex]
Based on the rate law:
a. Rate = K [ClO₂]²⁻ [OH⁻]¹
b. 0.0248 = K [0.06]² [0.03]¹
K = [tex]\dfrac {0.0248}{(0.06)^2 \times 0.03}[/tex] = 229.63 m⁻²s⁻¹
c. Rate = K [ClO]₂ [OH⁻]ⁿ
Rate = 229.62 x (0.2)² (0.05)¹
Therefore, the rate of the reaction is 0.45926 m/sec.
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Consider this reaction occurring at 298 K: N2O(g) + NO2(g) 2 3 NO(g) a. Show that the reaction is not spontaneous under standard conditions by calculating A Grx b. If a reaction mixture contains only N20 and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous? c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?
Answer:
A.) ΔG = 107.8 kJ/mol
ΔG rxn is positive implies that the reaction is non spontaneous.
B) P (NO) = 5.0 x 10^-7 atm
C) T > 923.4 K, therefore, the temperature should be 924K in order to make the reaction spontaneous.
Explanation:
N2O(g) + NO2(g) - 3NO(g)
A) ΔG = ΔG products - ΔG reactants
ΔGf (N2O) = 103.7 kJ/mol
ΔGf (NO2) = 51.3 kJ/mol
ΔGf (NO) = 87.6 kJ/mol
ΔG rxn = 3*87.60 - (103.7 + 51.3)
= 107.8 kJ/mol
ΔG rxn is positive implies that the reaction is non spontaneous.
(B). P (N2O) = P (NO2) = 1 atm
ΔG rxn = ΔG°rxn + RTln(K)
ΔG rxn = ΔG°rxn + RT×ln (P NO)^3 / [P(N2O) × P(NO2)]
When the reaction ceases to be spontaneous, then ΔG rxn = 0.
0 = 107.8 × 10^3 + 8.314 × 298 × ln (P NO)^3 / (1 * 1)
107.8 × 10^3 = - 8.314 × 298 × ln (P NO)^3
ln (P NO)^3 = - 43.51
3 × ln (P NO) = - 43.51
ln (P NO) = -14.50
P (NO) = e^(-14.50)
P (NO) = 5.0 x 10^-7 atm
(C). For a reaction to be spontaneous, ΔG rxn should be negative.
It is known that:
ΔG= ΔH - TΔS
ΔG= ΔH - TΔS < 0
ΔH< TΔS
T < ΔH / ΔD
For the reaction:
ΔH = 3×NO - (N2O + NO2)
= 3×(91.3) - (81.6 + 33.2)
= 159.1 kJ/mol
ΔS = 3×210.8 - (220.0 + 240.1)
= 172.3 J/mol.K
= 0.1723 kJ/mol.K
From the above expression.
T > 159.1 / 0.1723
T > 923.4 K
Therefore, the temperature should be 924K in order to make the reaction spontaneous.
Salad is a mixture because
the ingredients go
well together
the ingredients form
a pure substance
the ingredients are
not chemcically
combined
the ingredients are
chemically
combined
Answer:
are you asking if it's true or false?
Answer:
no
Explanation:
The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product.
Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
See the attached file for the structure
Explanation:
See the attached file
Standard reduction potentials for zinc(II) and copper(II) The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following: Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V
The question is incomplete, the complete question is:
Standard reduction potentials for zinc(II) and copper(II)
The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:
Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V
Part B
What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Express your answer to three decimal places and include the appropriate units.
Answer:
1.100 V
Explanation:
E∘cell= E∘cathode - E∘anode
E∘cathode= +0.337 V
E∘anode= −0.763 V
E∘cell= 0.337-(-0.763)
E∘cell= 1.1V
Which of the following solutions is a good buffer system?Which of the following solutions is a good buffer system?A solution that is 0.10 M Li OH and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 MLiCNA solution that is 0.10 MNaCl and 0.10 MHClA solution that is 0.10 M Li and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 M Li Cl
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN.
Explanation:
A buffer solution is a solution of a weak acid and its conjugate base or a solution of a weak base and its conjugate acid, so the pH of the solution changes in a very little range when an acid or base is added to the solution.
Let's evaluate each statement:
(a) A solution that is 0.10 M LiOH and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the acid conjugate of the LiOH, and also, the LiOH is a strong base and the HNO₃ is a strong acid.
(b) A solution that is 0.10 M HCN and 0.10 M LiCN
This is a buffer solution, with the following reaction:
HCN + H₂O ⇄ CN⁻Li⁺ + H₃O⁺
The acid HCN is a weak acid and the LiCN is its conjugate base. The fact that the concentrations are equal for both is appropriate for the buffer solution.
(c) A solution that is 0.10 M NaCl and 0.10 M HCl
This is not a buffer solution since the HCl is a strong acid. It dissociates in water to form Cl⁻ and H⁺.
(d) A solution that is 0.10 M Li and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the conjugate base of Li, the HNO₃ is a strong acid that dissociates in water to form H⁺ and NO₃⁻.
(e) A solution that is 0.10 M HCN and 0.10 M LiCl
This is not a buffer solution since the LiCl is not the conjugate base of the acid HCN.
Therefore, the correct option is: A solution that is 0.10 M HCN and 0.10 M LiCN.
I hope it helps you!
17. Of the four different substances, which of the following has the highest alkalinity?
Substance A pH = 1.0
Substance B pH = 3.0
Substance C pH = 7.0
Substance D pH = 12.0 *
Answer:
Explanation:
Substance A
In the laboratory you are given the task of separating Ag+ and Cu2+ ions in aqueous solution.
For each reagent listed below indicate if it can be used to separate the ions. Type "Y" for yes or "N" for no. If the reagent CAN be used to separate the ions, give the formula of the precipitate. If it cannot, type "No"
Y or N Reagent Formula of Precipitate if YES
1. NaI
2. K2S
3. K2CO3
Answer:
1. NaI Y AgI
2. K2S Y CuS
3. K2CO3 N
Explanation:
1. When we add NaI to the mixture, the reaction that takes place is:
NaI(aq) + Ag⁺(aq) → AgI(s) + Na⁺(aq)Such a reaction does not happen with Cu⁺².
2. When we add K₂S to the mixture, the reaction that takes place is:
K₂S(aq) + Cu⁺²(aq) → CuS(s) + 2K⁺(aq)Such a reaction does not happen with Ag⁺.
3. When we add K₂CO₃ to the mixture, the reactions that take place are:
K₂CO₃(aq) + 2Ag⁺(aq) → Ag₂CO₃(s) + 2K⁺(aq)K₂CO₃(aq) + Cu⁺²(aq) → CuCO₃(s) + 2K⁺(aq)This means both Cu⁺² and Ag⁺ would precipitate, thus they would not be separated.
Fatty acid oxidation occurs in the mitochondrial matrix. However, long-chain fatty acyl-CoA molecules cannot cross the inner membrane to enter the matrix. The carnitine shuttle system transfers the acyl group from CoA to carnitine, which can enter the mitochondrial matrix. Label the enzymes and compounds of the carnitine shuttle system. These abbreviations are used: intermembrane space, IMS; carnitine acyltransferase I, CAT1; mitochondrial carnitine acyltransferase II, CAT2; and carnitine/acylcarnitine translocase (carnitine carrier protein), CAT.
Answer:
Explanation:
see answer below in the attached file.
The carnitine shuttle system is an important system that helps in the fatty acid oxidation.
The functions of the enzymes and compounds of the carnitine shuttle system include the following:
Intermembrane space (IMS): The activities of the enzymes of the carnitine system takes place in this space within the cells.Carnitine acyltransferase I, (CAT1): This is also called Carnitine palmitoyltransferase 1. It converts long-chain acyl-CoA species to their corresponding long-chain acyl-carnitines for transport into the mitochondria.Mitochondrial Carnitine acyltransferase II, CAT2: This is an enzyme found inside the mitochondria that oxidizes long-chain fatty acids in the mitochondria.carnitine/acylcarnitine translocase (carnitine carrier protein), CAT: This is a membrane transporter that exchanges cytoplasmic acylcarnitine for mitochondrial carnitine.Learn more here:
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A chemistry student weighs out 0.120 g of acetic acid into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.
Answer:
28.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction between acetic acid and sodium hydroxide
CH₃COOH(aq) + NaOH(aq) = CH₃COONa(aq) + H₂O(l)
Step 2: Calculate the moles of acetic acid
The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.120 g are:
[tex]0.120g \times \frac{1 mol}{60.05g} =2.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of sodium hydroxide
The molar ratio of CH₃COOH to NaOH is 1:1. The reacting moles of NaOH are 2.00 × 10⁻³ mol.
Step 4: Calculate the volume of the 0.0700 M NaOH solution
[tex]2.00 \times 10^{-3} mol \times \frac{1L}{0.0700mol} =0.0286 L = 28.6 mL[/tex]
Answer:
We have to add 286 mL of NaOH
Explanation:
Step 1: Data given
Mass of acetic acid (CH3COOH)= 0.120 grams
Volume of acetic acid = 250 mL = 0.250 L
Molarity of NaOH = 0.0700 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles acetic acid
Moles acetic acid = mass / molar mass
Moles acetic acid = 0.120 grams / 60.05 g/mol
Moles acetic acid = 0.00200 moles
Step 4: Calculate molarity acetic acid
Molarity acetic acid = moles / volume
Molarity acetic acid = 0.00200 moles /0.250 L
Molarity acetic acid = 0.008 M
Step 5: Calculate the volume of solution the student will need to add to reach the equivalence point
C1*V1 = C2*V2
⇒with C1 = the molarity of acetic acid = 0.008 M
⇒with V1 = the volume of acetic acid = 0.250 L
⇒with C2 = the molarity of NaOH = 0.0700 M
⇒with V2 = the volume of NaOH neede = TO BE DETERMINED
0.008 M * 0.250 L = 0.0700 M * V2
V2 = (0.008M * 0.250 L) / 0.0700 M
V2 = 0.286 L = 286 mL
We have to add 286 mL of NaOH
In a redox reaction, oxidation is defined by the:
1.gain of electrons, resulting in an increased oxidation number.
2.loss of electrons, resulting in a decreased oxidation number.
3.gain of electrons, resulting in a decreased oxidation number.
4.loss of electrons, resulting in an increased oxidation number.
Answer:
Option 4. loss of electrons, resulting in an increased oxidation number.
Explanation:
Oxidation is a process involving loss of electron(s). When this happens the oxidation number of the atom being oxidised increases. This can be seen when calcium (Ca) reacts with chlorine (Cl2) to form calcium chloride (CaCl2) according to the equation given below:
Ca + Cl2 —> CaCl2
The oxidation number of calcium increases from 0 to +2. This implies that calcium is being oxidised as it loses its electrons. The oxidation number of chlorine decreases from 0 to - 1 as it gains electron.
Now, we can see that the oxidation of calcium i.e lose of electrons increased its oxidation number from 0 to +2.
From the simple illustrations above, we can see clearly that oxidation involves loss of electrons, resulting in an increased oxidation number.
Balance the redox reaction by inserting the appropriate coefficients.
H2O + Br- + Al3- = Al + BrO3- + H+
The balanced equation is given as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
Explanation:
H₂O + Br⁻ + Al³⁺ → Al + BrO₃⁻ + H⁺
Here the half reactions are:
Al³⁺→ Al [reduction]
Br⁻ → BrO₃⁻ [oxidation]
Now we have to balance the half reactions as,
Al³⁺ + 3e⁻ → Al
To balance H atoms we have to add water and electrons.
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to balance the electrons as,
2Al³⁺ + 6e⁻ → 2 Al
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to add both the equations as,
2Al³⁺ + 6e⁻ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺ + 6e⁻
6 electrons on both sides of the equation gets cancelled and so the balanced reaction can be written as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
How many liters would you need to make a 0.7 M solution if you have 3.34 mol of Potassium
Chloride?
The volume of Potassium chloride solution is 4.8 L.
Explanation:
Molarity is found by dividing the number of moles of the given substance by its volume in litres, and so the unit of molarity is mol/L.
Here number of moles is given as 3.34 mol
Molarity = 0.7 M
Volume = ? L
Molarity = [tex]$\frac{moles}{Volume (L) }[/tex]
To find the volume we have to rearrange the equation as,
Volume (L) = [tex]$\frac{moles}{molarity}[/tex]
Now, we have to plug in the values as,
Volume (L) = [tex]$\frac{3.34 mol}{0.7 mol/L}[/tex]
= 4.77 ≈ 4.8 L
So the volume of Potassium chloride solution is 4.8 L.
Isocyanates are good electrophiles that have been used for protein modification. However, they have limited stability in water. As a result isothiocyanates have been developed at less hydrolytically sensitive variants. Molecules like fluorescein isothiocyanate (FITC) have been used extensively to fluorescently modify proteins. Draw the product of a lysine side chain with FITC. (for the sake of simplicity we are just using a model for lysine's side chain butyl amine)
Find figure in the attachment
Answer:
Explanation:
solution to the question is found below
A σ bond arises from the straight-on overlap of two atomic orbitals. The electron density lies along the axis of the two bonded nuclei. Example: Sigma Bonding in methane, CH4 What atomic or hybrid orbital on the central I atom makes up the sigma bond between this I and an outer Br atom in iodine tribromide, IBr3 ?
In iodine tribromide, the I-Br sigma bond is formed through the straight-on overlap of sp3d hybrid atomic orbitals on iodine with p orbitals on bromine.
Explanation:In iodine tribromide (IBr3), the central iodine (I) atom is surrounded by three bromine (Br) atoms. The I-Br bond that forms is a sigma (σ) bond, which is the result of a straight-on overlap of atomic orbitals. Sigma bonding typically involves the hybridized orbitals. Specifically, in IBr3, the I atom undergoes hybridization and forms two lone pairs and three σ bonds with Br atoms using sp3d hybrid orbitals. So, the hybrid orbital on the iodine atom that makes up the σ bond with a bromine atom is the sp3d hybrid orbital.
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The sigma bond in IBr3 between the central I atom and an outer Br atom is formed by the overlap of a sp3d hybrid orbital from iodine with a p orbital from bromine to accommodate the T-shaped geometry of the molecule.
In iodine tribromide (IBr3), the sigma bond between the central iodine (I) atom and an outer bromine (Br) atom is typically formed by the overlap of a hybrid orbital from the iodine with a p orbital from the bromine. When we consider the valence shell electronic configuration of iodine (5s2 5p5), upon forming IBr3, three of the p orbitals would be used to form sigma bonds with the bromine atoms, and this would likely involve hybridization to accommodate the molecular geometry of IBr3. The exact nature of the hybrid orbitals would depend on the geometry of the molecule, which, due to the presence of only three bond pairs and two lone pairs, adopts a T-shaped geometry, pointing towards sp3d hybridization. However, the exact details of this hybridization can only be confirmed with molecular orbital calculations
When 102 g of water at a temperature of 22.6 °C is mixed with 66.9 g of water at an unknown temperature, the final temperature of the resulting mixture is 48.9 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.) Initial temperature = °C
Answer:
[tex]T_{2,H_2O}=89^oC[/tex]
Explanation:
Hello,
In this case, we notice that the energy gained by the first sample of water is lost by the second sample of water as they are heated and cooled respectively, therefore, in terms of heats:
[tex]Q_{1,H_2O}=-Q_{2,H_2O}[/tex]
Which in terms of masses, heat capacities and temperatures is:
[tex]m_{1,H_2O}*Cp_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*Cp_{2,H_2O}*(T_{EQ}-T_{2,H_2O})[/tex]
In such a way, as the heat capacity is the same and the initial temperature of the cold water is required, we solve for it via:
[tex]m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*(T_{EQ}-T_{2,H_2O})\\T_{EQ}-T_{2,H_2O}=\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{-m_{2,H_2O}} \\\\T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}[/tex]
With the given data we obtain:
[tex]T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}\\\\T_{2,H_2O}=48.9^oC+\frac{102g*(48.9^oC-22.6^oC)}{66.9g} =48.9^oC+40.1^oC\\T_{2,H_2O}=89^oC[/tex]
Which means the second sample was hot.
Regards.
Answer:
The initial temperature of the second sample of water is 8.8 °C
Explanation:
Step 1 :Data given
Mass of water = 102 grams
Initial temperature of water = 22.6 °C =
Mass of other water = 66.9 grams
The final temperature is 48.9 °C
The specific heat capacity of liquid water is 4.184 J/g *K = 4.184 J/g°C
Step 2: Calculate the initial temperature of the second sample water
Heat gained = heat lost
Qgained = -Q lost
Q = m* C* ΔT
m(water1)*C(water)*ΔT(water1) = -m(water2) * C(water) *ΔT(water2)
⇒with m(water1) = the mass of the water at 22.6 °C = 102 grams
⇒with C(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT = the change of water of the first sample = 48.9 - 22.6 = 26.3 °C
⇒with m(water2) = the mass of the other unknown sample water = 66.9 grams
⇒with with C(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT = the change of water of the second sample = 48.9 - T1
102 * 4.184 * 26.3 °C = 66.9 * 4.184 * (48.9 - T1)
102 * 26.3 = 66.9 * (48.9 - T1)
2682.6 = 3271.4 - 66.9 T1
-588.8 = -66.9 T1
T1 = 8.8 °C
The initial temperature of the second sample of water is 8.8 °C
18.67 Consider the reaction Given that DG8 for the reaction at 258C is 173.4 kJ/mol, (a) calculate the standard free energy of formation of NO, and (b) calculate KP of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 11008C, estimate KP for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why
Answer:
a. 86.7KJ/mol
b. 2.5 * 10^30
c. 1.42 * 10^55
d. Lightning produces more amount of NO from nitrogen and oxygen thus giving a better crop as NO is essential for crop growth
Explanation:
Complete question is as follows;
Consider the reaction N2(g)+ O2(g) —-> 2NO(g)
Given that ΔG0 for the reaction at 25 degrees celsius is 173.4 kJ/mol a) calculate the standard freeenergy of formation of NO and B) calculate KP ofthe reaction. C) One of the starting substances in smog formationis NO. Assuming that the temperature in a running automobile engine is 1100 degreesC, estimate KP of the above reaction.D) As farmers know, lighting helps to produce a better crop.Why?
solution
Please check attachment for complete solution and step by step explanation
Two friends, Pamela and Elaine, compared snack items they planned to enjoy during a work break. Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein. Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient Fuel value (kJ/g):Carbohydrate 17, Fat 38, Carbs 17
(a) Using the fuel values provided, calculate the number of calories in each person's snack. Pamela's snack Cal Elaine's snack Cal
(b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 calories and Elaine expended 48 calories, who had the smaller net gain of calories after snacking and walking?
Answer:
a) Pamela's shortbread cookies has 117,471.3 calories = 117.5 kcal
Elaine's popcorn has 166,515.3 calories = 166.5 kcal
b) Pamela had the smaller net gain of calories after snacking and walking.
Explanation:
- Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein.
- Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient | Fuel value (kJ/g):
Carbohydrate | 17
Fat | 38
Protein | 17
Noting that instead of Carbs, Protein was intended to be written.
a) For Pamela
Carbohydrates = (21 g) × (17 KJ/g) = 357 KJ
Fats = (2.6 g) × (38 KJ/g) = 98.8 KJ
Protein = (2.1 g) × (17 KJ/g) = 35.7 KJ
Total fuel value = 357 + 98.8 + 35.7 = 491.5 KJ = 491,500 J
Note that 1 cal = 4.184 J
491,500 J = (491,500/4.184) = 117,471.3 calories = 117.5 kcal
Carbohydrates = (31 g) × (17 KJ/g) = 527 KJ
Fats = (2.9 g) × (38 KJ/g) = 110.2 KJ
Protein = (3.5 g) × (17 KJ/g) = 59.5 KJ
Total fuel value = 527 + 110.2 + 59.5 = 696.7 KJ
Note that 1 cal = 4.184 J
696,700 J = (696,700/4.184) = 166,515.3 calories = 166.5 kcal
b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 kcal and Elaine expended 48 kcal, who had the smaller net gain of calories after snacking and walking?
Pamela gains 117.5 kcal
And loses 43 kcal
Net gain = 117.5 - 43 = 74.5 kcal
Elaine gains 166.5 kcal
And loses 48 kcal
Net gain = 166.5 - 48 = 118.5 kcal
Pamela had the smaller net gain of calories after snacking and walking.
Hope this Helps!!!
Ampicillin is a bacteriostatic antibiotic. (Instead of killing bacteria, it inhibits their growth.) Ramon plated bacteria transformed with a plasmid that confers ampicillin resistance. He plated the bacteria on Thursday and left them in a 37 oC incubator overnight. On Friday, he observed moderately sized bacterial colonies. He decide to leave them in the incubator a little longer before picking the colonies that he wanted to work with further. Unfortunately, he forgot about the plate and left it in the incubator over the weekend. On Monday, his plate had large colonies that were each surrounded by very small colonies
A) Interpret and explain Ramon’s observations.
B) Predict what he will find when he picks some large and some small colonies and follows the plasmid isolation protocol for each of the colonies.
Answer:
A) That resistance in bacteria is produced due to inactivation of ampicillin by the beta lactamase enzyme. This enzyme is expressed by the bla gene found in the plasmid. This enzyme is secreted into the culture medium, thereby inactivating ampicillin. Thanks to this inactivation, the bacteria colonies will be able to develop. After a day of incubation, only those bacteria that took the plasmid that gives them resistance to ampicillin will grow after transformation. After prolonged incubation, two types of colonies can be observed in the culture medium. One large colony with ampicillin resistance, and another small colony, both of which are sensitive to ampicillin.
B) Large colonies are characterized by being resistant to ampicillin. When Ramón isolates the plasmid, he will have the gene that provides resistance to antibiotics. Said plasmid can be used again on those bacteria that are sensitive to ampicillin.
On the other hand, satellite colonies are sensitive to ampicillin. These types of colonies do not have the plasmid that contains the gene that gives ampicillin resistance. It is not possible to isolate any plasmids from these satellite colonies. These satellite bacteria will not be able to grow if they are transferred to a plate containing fresh ampicillin, while large colonies, which possess the plasmid that gives them resistance to ampicillin, will be able to grow on that plate.
Explanation:
You have been asked to prove that the Oil Spill Eater speeds up the reaction as you found above. Describe the basic set up of a laboratory experiment to measure the effect of enzyme on an oil spill in the ocean. Select all necessary test reactions. Group of answer choices Run a test reaction of crude oil with ocean water over time without Oil Spill Eater present. Run a test reaction of crude oil at a cool temperature with Oil Spill Eater present Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present. Run a test reaction of crude oil at a warm temperature with Oil Spill Eater present Run a test reaction of crude oil at a warm temperature without Oil Spill Eater present Run a test reaction of crude oil at a cool temperature without Oil Spill Eater present
Answer:
The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present
Explanation:
In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.
We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.
A typical refrigerator is kept at 4˚C, and a soda can has a pressure of
1.18 atm. The inside of a car can reach up to 60˚C (140˚F) when it is left in
direct sunlight on a hot day. If you left the soda can in the car, what would
be the new pressure if it reached 60˚C?
Answer: The new pressure will be 1.42 atm
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=1.18atm\\T_1=4^0C=(4+273)K=277K\\P_2=?\\T_2=60^0C=(60+273)K=333K[/tex]
Putting values in above equation, we get:
[tex]\frac{1.18}{277}=\frac{P_2}{333}\\\\P_2=1.42[/tex]
Hence, the new pressure will be 1.42 atm
A salt bridge: Question 3 options: provide a pathway through which the electrons travel from the cathode to the anode. provide a pathway through which the electrons travel from the anode to the cathode. provide a source of counterions to prevent the build-up of charge at both the cathode to the anode. provide a physical connection between the two cells, allowing the solutions in both cells to slowly mix.
Answer:
The correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
Explanation:
A salt bridge is a U-shaped glass tube that is used in a voltaic cell or galvanic cell to connect the oxidation and reduction half-cells and complete the electric circuit.
It allows the ions to pass through it, thus preventing the accumulation of charge on the anode and cathode as the chemical reaction proceeds.
Therefore, the correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
The student realizes that the precipitate was not completely dried and claims that as a result, the calculated Na2CO3 molarity is too low. Do you agree with the student’s claim? Justify your answer
Answer:I disagree, the molarity is meant to be higher if all other factors remain constant.
Explanation:
Molarity=amount of substance/volume of the liquid
So,the lesser,the volume, the higher the molarity and vice versa.
The molarity calculation of Na2CO3 could indeed be lower if the precipitate was not adequately dried. Excess moisture adds to the mass, misrepresenting the actual quantity of Na2CO3, which may lead to a lower calculated molarity.
Explanation:In the field of chemistry, the student's claim about the molarity of Na2CO3 (sodium carbonate) being inaccurately low due to incomplete drying of the precipitate can be valid. As the drying process helps in removing water molecules from the precipitate, any remaining moisture can add extra mass, causing the quantity of pure Na2CO3 to be undervalued. While calculating the molarity, this would imply more volume per mole, hence a lower molarity. This is a common issue in stoichiometry where proper handling and processing of chemical substances significantly affect the outcome of calculations.
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What is the Kelvin temperature of the air in a tire if the pressure inside the tire is 188 kPa at 32°C. While driving under perilous road conditions on a hot road; the pressure in the tire has increased to 225 kPa.
Answer:
414KExplanation:
The question is not completely clear because some missing parts or grammar (syntax) errors.
Interpreting it, the question is what is the Kelvin temperatue of the air in a tire, when the pressure in the tire has increased to 225 kPa, if the initial conditions of the air inside the tire were 188kPa of pressure and a temperature of 32ºC.
To solve this, you can assume constant volume and use the law of Gay-Lussac for ideal gases:
[tex]\dfrac{P_1}{T_{1}}=\dfrac{P_2}{T_2}[/tex]
That equation works with absolute temperatures, i.e. Kelvin.
32ºC = 32 + 273.15 K = 305.15KThen solve for T₂, substitute and compute:
[tex]T_2=P_2\times \dfrac{T_1}{P_{1}}=225kPa\times \dfrac{305.15K}{188kPa}[/tex]
[tex]T_2=413.90K\approx 414K\longleftarrow answer[/tex]
Be sure to answer all parts. Use MO diagrams to place C2−, C2, C2+ in order of the following properties: (a) increasing bond energy C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2 (b) increasing bond length C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2
Answer:
Bond Order = ½ [# bonded e--# antibonded e-]
C2= ½ [6-2] = 2
C2+= ½ [5-2] = 1.5
C2- = ½ [7-2] = 2.5
Bond Length: C2+> C2> C2-
Bond Energy: C2-> C2> C2+
Explanation:
Bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule.
How to calculate bond order
1. Draw the Lewis structure.
2. Count the total number of bonds.
3. Count the number of bond groups between individual atoms.
4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.
Bond energy (E) is defined as the amount of energy required to break apart a mole of molecules into its component atoms. It is a measure of the strength of a chemical bond.
The higher the bond order, the shorter the bond length and the greater the bond energy.
The molecular orbital configuration of a chemical specie shows the arrangement of electrons in such a specie into molecular orbitals in accordance with the Aufbau principle.
For C2;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2
Bond order = 1/2(8 - 4) = 2
For C2^-;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2 σ2px1
Bond order = 1/2(9 - 4) = 2.5
For C2^+;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz1
Bond order = 1/2(7 - 4) = 1.5
We must recall that the higher the bond order, the shorter the bond length and the greater the bond energy. Hence;
a) In order of increasing bond energy; C2^+ < C2 < C2^-
b) In order of increasing bond length; C2^- < C2 < C2^+
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When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the skin. This makes it possible for the metal of the electrode to form metallic ions that move into the electrolyte gel. Often, after prolonged use, this electrolyte gel begins to dry out and change the characteristic of the electrodes. Draw an equivalent circuit for the electrode while the electrolyte gel is fresh. Then discuss and illustrate the way you expect this equivalent circuit to change as the electrolyte gel dries out. In the extreme case where there is no electrolyte gel left, what does the equivalent circuit of the electrode look like
Answer:
The equivalent circuit for the electrode while the electrolyte gel is fresh
From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin
Now as the electrolyte gel start to dry out the resistance [tex]R_s[/tex] of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out the resistance of the gel [tex]R_s[/tex] then increases to infinity and this in turn cut off flow of current.
The diagram illustrating this is shown on the second uploaded image
Explanation:
Write a balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Use the pull-down boxes to specify states such as (aq) or (s).
The solubility of Cu(OH)2(s) increases in the presence of a strong acid due to the dissolution of the salt and formation of Cu2+ ions and water. The balanced net ionic equation for this process is Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l). The equilibrium constant (K) for the reaction can be calculated using the concentrations of the reactants and products at equilibrium.
Explanation:The balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid is:
Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)
Cu(OH)2 is a sparingly soluble salt, meaning it has low solubility in water. However, in the presence of a strong acid, the concentration of H+ ions increases, leading to the dissolution of Cu(OH)2 and the formation of Cu2+ ions and water.
The equilibrium constant for this reaction can be calculated by using the concentrations of the reactants and products at equilibrium. The equation for calculating the equilibrium constant (K) is:
K = [Cu2+][H2O2]/[Cu(OH)2][H+]2
Note: The square brackets indicate the concentrations of the species in the solution.
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