Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
Consider the following reaction: 2Fe2+(aq) + Cu2+ --> 2Fe3+(aq) + Cu. When the ion concentrations change to the point where the reaction comes to equilibrium, what would be the cell voltage?
Answer:
At equilibrium, the cell voltage is zero volts.
Explanation:
During an electrochemical reaction, electrical energy is produced. The reaction continues to produce electrical energy until a point is reached in which the reaction attains equilibrium.
Before the reaction attains equilibrium, the cell voltage continues to decrease progressively as the reaction progresses. At equilibrium, the cell voltage becomes zero and the read out voltmeter records 0 V.
Hence, at equilibrium, the cell voltage is zero volts.
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
Answer:
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Explanation:
Given that:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.
If the radiation wavelength is 12.5 cm,
density of water = 1g/cm³
volume of the container = 0.250 L = 250 cm³
density = mass/volume
mass of the water = density × volume
mass of the water = 1g/cm³ × 250 cm³
mass of the water = 250 g
specific heat capacity of water = 4.182 J/g°C
The change in temperature was from 20.0° C to 99° C
ΔT =( 99 -20.0)° C
ΔT = 79.0° C
The heat absorbed in the process is calculated by using the formula,
q = mcΔT
q = 250 g × 4.182 J/g°C × 79.0° C
q = 82594.5 Joules
Recall that the radiation wavelength λ = 12.5 cm = 0.125 m
The amount of energy of one photon of the radiation wavelength is determined by using the formula:
E = hv
since v = c/λ
E = hc/λ
where;
h = Planck's constant = 6.626 × 10⁻³⁴ J.s
c = velocity of light = 3.0 × 10⁸ m/s
∴
E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m
E = 1.59024⁻²⁴ Joules
The total photons required for this radiation = total heat energy/energy of radiation
The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.
Answer:
466 torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperatureStep 2: Calculate the final pressure
Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 626 torr × 5.00 L / 6.72 L
P₂ = 466 torr
When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.
Explanation:
An atom undergoes alpha decay by losing a helium atom.
So when bismuth undergoes alpha decay, we have;
²¹⁰₈₃Bi --> ⁴₂He + X
Mass number;
210 = 4 + x
x = 206
Atomic number;
83 = 2 + x
x = 81
The element is Thallium. The symbol is Ti.
For the second part;
X --> ⁴₂He + ²³⁴₉₀Th
Mass number;
x = 4 + 234 = 238
Atomic Number;
x = 2 + 90 = 92
The balanced nuclear equation is;
²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th
Which of the following pieces of information is given in a half-reaction?
O A. The number of electrons transferred in the reaction
B. The compounds that the atoms in the reaction came from
C. The state symbol of each compound in the reaction
D. The spectator ions that are involved in the reaction
Answer:
The number of electrons transferred in the reaction
Explanation:
Answer:
A
Explanation:
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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Write a balanced nuclear equation for the following: The nuclide thorium-234 undergoes beta decay to form protactinium-234 .
Answer:
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e
Explanation:
thorium-234 = ²³⁴₉₀Th
beta decay = ⁰₋₁e
protactinium-234 = ²³⁴₉₁Pa
The balanced nuclear equation is given as;
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e
A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
Which of the following forms a molecular solid? Which of the following forms a molecular solid? C10H22 CaO gold C, graphite
Answer:
C10H22
Explanation:
Graphite is known as an allotrope of carbon. Its characteristics include being black and slippery and as used as lubricants.
Gold (Au) is an element on the periodic table with atomic number 79 and a mass number 197 which exists as a metal due to its hydrogen bonds.
C10H22 which is also known as decane belongs to the Alkane family.The General forces of attraction between the alkane family are weak but in the case of decade there is Van der waal force which makes Decane C10H22 a Molecular Solid.
A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell
Answer:
The answer is D) Electrolytic cell
Explanation:
An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.
When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.
These cells are the closest thing to a galvanic battery.
Answer:
b. voltaic cell
Explanation:
Founders Education answer. had to take this quiz 4 times
Write the IUPAC and common names, if any, of the carboxylate salts produced in the reaction of each of the following carboxylic acids with NaOH: 2-methylhexanoic acid
Part A
2-bromopropanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Part B
2-methylhexanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Answer:
Following are the explanation to this question:
Explanation:
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Following are the description of the given reaction:
In reaction A:
2-Bromopropanoic acid= [tex]C_3H_5BrO_2[/tex]
[tex]C_3H_5BrO_2+NaOH[/tex]⇄ [tex]C_3H_4BrNaO_2 +H_2O[/tex]
The IUPAC name is Sodium-2-Bromopropanate
In reaction B:
2-Methylhexanoic acid= [tex]C_7H_{14}O_2[/tex]
[tex]C_7H_{14}O_2+NaOH[/tex]⇄ [tex]C_7H_{13}NaO_{2}+H_2O[/tex]
The IUPAC name is Sodium-2-Methyl hexanoate
The IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the compound carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.
Thus, the IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
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2. Which one is the odd one
out and why?
o Water
• Hydrogen
Chlorine
o Aluminum
Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but addition of (SO4)2- ion resulted in a precipitate. Which cation is present
Answer:
We can have: Calcium, strontium, or barium
Explanation:
In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:
Sulfate
All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.
Chloride
All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.
If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:
"Calcium, strontium, or barium".
I hope it helps!
An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Thats all i know
g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined
Answer:
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
Explanation:
Yes! A reactiin occurs between barium hydroxide and auminium sulphate.
barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.
The reaction is given by the equation below;
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.25 kJ·mol−1 at 2000. K and −63.12 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature.
Answer:
[tex]K^{2000K}=0.774\\\\K^{3000K}=12.56[/tex]
Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:
[tex]\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}[/tex]
Thus, at 2000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol[/tex]
And at 3000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol[/tex]
Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]K=exp(-\frac{\Delta _rG}{RT} )[/tex]
Thus, at each temperature we obtain:
[tex]K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56[/tex]
In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
Best regards.
3. Identify the reagents you would use to convert 1-bromopentane into each of the following compounds: (a) Pentanoic acid (b) Hexanoic acid (c) Pentanoyl chloride (d) Hexanamide (e) Pentanamide (f) Ethyl hexanoate
Answer:
Explanation:
a )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH
b )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .
c )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl chloride ) ⇒ CH₃CH₂CH₂CH₂COCl
d )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂
e )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒
CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )
Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O
Answer:
0.423 m.
Explanation:
The following data were obtained from the question:
Mass of glycine (NH2CH2COOH) = 141.5 g
Mass of water = 4.456 kg
Molality =.?
Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.
This is illustrated below:
Mass of glycine (NH2CH2COOH) = 141.5 g
Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol
Mole of glycine (NH2CH2COOH) =.?
Mole = mass /Molar mass
Mole of glycine (NH2CH2COOH) = 141.5/75
Mole of glycine (NH2CH2COOH) = 1.887 moles
Finally, we shall determine the molality of the solution as follow:
Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:
Molality = mole / mass (kg) of water
With the above formula, we can obtain the molality of the solution as follow:
Mole of glycine (NH2CH2COOH) = 1.887 moles
Mass of water = 4.456 kg
Molality =.?
Molality = mole /mass (kg) of water
Molality =1.887/4.456
Molality = 0.423 m
Therefore, the molality of the solution is 0.423 m
Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best regards.
[tex]pH = -log10[H+ ][/tex]. If [tex][H^+][/tex] is [tex]1.2 * 10^-^6[/tex], what is the [tex]pH[/tex]? (F5 to refresh page if you can't see it)
whts the ph of po4 9.78
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b. Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) Trypsin
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) Chymotrypsin
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.
Determine whether the following statement about equilibrium is true or false.
(a) When a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants.
(b) When a system is at equilibrium, Keq = 1.
(c) At equilibrium, the rates of the forward reaction and the reverse reaction are equal.
(d) Adding a catalyst to a reaction system will shift the position of equilibrium to the right so there are more products at equilibrium than if there was no catalyst present.
Answer:
(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants
Determining whether the statements about equilibrium is True or False
A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE
B) When a system is at equilibrium, Keq = 1 : TRUE
C) The rates of the forward reaction and the reverse reaction are equal at equilibrium : TRUE
D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE
Reaction at equilibriumIn a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.
A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).
Hence we can conclude that the answers to your questions are as listed above.
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What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
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. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
