During which phase do the centromeres split, allowing the two linked chromatids to separate?

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

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.

Answer:

25.7 mL

Explanation:

Step 1: Given data

Step 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

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.

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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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:

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.

Learn more about the ionic compound containing a transition metal.

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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.

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.

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.

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.

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.

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.

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.

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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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.

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.

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:

The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.

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.

Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.

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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:

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.

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!

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.

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.

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.

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.

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.

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!

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:

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:

In this case, we can then determine that a reaction will shift to the right if Q<K

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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