Given the DNA template strand 3' GCATTCAAG 5', write the amino acid sequence in the N‑terminal to C‑terminal direction. Note: Enter the amino acids using their three-letter designations. Put a hyphen between each amino acid. (for example, Glu‑Asp‑Val).

Answer: [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

[tex]\Delta S[/tex] is positive when randomness increases and [tex]\Delta S[/tex] is negative when randomness decreases.

a) [tex]N_2(g)+3H_2(g)\longrightarrow 2NH_3(g)[/tex]

4 moles of gas are converting to moles of gas, hence the entropy decreases.

b) [tex]Br_2(g)\longrightarrow Br_2(l)[/tex]

1 mole of gas is converting to 1 mole of liquid, hence the entropy decreases.

c) [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex]

1 mole of gas is converting to 2 moles of gas, hence the entropy increases.

d) [tex]NaOH(aq)+CO_2(g)\longrightarrow NaHCO_3(aq)[/tex]

1 mole of gas is disappearing , hence the entropy decreases.

Thus [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex] has positive entropy change.

Answer:

(a) X electrode

(b) Y electrode

(c) Y electrode

(d) X electrode

(e) Y electrode

Explanation:

A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.

In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.

Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻

Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)

Classify the descriptions by whether they apply to the X or Y electrode.

(a) anode. Is where the oxidation takes place (X electrode).

(b) cathode. Is where the reduction takes place (Y electrode).

(c) electrons in the wire flow toward. Electrons in the wire flow toward the cathode (Y electrode).

(d) electrons in the wire flow away. Electrons in the wire flow away from the anode (X electrode).

(e) cations from salt bridge flow toward. Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.

In a galvanic cell, the more reactive metal X is the anode where electrons flow away from, and the less reactive metal Y is the cathode where electrons flow toward. Cations from the salt bridge migrate toward the anode, maintaining electrical neutrality.

In a galvanic (voltaic) cell, the generic metal X is more reactive than Y, indicating that X acts as the anode where oxidation takes place, and Y acts as the cathode where reduction occurs. The characteristics of the electrodes in a voltaic cell are as follows:

These characteristics are essential to the proper functioning of the galvanic cell, ensuring the flow of electrons from the anode to the cathode, and allowing the cell to generate electricity through spontaneous redox reactions.

Answer:

D) 2 NOCl(g) ⇄ 2 NO(g) + Cl₂(g); Kp = 1.7 × 10⁻²

Explanation:

In order to compare the degree of decomposition of these reactions, we have to compare the equilibrium constant Kp. Kp is equal to the partial pressure of the products raised to their stoichiometric coefficients divided by the partial pressure of the reactants raised to their stoichiometric coefficients. The higher the Kp, the more products and fewer reactants at equilibrium. Among these reactions, D is the one that has the highest Kp, therefore the one experiencing the largest degree of decomposition.

You may find bellow the net ionic equations.

Explanation:

Net ionic equation of triethylamine in water:

(C₂H₅)₃N (l) + H₂O (l) → (C₂H₅)₃NH⁺ (aq) + OH⁻ (aq)

where:

(l) - liquid

(aq) - aqueous

You may find in the attached picture the ionization of 2-methylpiperidine in water.

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Triethylamine and 2-methylpiperidine act as weak bases in water, accepting protons to form hydroxide ions and their respective conjugate acids. This is an example of base ionization. It was not possible to provide an exact equation for 2-methylpiperidine without a specific structure.

When triethylamine, also a weak nitrogen base like ammonia, is dissolved in water, it accepts a proton from a water molecule to form a hydroxide ion (OH-) and an ethylammonium ion ((C2H5)3NH+). This is represented by the equation: (C2H5)3N(aq) + H2O(l) ⇌ (C2H5)3NH+(aq) + OH-(aq).

Similarly, 2-methylpiperidine being a weak base, will also react with water to form hydroxide ions (OH-) and 2-methylpiperidinium ions. However, without a specific structure for 2-methylpiperdine, it's challenging to write the exact equation.

These are examples of base ionization, where a base reacts with water to produce hydroxide ions and a conjugate acid. Both triethylamine and 2-methylpiperidine act as Brønsted bases, accepting protons from water molecules.

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Explanation:

According to the first law of thermodynamics, energy can neither be created nor it can be destroyed. It can only be converted from one form to another.

Formula given by first law of thermodynamics is as follows.

               U = q + w

The given values are as follows.

       q = +563 J

        w = -498 J      (as work done by the system is negative)

Hence, putting the given values into the above formula and calculate the internal energy as follows.

                   U = q + w

                       = 563 J + (-498 J)

                       = 65 J

As internal energy is represented by [tex]\Delta U[/tex] or [tex]\Delta E[/tex]. Therefore, the value of [tex]\Delta E[/tex] for the given system is 65 J.

Answer:

D.) Rate-Determining Step

Explanation:

Rate - Determining Step -

The slowest step of any of the chemical reaction helps to find the overall rate of the chemical reaction , and , hence is known as the rate - determining step .

There are two type of reaction possible , one is elementary reaction and a complex reaction .

Elementary reaction is a single step reaction , and hence , that very step determined the rate of the reaction and hence is known as the rate determining step .

And ,

In case of a complex reaction , the reaction is preceded by many steps , and hence , the slowest step among other steps is known as the rate determining step of the reaction .

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

Answer:

greater number of occupied sublevels in the lithium atom

Explanation:

Although Hydrogen has only one electron, the atom has several different energy levels. When the electron transitions from a higher energy level to a lower one,  photons are released. The photons absorb light at different wavelengths and colors.then they appear as lines on the spectrum of the hydrogen atom. Lithium has a greater number of occupied sub-levels which means more electrons will transition through the energy levels leading to more lines appearing on its spectrum .

Answer:

0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms

Explanation:

0.20 mol Ba(OH)2 and 0.20 mol H2SO4

In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.

Number of O atoms in 0.20 moles Ba(OH)2 = 2*0.20 = 0.40 moles O atom

In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.

Number of O atoms in 0.20 moles H2SO4 = 4*0.20 = 0.80 moles O atom

⇒ 0.20 mol Ba(OH)2 and 0.20 mol H2SO4 do not have the same number of oxygen atoms.

0.20 mol Br2O and 0.20 mol HBrO

In Br2O there is 1 mol of O atoms in every mol Br2O

Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom

In HBrO there is 1 mol of O atom in every mol HBrO

Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom

⇒ in 0.20 moles Br2O and 0.20 moles HBrO we have the same number of oxygen atoms

0.20 moles of Oygen contains: = 0.20 * 6.022*10^23 = 1.2 *10^23 O atoms

0.10 mol Fe2O3 and 0.50 mol BaO

In Fe2O3 there are 3 moles of O atoms in every mol of Fe2O3

Number of O atoms in 0.10 moles Fe2O3 = 0.10 * 3 = 0.30 moles O atom

In BaO there is 1 mol of O atoms in every mol BaO

Number of O atoms in 0.50 mol BaO = 1*0.50 = 0.50 moles O atom

⇒ 0.10 mol Fe2O3 and 0.50 mol BaO do not have the same number of oxygen atoms.

0.10 mol Na2O and 0.10 mol Na2SO4

In Na2O there is 1 mol of O atoms in every mol of Na2O

Number of O atoms in 0.10 mol Na2O = 1*0.10 = 0.10 moles O atom

In Na2sO4 there are 4 moles of O atoms in every mol of Na2SO4

Number of O atoms in 0.10 mol Na2SO4 = 4*0.10 = 0.40 moles O atom

⇒ 0.10 mol Na2O and 0.10 mol Na2SO4 do not have the same number of oxygen atoms.

The pair of samples that contains the same number of oxygen atoms in each compound is 0.20 mol H2SO4 and 0.20 mol HBrO.

To determine which pair of samples contains the same number of oxygen atoms in each compound, we need to calculate the number of oxygen atoms in each sample.

Therefore, the pair of samples that contains the same number of oxygen atoms in each compound is 0.20 mol H2SO4 and 0.20 mol HBrO.

The molarity of the nitric acid solution can be calculated using the principles of stoichiometry and titration. By knowing the molarity and volume of strontium hydroxide, we can determine the moles of nitric acid due to a 1:1 stoichiometry in the balanced equation. The molarity of nitric acid is found to be 1.13 M.

The subject of the question is related to a concept in chemistry known as titration which is used to determine the concentration of an unknown solution (in this case, nitric acid) using a known solution (strontium hydroxide). Given that the volume of nitric acid is 9.96 mL and the volume and molarity of strontium hydroxide are 13.25 mL and 0.085 M respectively, we can determine the molarity of the nitric acid.

The balanced chemical equation for this acid-base reaction is HNO3(aq) + Sr(OH)2(aq) -> Sr(NO3)2(aq) + 2H2O(l). From this, we can see that one molecule of nitric acid reacts with one molecule of strontium hydroxide. Hence, we can apply the principle of stoichiometry and say that the moles of strontium hydroxide used in the titration are equal to the moles of nitric acid present in the solution.

The moles of strontium hydroxide can be calculated as Molarity x Volume (in L), which gives 0.085 mol/L * 13.25 mL = 0.01126 mol. Since the stoichiometry between strontium hydroxide and nitric acid is 1:1, this is also the moles of nitric acid in our sample. The molarity of nitric acid is then calculated as moles of solute/volume of solution in L, providing us with the molarity. Molarity = 0.01126 mol / 0.00996 L = 1.13 M.

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Explanation:

The given data is as follows.

     [tex]P_{1}[/tex] = 1 atm,           [tex]P_{2}[/tex] = 2 atm

     [tex]V_{1}[/tex] = 250 ml,        [tex]V_{2}[/tex] = 500 ml

Total volume = [tex]V_{1} + V_{2}[/tex] = 750 ml

Therefore, total pressure will be as follows.

           [tex]P_{total} V_{total} = P_{1}V_{1} + P_{2}V_{2}[/tex]

       [tex]P_{total} = \frac{P_{1}V_{1} + P_{2}V_{2}}{V_{total}}[/tex]

                 = [tex]\frac{1 atm \times 250 ml + 2 atm \times 500 ml}{750 ml}[/tex]

                = 1.66 atm

Thus, we can conclude that total pressure of the given mixture is 1.66 atm.

Intramolecular forces describe those forces that are found within the molecules and are shared by the electronic bonds. They hold the atoms and molecules together.

Hence the option C is correct.

Learn more about the statement about intermolecular forces is true.

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Answer: a)  [tex]68.3\times 10^6cal[/tex]

b) [tex]log_{10}(68.3\times 10^6cal)=7.83[/tex]

Explanation:

Heat is defined as a spontaneous flow of energy from one object to another. It is measured in Joules, calories, kilo Joules etc.

These units of energy are inter convertible.

We are given:

a) Energy absorbed by limestone = [tex]2.97\times 10^6kJ[/tex]

Converting this unit of temperature into [tex]calories[/tex] by using conversion factor:

1 kJ = 239.006 calories

[tex]2.97\times 10^6kJ=\frac{239.006}{1}\times 2.97\times 10^6=68.3\times 10^6cal[/tex]

Thus the energy in calories is [tex]68.3\times 10^6cal[/tex]

b) The value of log base 10 of [tex]68.3\times 10^6cal[/tex]  is:

[tex]log_{10}(68.3\times 10^6cal)=7.83[/tex]

Final answer:

The concentration of silver in the unknown solution can be determined using the Nernst equation with the given cell potential of +0.045 V. By rearranging and substituting known values into the Nernst equation, we can calculate the unknown silver ion concentration at the anode.

Explanation:

To determine the concentration of silver in the unknown AgNO3 solution using a concentration cell, we can apply the Nernst equation. The reaction in a concentration cell is Ag+ (aq) → Ag+ (aq), which proceeds with a transfer of electrons from one compartment to the other. Since there is a potential difference (E) of +0.045 V, we can calculate the unknown concentration at the anode using the Nernst equation:

E = E°_cell - (RT/nF) ln(Q)

Where:

Rearranging the Nernst equation and solving for Q will provide us the ratio of the Ag+ concentration in the anode to that in the cathode. By substituting the measured E value, R, T, n, and F into the Nernst equation and knowing that [Ag+] cathode is 1.0 M, we can solve for the unknown [Ag+] anode concentration. Finally, by taking the anti-logarithm of the result, we can find the concentration of silver in the unknown solution.

The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed. This is indicative of an exothermic reaction.

In the given reaction, CO + 2 H2 ---> CH3OH ΔH < 0, this is an exothermic reaction as indicated by the negative delta H value. The negative ΔH value signifies that the total energy released when the new bonds are formed in the product (CH3OH) is more than the total energy absorbed to break the bonds in the reactants (CO and H2).

Therefore, the correct statement is: The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed.

This corresponds to the fact that in an exothermic reaction, the products are at a lower energy level than the reactants, therefore, energy is released in the reaction.

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The correct statement is that the energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed, indicating an exothermic reaction.

The correct statement is: The energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed.

In this specific reaction, energy is required to break a C-O triple bond and two H-H single bonds in the reactants. Energy is then released as three C-H single bonds, a C-O single bond, and an O-H single bond are formed in the product, CH₃OH.

Since ΔH is negative, it indicates that the total energy released in bond formation is greater than the total energy absorbed in bond breaking, reaffirming that the reaction is exothermic.

Answer:

The student has to weigh 7.9 ±0.1 grams

Explanation:

Density of acetone : 0.790 g/cm³

If the student has to measure 10.0 mL of acetone, he can weigh an specific mass.

Density of acetone = Mass of acetone / Volume of acetone

cm³ = mL

0.790 g/cm³ = Mass of acetone / 10 cm³

0.790 g/cm³  . 10cm³ = 7.9 g

Final answer:

To find the mass of acetone needed for the experiment, the student multiplies the density of acetone (0.790 g/mL) by the required volume (10.0 mL) to get 7.90 g, considering significant figures.

Explanation:

The student needs to calculate the mass of acetone required for their experiment. Given that the density of acetone is 0.790 g/cm³ (which is equivalent to 0.790 g/mL since 1 cm³ is equal to 1 mL), and the volume needed is 10.0 mL, they can use the formula density = mass/volume to find the mass.

To find the mass, the student should rearrange the formula to mass = density × volume. Using the given values, the calculation is as follows:

mass = 0.790 g/mL × 10.0 mL = 7.90 g

Therefore, the student should weigh out 7.90 g of acetone for their experiment, making sure to use the correct number of significant digits as provided in the question.

Answer:

The specific heat of copper is 0.37 J/g°C

Explanation:

Step 1: Data given

Mass of copper = 19.5 grams

Initial temperature of copper = 98.27 °C

Mass of water = 76.3 grams

Initial temperature of water = 24.05 °C

Final temperature of water and copper = 25.69 °C

Step 2: Calculate specific heat of copper

Qgained = -Qlost

Q = m*c*ΔT

Qwater = -Qcopper

m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)

⇒ with m(water) = 76.3 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64

⇒ with m(copper) = 19.5 grams

⇒ with c(copper) = TO BE DETERMINED

⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58

76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58

523.552 = 1415.31 * c(copper)

c(copper) = 0.37 J/g°C

The specific heat of copper is 0.37 J/g°C

Answer:

0.210 M

Explanation:

A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.

In order to find out the resulting concentration (C₂) we will use the dilution rule.

C₁ × V₁ = C₂ × V₂

1.70 M × 75.0 mL = C₂ × 278 mL

C₂ = 0.459 M

A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.

Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.

C₁ × V₁ = C₂ × V₂

0.459 M × 139 mL = C₂ × 304 mL

C₂ = 0.210 M

Answer:

The correct answer is 0.21 M

Explanation:

We have an initial solution of concentration 1.70 M and we dilute it twice. In each dilution we can calculate the final concentration (Cf) from the initial concentration (Ci) and the final and initial volumes respectively (Vf and Vi) as follows:

Cf x Vf= Ci x Vi

Cf= Ci x Vi/ Vf

In the first dilution, Ci is 1.70 M, the initial volume we take is 75 ml and the final volume is 278 ml.

Cf= 1.70 M x 75 ml / 278 ml = 0.46 M

In the second dilution, the initial concentration is the previously obtained (Ci= 0.46 M), the initial volume is Vi= 139 ml and the final volume is the addition of 139 ml and 165 ml (because we add 165 ml to 139 ml). Thus:

Cf= (0.46 M x 139 ml)/ (139 ml + 165 ml) = 0.21 M

Answer:

(d) All spontaneous reactions have a negative free-energy change

Explanation:

All spontaneous reactions releases free energy which can be used later for the work to be done. A reaction with a negative value for ΔG releases free energy and is thus spontaneous

Among the options listed, the correct statement is that all spontaneous reactions result in a negative free-energy change, reflecting the energy released during the process. So the correct option is d.

The true statement among the options provided is:

(d) All spontaneous reactions have a negative free-energy change.

This is based on the principles of thermodynamics. The second law of thermodynamics states that the entropy of the universe increases in the course of spontaneous reactions. Nevertheless, a characteristic of spontaneous processes is that they release free energy, leading to a negative change in Gibbs free energy (ΔG). Therefore, not all spontaneous reactions have a negative enthalpy change (ΔH), nor do they always have a positive entropy change (ΔS), but they must result in a decrease in Gibbs free energy (negative free-energy change).

Answer:

The correct answer is B the total bond energies of the product is greater than the total bond energies of the reactants.

Explanation:

3H2+N2=2NH3

This reaction is an endothermic reaction that means that reaction consume heat energy.

   Due the above reaction is endothermic the enthalpy change of the reaction is negative(-ve).

  During endothermic reaction the total bond energies of the product is greater than the total bond energies of the reactants.

Answer:

114 kPa

Explanation:

By Bernoulli's equation when a fluid flows steadily through a pipe:

P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.

By the continuity equation, the liquid flow must be constant in the pipe, and then:

A1*v1 = A2*v2

Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

0.007v2 = 0.027

v2 = 3.9 m/s

Then:

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1

ρ*g*Δy + v1² - v2² = ΔP

ΔP = 1290*9.8*9.01 + 9.91² - 3.9²

ΔP = 113,987.42 Pa

ΔP = 114 kPa

Calculate the total energy for individual process, including energy to vaporize Ca(s), first and second ionization energies of Ca, bond energy of Cl2 and electron affinity of Cl. The lattice energy is the result of subtracting standard heat of formation of CaCl2 from the total energy calculated, which equals to 2635.9 kJ.

The lattice energy for CaCl2 can be calculated using the given information and by understanding that the process of forming a salt like CaCl2 involves several different energies. First, there is the energy required to vaporize Ca(s), then the energy for first and second ionization of the calcium atom, followed by the electron affinity for Cl and the bond energy of Cl2.

To start calculating, first add the energy to vaporize Ca(s), the first and second ionization energies for Ca: 192kJ + 589.5kJ + 1146kJ = 1927.5kJ. When 2 Cl react with Ca, this process includes the bond energy for Cl2 and twice the electron energy for Cl: 242.6kJ + 2 * -348kJ = -453.4kJ. To calculate the energy required to form lattice, subtract the standard heat of formation of CaCl2 from the total energy calculated: 1927.5kJ - (-453.4kJ + -795kJ) = 2635.9 kJ

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Answer:

0,051g of O₂ are consumed

Explanation:

The reaction of precipitation of Fe(II) is:

4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

The moles of Fe(II) you have in 85ml of 0.075 M Fe(II) are:

0,085L×0,075M = 6,4x10⁻³ moles of Fe(II)

For the reaction, you require 4 moles of Fe(II) and 1 mol of O₂(g) to precipitate the iron. Thus, moles of O₂(g) you require are:

6,4x10⁻³ moles of Fe(II)×[tex]\frac{1molO_2}{4molesFe(II)}[/tex] = 1,6x10⁻³ moles of O₂(g)

These moles are:

1,6x10⁻³ moles of O₂(g)×[tex]\frac{32g}{1mol}[/tex] = 0,051g of O₂ are consumed

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Answer:

Yogurt displays characteristics of an acid.

Explanation:

The characteristics of an Acid are as follows:

Only Yogurt displays one of these properties ie. sour taste, therefore it is the only material having characteristic of acid.

Bittery taste and slippery touch are characteristics of bases.

Answer:

its c

Explanation:

Answer:

5.28 %

Explanation:

To solve this problem we are going to calculate the moles of NaOH required to titrate the vinegar. Since this reaction is one mole NaOH to 1 mol acetic acid we will have the moles of acetic acid  present  in the 5 mL sample . By multiplying the number of moles of acetic acid by its molecular weight we obtain the grams of acid.

Now since we are being asked the w/w % we need the weight of solution which is not given in the problem. However we can assume that its value will be close to that of pure water and give it the value of 5 g since the density of water is one g/mL.

moles NaOH = 44.5 mL x 1L/1000 mL x 0.098 mol/L = 0.00440

mol acetic acid = 0.00440

mass acetic acid = 0.00440 mol x 60.04 g/mol = 0.264 g acetic acid

percent (w/w) = (0.264 g/ 5 g  ) x 100 = 5.28 %

Answer:

Reaction A, B, E and F are exothermic reactions.

Explanation:

Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system.

The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system.

The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be positive.

So, from the given option reaction which are exothermic are with negative value [tex](\Delta H)[/tex] that enthalpy of reaction and those are:

1) [tex]2Mg(s) + O_2( g )\rightarrow 2MgO (s)[/tex],ΔH = -1203 kJ/mol

2)[tex]NH_3 (g) + HCl (g)\rightarrow NH_4Cl (s)[/tex],ΔH =-176 kJ/mol

3) [tex]C(graphite) + O_2 (g)\rightarrow CO_2 (g) [/tex],ΔH = -393.5 kJ/mol

4) [tex]CH_4 (g) + 2O2 (g)\rightarrow CO_2 (g) + 2H_2O (l) [/tex],ΔH =-891 kJ/mol

Reactions A, B, and E are exothermic because they have negative ΔH values, indicating that energy is released during these reactions.

Exothermic reactions are characterized by the release of heat, which is indicated by a negative change in enthalpy (ΔH). In the provided reactions, reactions A (2Mg(s) + O2( g ) -----> 2MgO(s) ΔH = -1203 kJ/mol), B (NH3(g) + HCl(g) -----> NH4Cl (s) ΔH = -176 kJ/mol), and E (C(graphite) + O2(g) -----> CO2(g) ΔH = -393.5 kJ/mol) are exothermic because they have negative ΔH values, indicating that energy is released during these reactions. Conversely, reactions C and D are endothermic (they absorb heat), as indicated by their positive ΔH values.

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Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

Final answer:

The question requests the calculation of the change in standard molar entropy for a specific chemical reaction, but specific entropy values are not provided, making it impossible to complete the calculation as asked.

Explanation:

The question asks to calculate the change in standard molar entropy (ΔS°) for the reaction of ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O) at 298.0 K and standard pressure. The formula to use is ΔS° = ∑S°(products) - ∑S°(reactants). Unfortunately, the standard molar entropies (S°) for the reactants and products are not provided in the question. Normally, these values would be found in a standard thermodynamic table. Once you have these values, you would multiply the entropy of each substance by its coefficient in the balanced equation, sum these values for both reactants and products, and then subtract the sum for reactants from the sum for products. This will give you the change in entropy for the reaction.

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

Step 1: Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

Step 2: Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+  in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed  in the back titration with Ca2+:

Step 3: Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

Step 4: Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

Step 5: Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction  with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

Step 6: Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

Step 7: Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Answer:

The mass percent = 15.79 %

Explanation:

Step 1: Data mass

Mass KCl = 30 grams

volume of water = 160 mL

density of water = 1.00 g/mL

Step 2: Calculate mass of water

mass water = volume * density of water

mass water = 160 mL  / 1.00 g/mL

mass = 160 grams

Step 3: Calculate total mass of solution

30 grams + 160 grams = 190 grams

Step 4: Calculate mass % :mass KCl / Mass of solution

% mass = (30 grams / 190 gram) *100%

% mass = 15.79 %

The mass percent = 15.79 %

Answer:

The mean free path is [tex]0.0000373631 m[/tex]

Explanation:

The formula for mean free path is :

λ = [tex]\frac{V}{\sqrt{2}\pi d^{2}N  }[/tex]

where,

λ - is the mean free path distance

V - volume of the gas

d - the diameter of the molecule

N - number of molecules.

now ,

density [tex]D[/tex] = [tex]\frac{mass}{volume}[/tex] = [tex]\frac{M}{V}[/tex] ;

mass of the gas = (number of molecules)[tex]*[/tex](mass of one molecule) ;

as it's atomic hydrogen

[tex]M = N*m \\m=1.66*10^{-24}\\M=N*1.66*10^{-24}[/tex]

∴  

[tex]D[/tex] = [tex]\frac{N*1.66*10^{-24}}{V}[/tex]

∴

[tex]\frac{V}{N*1.66*10^{-24}} = \frac{1}{ D}[/tex]

⇒ λ = [tex]\frac{1}{\sqrt{2}\pi d^{2}D  }[/tex]

⇒ λ = [tex]\frac{1.66*10^{-24}}{\sqrt{2}\pi (100*10^{-12})^{2}*1    }[/tex]

⇒ λ = [tex]0.0000373631 m[/tex]