Answer:
Anaphase
Explanation:
The centromere splits during the anaphase of the cell division. Thus, allowing the two linked chromatids to separate.
A typical chromosome is made up of two sister chromatids joined together by a structure known as the centromere. During cell division - at the metaphase stage - the chromosomes align at the equator of the cell, forming the metaphase plate. The spindle from the opposing ends of the cell engages each chromosome at the kinetochore of the centromere.
At the anaphase stage, the centromere splits, leading to the separation of the sister chromatids of each chromosome. The sister chromatids of the same then start migrating in the opposite direction as a result of the shortening of the spindle fiber.
20. What volume of 0.350M KMnO4 solution must be diluted to prepare 600. mL of
0.150M KMnO4 solution?
Answer:
25.7 mL
Explanation:
Step 1: Given data
Initial volume (V₁): ?Initial concentration (C₁): 0.350 MFinal volume (V₂): 600 mLFinal concentration (C₂): 0.150 MStep 2: Calculate the volume of the initial solution
We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.150 M × 600 mL / 0.350 M
V₁ = 25.7 mL
Which one of the following compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide?
a. butanal
b. 2-methylbutanal
c. 3-methylpentanal
d. 2, 2-dimethylbutanal
The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.
What is aldol reaction?The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.
For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.
Butanal, 2-methylbutanal and 3-methylpentanal will have aplha hydrogens in it so that it takes part in the aldol reaction.Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.
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When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have the greatest effect on the pH of the solution is:______.
a. the acid.
b. the base.
c. the conjugate acid.
d. the conjugate base.
Answer:
d. the conjugate base.
Explanation:
The general reaction of a weak acid, HA, with a strong base YOH, is:
HA + YOH → A⁻ + H₂O + Y⁻
Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.
That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:
A⁻ + H₂O ⇄ HA + OH⁻
As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base
Right option:
d. the conjugate base.If we want to change a gas to its liquid state, should we add or remove energy from the gas?
Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of ______kJ/mol
Answer:
21 KJ/mol
Explanation:
For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).
Additionally, we have several energy values for each interaction present in the Newman structures:
-) Methyl-methyl gauche: 3.8 KJ/mol
-) Methyl-H eclipse: 6.0 KJ/mol
-) Methyl-methyl eclipse: 11.0 KJ/mol
-) H-H eclipse: 4.0 KJ/mol
Now, we can calculate the energy for each molecule.
Molecule A
In this molecule, we have 2 Methyl-methyl gauche interactions only, so:
(3.8x2) = 7.6 KJ/mol
Molecule B
In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:
(11)+(6)+(4) = 21 KJ/mol
Molecule C
In this molecule, we have 1 Methyl-methyl gauche interaction only, so:
3.8 KJ/mol
Molecule D
In this molecule, we have three Methyl-H eclipse interaction, so:
(6*3) = 18 KJ/mol
Molecule E
In this molecule, we have 1 Methyl-methyl gauche interaction only, so:
3.8 KJ/mol
Molecule F
In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:
(11)+(6)+(4) = 21 KJ/mol
The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).
I hope it helps!
What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
To calculate changes in concentration for a system not at equilibrium, the first step is to determine the direction the reaction will proceed. To do so, we calculate Q and compare it to the equilibrium concentration, K. We can then determine that a reaction will shift to the right if:__________
Answer:
We can then determine that a reaction will shift to the right if Q<K
Explanation:
Comparing Q with K allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium. If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant. In this way, the concentrations of the products will decrease and the concentrations of the reagents will increase. In other words, the reverse reaction is favored to achieve equilibrium. Then the system will evolve to the left (ie products will be consumed and more reagents will be formed).If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. This implies that the concentrations of the products will increase and those of the reagents will decrease. In other words, to achieve balance, direct reaction is favored. Then the reaction will shift to the right, that is, reagents will be consumed and more products will be formed.In this case, we can then determine that a reaction will shift to the right if Q<K
Atoms are indivisible spheres. 1.plum pudding model 2.Dalton model 3.Bohr model
Answer: 2. Dalton Model
Explanation:
John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made a similar statement in the past ( all matter are made up of small indivisible particle called atom).
As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid if 21.5 mL of the NaOH solution is required to neutralize the sample?
Answer:
[tex]MM_{acid}=140.1g/mol[/tex]
Explanation:
Hello,
In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:
[tex]n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}[/tex]
Thus, solving for the moles of the acid, we obtain:
[tex]n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol[/tex]
Then, by using the mass of the acid, we compute its molar mass:
[tex]MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol[/tex]
Regards.
I need to name an ionic compound containing a transition metal cation and a halogen anion. Below are the rules I should follow to write the correct name for such compound, but one of the options is incorrect: identify and select it.
a. Identify the metal and write its name first
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
c. From the charge of the anion work out the charge of cation as Roman number in parenthesis: specify this charge in the name as a Roman number in parenthesis.
d. Write the number of the anion after the name of the metal
Answer:
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
Explanation:
The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).
Among the options, the incorrect option is option B.
This is because transition metals d not have a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.
The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.
Option B use the ability to check and to work out the charges (oxidation number) of the transition metal as per the group given in the table. The problem with the keyword is transition metal.Learn more about the ionic compound containing a transition metal.
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The following reaction, catalyzed by iridium, is endothermic at 700 K: CaO(s) + CH4(g) + 2H2O (g) → CaCO3 (s) + 4H2 (g) For the reaction mixture above at equilibrium at 700 K, how would the following changes affect the total quantity of CaCO3 in the reaction mixture once equilibrium is re-established?
a. Increasing the temperature
b. Adding calcium oxide (CaO)
c. Removing methane (CH4)
d. Increasing the total volume
e. Adding iridium
Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation:
When 2 moles of NH3(g) react with N2O(g) to form N2(g) and H2O(g) according to the following equation, 880 kJ of energy are evolved. 2NH3(g) 3N2O(g)4N2(g) 3H2O(g) Is this reaction endothermic or exothermic
Answer:
Explanation:
This is a bit of a trick question.
Usually an exothermic reaction is written as
A + B - heat = C + D
The meaning of this equation is that when the bonds of the reactants break, heat has to be given away to the environment. On the left, exothermic means that heat has to be given.
The wording on this question means that heat is a product
A + B = C + D + heat.
In other words heat is given up to the environment. So this reaction is exothermic.
An aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.
Answer:
True
Explanation:
The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:
Percentage by mass = mass of substance in solution/mass of solution x 100
In this case;
mass of KBr = 4.34 grams
mass of water = 17.4 grams
mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74
Percentage by mass of KBr = 4.34/21.74 x 100
= 19.96 %
19.96 is approximately 20%.
Hence, the statement is true.
A laboratory assistant needs to prepare 217 mL of 0.246 M solution. How many grams of calcium chloride will she need
Answer:
5.92 g
Explanation:
Convert milliliters to liters.
217 mL = 0.217 L
Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.
0.217 L × 0.246 M = 0.05338 mol
Now, convert the moles to grams using the molar mass. The molar mass of calcium chloride is 110.98 g/mol.
0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g
You will need 5.92 g of calcium chloride.
Consider the following reaction at 298K.
I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-
Answer:
Eªcell > 0; n = 2
Explanation:
The reaction:
I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)
Is product favored.
A reaction that is product favored has ΔG < 0 (Spontaneous)
K > 1 (Because concentration of products is >>>> concentration reactants).
Eªcell > 0 Because reaction is spontaneous.
And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:
Eªcell > 0; n = 2Draw the major condensation product obtained by treatment of ethyl 3-methylbutanoate with sodium ethoxide in ethanol.
Answer:
ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate
Explanation:
In this case, we have a very strong base (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a carbanion.
This carbanion, can attack another ethyl 3-methyl butanoate molecule on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.
Then the ethanol can protonate the molecule generating an "OH" group and the ethoxide.
See figure 1
I hope it helps!
Calculate the moles of Iron (Fe) in 3.8 x 10^{21} atoms of Iron. Please show your work
Answer: 6.31×10⁻³ moles Fe
Explanation:
To calculate moles when given atoms, we need to use Avogadro's number.
Avogadro's number: 6.022×10²³ atoms/mol
[tex](3.8*10^2^1 atoms)*\frac{mol}{6.022*10^2^3 atoms} =6.31*10^-^3 mols[/tex]
The atoms cancel out, and we are left with moles. There are 6.31×10⁻³ moles Fe.
Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
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Predict the sign and calculate ΔS° for a reaction. Close Problem Consider the reaction H2CO(g) + O2(g)CO2(g) + H2O(l) Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . Using standard thermodynamic data (in the Chemistry References), calculate Sºrxn at 25°C. Sºrxn = J/K•mol
Answer:
[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]
Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative
Explanation:
Consider the reaction:
H2CO(g) + O2(g) --------> CO2(g) + H2O(l)
Using standard thermodynamic data;
Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol
At standard thermodynamic data
[tex]\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}[/tex]
[tex]S^0(CO_2)[/tex] = 213.79 J/mol.K
[tex]S^0(H_2O)=[/tex] 69.95 J/mol.K
[tex]S^0 ({H_2CO}) =[/tex] 218.95 J/mol.K
[tex]S^0 (O_2)[/tex] = 205.2 J/mol.K
[tex]\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K - (218.95+ 205.2) J/mol.K}[/tex]
[tex]\mathtt{S^0_{rxn} = (283.74) J/mol.K - (424.15) J/mol.K}[/tex]
[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]
Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative
Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)
Explanation:
C5H10O2 + NaOH = C2H5COONa + C2H5OH
your result are : sodium propanoate and ethanol
A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.
What is hydrolysis?Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.
The base hydrolysis of methyl butanoate is shown as,
C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH
Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).
Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.
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At standard temperature and pressure conditions, the volume of an ideal gas contained in a jar is 55.3 L. How many molecules are in the jar. This question is to be answered in scientific notation.(eg. 1.5 e5)
Answer:
1.49e24
Explanation:
Standars temperature and pressure are 273.15K and 1atm, respectively.
Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:
PV = nRT
PV / RT = n
Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.
Replacing:
PV / RT = n
1atm*55.3L / 0.082atmL/molK*273.15K = n
2.47 moles = n
Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.
As we have 2.47 moles:
2.47 mol × (6.022x10²³ molecules / 1 mole) =
1.49x10²⁴ molecules that are in the jar
In scientific notation:
1.49e24Determine whether the following statement about reaction rates is true or false. If the statement is false, select the reason why.
Increasing the temperature of a reaction system decreases the activation energy of the reaction.
A. True
B. False
Answer:
False
Explanation:
The reaction rate increases if the temperature also increases, as does the concentration of ractives or the presence of catalysts.
The reaction rate talks about how reagents are converted into products as a function of time, this process can take less or more depending on the factors to which the reaction is exposed.
The increasing temperature generates an increase in the kinetic energy of the particles, promoting their proximity and their reaction between them to be able to give the final product, so a faster reaction occurs, which is why it has promoted when the particles collide.
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.
Half-cells were made from a nickel rod dipping in nickel sulfate solution and a copper rod dipping in copper sulfate solution. The cells were combined to construct a voltaic electrochemical cell. Sketch the cell and label anode and cathode with charges, electrode material and electrolyte solutions, half-reactions and overall reaction, give direction of electron flow and movement of ions.
Answer:
Check the Attachment.
Half-reactions:
Anode: (OXIDATION) Ni --> Ni2+ + 2e-
Cathode: (REDUCTION) Cu2+ +2e- --> Cu
Overall reaction: Ni + Cu2+ --> Ni2+ + Cu
Explanation:
Overall, reaction is basically Anode + Cathode, where electrons on both sides cancel out (if not, you need to multiply the equation in a way you can cancel them out).
Hope this helps.
3. Write the following isotope in hyphenated form (e.g., "carbon-14”): Kr
a. Krypton-109
b. Krypton -37
c. Krypton -36
d. Krypton -73
Answer:
Krypton -73
Explanation:
There are 33 known isotopes of krypton (36Kr) with atomic mass numbers from 69 through 101.
Good luck!
Answer:
D. Krypton-73
Explanation:
An isotope of an element has the same atomic number and the same number of protons but a different number of neutrons and a different atomic weight. Krypton is the 36th element on the periodic table. It has 36 protons and 48 neutrons. Krypton-73 is one of 33 known isotopes of Krypton and is the only one that actually exists from the list of choices.
Hope that helps.
A chemist prepares a solution of sodium nitrate by measuring out of sodium nitrate into a volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's sodium nitrate solution. Round your answer to 3 significant digits.
Answer:
5.74M or 5.74 mol/L (to 3 sign. fig.)
Explanation:
The molar mass of NaNO3 is 85g/mol, which means that:
1 mole of NaNO3 - 85g
? moles - 122.0g
= 122/85 = 1.44 moles
Concentration in mol/L = no. of moles (moles) ÷ volume (L)
[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)
I hope the steps are clear and easy to follow.
A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature, the rate constant increases by a factor of 5.90. Calculate the higher temperature (in °C).
Answer:
Explanation:
We shall apply Arrhenius equation which is given below .
[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]
K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .
Putting the given values
[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]
[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]
T₂ = 335.27 K
= 62.27 °C
The higher temperature is 62.27°C.
Calculating the higher temperature:Given that the activation energy of the reaction is:
Eₐ = 39.5 kJ/mol
initial temperature T₁ = 25°C = 298K
Let the final temperature be T₂
The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.
According to the question: K₂/K₁ = 5.9
Now, applying the Arrhenius equation we get:
[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]
T₂ = 335.27K
T₂ = 335.27 -273
T₂ = 62.27°C
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What is Non Metal?
help me find
The element which can not loose electron easily and having electronagtive character is called non-metal it has following property-
1. it can not conduct heat and electricity
2. it is netiher ductile not malleable
3. it is not lsuturous and also not sonorous
Explanation:
a nonmetal (or non-metal) is a chemical element that mostly lacks the characteristics of a metal. Physically, a nonmetal tends to have a relatively low melting point, boiling point, and density. A nonmetal is typically brittle when solid and usually has poor thermal conductivity and electrical conductivity. Chemically, nonmetals tend to have relatively high ionization energy, electron affinity, and electronegativity. They gain or share electrons when they react with other elements and chemical compounds. Seventeen elements are generally classified as nonmetals: most are gases (hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon and radon); one is a liquid (bromine); and a few are solids (carbon, phosphorus, sulfur, selenium, and iodine). Metalloids such as boron, silicon, and germanium are sometimes counted as nonmetals.
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.
