Which attacking species would favor an E2 reaction over an SN2 reaction? Looking for explanation for this & two more.A. IB. IIC. IIID. IVE. V

The solution containing glycine and alanine, the possible dipeptide products that can form are Glycine-Glycine: Gly-Gly Glycine-Alanine Gly Ala Alanine Glycine Ala Gly Alanine-Alanine: Ala-Ala These combinations represent the possible dipeptide products using the three-letter codes for glycine Gly and alanine Ala.

Glycine and alanine can form several dipeptides such as glee Cy alanine guy ala, alanyl glycine ala gly, and diglycine glegly the different peptides that can be made from alanine & glycine are Alpha Alanyl Glycine, Beta Alanyl Glycine, Glycidyl Alpha Alanine and Glycidyl Beta Alanine.How many dipeptides are possible by the reaction of glycine and alanine? Glycine has no chiral centre. Alanine has one chiral centre. The products are alanylglycine and glycinylalanine (I not sure whether this is the correct name or not. Please excuse), and I got a total of 4 isomers for each,

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When 2-pentanone (also known as methyl propyl ketone) reacts with NaOH (sodium hydroxide), it undergoes nucleophilic addition and subsequent dehydration to form two possible products: 3-pentanol and 2-pentene. The product that contains one chiral center is 3-pentanol.

The product that contains one chiral center is formed by Aldol condensation.

1. 2-pentanone is deprotonated by NaOH, resulting in the formation of an enolate ion.
2. The enolate ion attacks another molecule of 2-pentanone at the carbonyl carbon, creating a new carbon-carbon bond.
3. A hydroxide ion from NaOH then protonates the oxygen of the newly formed alkoxide ion.
4. The product formed is a β-hydroxy ketone, which contains one chiral center.

The structure of the product containing one chiral center is:

  O
  ‖
H3C-C-CH2-CH(OH)-CH3

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4.17 grams of [tex]NiCl_2.6H_2O[/tex] should be used to prepare 500 mL of 0.0350 M [tex]NiCl_2[/tex] solution.

To calculate the mass of [tex]NiCl_2.6H_2O[/tex] required to prepare 500 mL of 0.0350 M [tex]NiCl_2[/tex]solution, we first need to calculate the moles of [tex]NiCl_2[/tex]required:

moles of [tex]NiCl_2[/tex]= volume of solution (L) x concentration of [tex]NiCl_2[/tex](M)

moles of [tex]NiCl_2[/tex]= 0.5 L x 0.0350 mol/L

moles of [tex]NiCl_2[/tex]= 0.0175 mol

[tex]NiCl_2.6H_2O[/tex] has a molar mass of 237.69 g/mol. Therefore, the mass of [tex]NiCl_2.6H_2O[/tex] required is:

mass of [tex]NiCl_2.6H_2O[/tex] = moles of [tex]NiCl_2.6H_2O[/tex] x molar mass of [tex]NiCl_2.6H_2O[/tex]

mass of [tex]NiCl_2.6H_2O[/tex] = (0.0175 mol [tex]NiCl_2[/tex]) x (237.69 g/mol [tex]NiCl_2.6H_2O[/tex])

mass of [tex]NiCl_2.6H_2O[/tex] = 4.17 g

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The brass is getting hotter as a result of energy passing from the water to it. Instead of making the brass colder, the water is transferring energy to the brass, which lowers its temperature.

The water is transferring energy to the brass, which causes the brass to gain heat while losing heat from the water, making the water hotter while the brass becomes colder.

The metal will eventually cool while the water warms up. The temperatures of the two things will eventually be equal. When this occurs, it is said that they are in thermal equilibrium with one another. The hot metal is transferring energy to the water in the interim.

Heat is transferred by a fluid, such as water or air, by convection. As the fluid (liquid or gas) flows from one place to another, it also transfers heat. A current is a term used to describe the movement of heated water or air. Heat is transferred by electromagnetic waves or radiation.

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To find the number of moles of sucrose in 250ml of 0.150m solution, multiply the concentration (0.150m) by the volume in liters (0.25L) to get 0.0375 moles.

To calculate the number of moles of sucrose dissolved in 250ml of solution with a concentration of 0.150m, you can use the formula:  moles = concentration x volume
First, convert the volume of the solution from milliliters to liters by dividing by 1000:

250ml / 1000ml/L = 0.25L
Next, plug in the values for concentration and volume:
moles = 0.150m x 0.25L
moles = 0.0375 moles
Therefore, there are 0.0375 moles of sucrose dissolved in 250ml of solution with a concentration of 0.150m.

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Answer:

hydrogen bonds

Explanation:

The concentration of SO[tex]_{2}[/tex][tex].^{-3}[/tex] in equilibrium with Ag[tex]_{2}[/tex]SO[tex]_{3}[/tex](s) and 7.30×1[tex]0^{-3}[/tex] M Ag is 2.25×1[tex]0^{-14}[/tex] M.

To find the concentration of SO[tex]_{2}[/tex][tex].^{-3}[/tex]  in equilibrium with Ag[tex]_{2}[/tex]SO[tex]_{3}[/tex](s) and 7.30×1[tex]0^{-3}[/tex] M Ag, we first need to write out the balanced equation for the dissolution of Ag[tex]_{2}[/tex]SO[tex]_{3}[/tex]:

Ag[tex]x_{2}[/tex]SO[tex]_{3}[/tex](s) ⇌ 2Ag+(aq) + SO[tex]_{2}[/tex][tex].^{-3}[/tex](aq)

The equilibrium constant expression for this reaction is:

Ksp = [Ag+]²[SO[tex]_{2}[/tex][tex].^{-3}[/tex]]

We are given the value of Ksp for Ag[tex]^{2}[/tex]SO[tex]_{3}[/tex], which is 1.5×1[tex]0^{-8}[/tex]. Using this value and the concentration of Ag+, we can solve for the concentration of SO[tex]_{2}[/tex][tex].^{-3}[/tex]  :

1.5×1[tex]0^{-8}[/tex] = (7.30×10−3)²[SO[tex]_{2}[/tex][tex].^{-3}[/tex] ]
[SO[tex]_{2}[/tex][tex].^{-3}[/tex] ] = 2.25×1[tex]0^{-14}[/tex] M

Therefore, the concentration of SO[tex]_{2}[/tex][tex].^{-3}[/tex]  in equilibrium with Ag[tex]^{2}[/tex]SO[tex]_{3}[/tex](s) and 7.30×1[tex]0^{-3}[/tex] M Ag is 2.25×1[tex]0^{-14}[/tex] M.

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a) The pH of the solution is: 2.58

b) the pH of the solution is: 11.20

c) the pH of the solution is: 5.20

(a) To solve this problem, we need to use the equilibrium expression for the ionization of proprionic acid:

C2H5COOH + H2O ⇌ C2H5COO- + H3O+

The acid dissociation constant, Ka, for proprionic acid is 1.3 x 10^-5 at 25°C. The concentration of proprionate ion, C2H5COO-, can be calculated from the concentration of potassium propionate, C2H5COOK, since potassium propionate completely dissociates in water:

C2H5COO- = C2H5COOK = 0.060 M

Now, we can set up an ICE table and use the Ka expression to solve for the pH:

C2H5COOH + H2O ⇌ C2H5COO- + H3O+

I 0.085 M 0 M 0.060 M 0

C -x +x +x +x

E 0.085-x x 0.060+x x

Ka = [C2H5COO-][H3O+]/[C2H5COOH]

1.3 x 10^-5 = (0.060+x)x/(0.085-x)

Solving for x, we get:

x = 2.61 x 10^-3 M

Therefore, the pH of the solution is:

pH = -log[H3O+]

pH = -log(2.61 x 10^-3)

pH = 2.58

(b) Trimethylamine acts as a base and reacts with water to produce hydroxide ion and trimethylammonium ion:

(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-

The equilibrium constant for this reaction, Kb, is 6.4 x 10^-5 at 25°C. The concentration of hydroxide ion, OH-, can be calculated from the concentration of trimethylammonium chloride, (CH3)3NHCl, since it completely dissociates in water:

OH- = (CH3)3N = 0.075 M

Now, we can set up an ICE table and use the Kb expression to solve for the pOH:

(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-

I 0.075 M 0 M 0 0

C -x +x +x +x

E 0.075-x x x x

Kb = [(CH3)3NH+][OH-]/[CH3)3N]

6.4 x 10^-5 = x^2/(0.075-x)

Solving for x, we get:

x = 1.57 x 10^-3 M

Therefore, the pOH of the solution is:

pOH = -log[OH-]

pOH = -log(1.57 x 10^-3)

pOH = 2.80

Finally, we can calculate the pH:

pH = 14 - pOH

pH = 14 - 2.80

pH = 11.20

(c) The solution is a buffer solution made by mixing a weak acid, acetic acid, and its conjugate base, sodium acetate. The equilibrium expression for the ionization of acetic acid is:

CH3COOH + H2O ⇌ CH3COO- + H3O+

Follow the same procedure as above.

we get pH = 5.20

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The reason why the  the reaction between metal Y and excess dilute sulfuric acid stopped even though there was solid metal Y left is because the solid was not ground into powder

The surface area of a solid reactant can have a significant effect on the rate of a chemical reaction. This is because reactions typically occur at the interface between the reactants, where they come into contact with one another.

The larger the surface area of a solid reactant, the more area is available for the other reactant(s) to come into contact with, leading to an increase in the rate of reaction.

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The new ΔH for the given equation between HCL and NAOH  is -232 kJ, rounded to the nearest ten.

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The initial reaction is:

HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l), with ΔH = -58 kJ.

The new equation is: 4HCl(aq) + 4NaOH(aq) ⟶ 4NaCl(aq) + 4H2O(l). To find the new ΔH, simply multiply the original ΔH by the number of moles of the reactants, which is 4 in this case.

New ΔH = 4 × (-58 kJ) = -232 kJ

Hence, the standard enthalpy change of the reaction is -232 kJ

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The correct answer to your question about heavy metals is: a. their specific gravity exceeds that of water by five or more times.

Heavy metals are generally defined by their high specific gravity, which is a measure of their density compared to water. While some heavy metals are essential for life in small amounts and not all of them are toxic at trace levels, it is their specific gravity that defines them as heavy metals.

Some examples of these elements are lead, mercury, arsenic. Heavy metals consisting of elevated atomic weight with a certain gravity that helps to exceed the specific gravity of H2O by 5 or more times at 4°C temperature.

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6.31 kJ of heat are required in total to melt 18.7 g of ice and raise its temperature to 4.7 °C.

The amount of heat needed to melt 18.7 g of ice (H₂O) at 0°C is given by:

q₁ = nΔHfus = (18.7 g / 18.015 g/mol) × 6.01 kJ/mol = 6.27 kJ

where ΔHfus is the heat of fusion of water, which is 6.01 kJ/mol.

The amount of heat needed to bring 18.7 g of liquid water from 0°C to 4.7°C is given by:

q₂ = nCpΔT = (18.7 g / 18.015 g/mol) × 4.184 J/(g°C) × (4.7°C) = 43.9 J

where Cp is the specific heat of water, which is 4.184 J/(g°C).

To convert J to kJ, we divide by 1000:

q₂ = 43.9 J / 1000 J/kJ = 0.0439 kJ

Therefore, the total amount of heat needed to melt 18.7 g of ice and bring it to a temperature of 4.7°C is:

q_total = q₁ + q₂ = 6.27 kJ + 0.0439 kJ = 6.31 kJ

Rounded to 3 significant digits, the answer is 6.31 kJ.

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Answer:

morals or values

Explanation:

One limitation about science is that it cannot answer questions about values or morality. Science is limited to explaining natural phenomena and exploring empirical evidence. Questions about what is right or wrong, good or bad, ethical or unethical are beyond the scope of science and fall under the domain of philosophy, ethics, and other fields.

The concentration of Ag remaining in the solution after the titration is 0 M, which is not sufficient to precipitate any silver chromate.

The balanced chemical equation for the reaction is:
2[tex]AgNO_{3}[/tex] + [tex]Na_{2}CrO_{4}[/tex]+ 2NaCl → [tex]Ag_{2}CrO_{4}[/tex] + 2[tex]NaNO_{3}[/tex] + 2NaCl

From the equation, we can see that 2 moles of [tex]AgNO_{3}[/tex]  react with 1 mole of [tex]Na_{2}CrO_{4}[/tex] and 2 moles of NaCl to form 1 mole of  [tex]Ag_{2}CrO_{4}[/tex] and 2 moles of [tex]NaNO_{3}[/tex]  and 2 moles of NaCl.

1. Calculate the moles of [tex]Cl^{-}[/tex] and [tex]Ag^{+}[/tex] ions in the solution:
Moles of [tex]Cl^{-}[/tex] = (32.00 [tex]cm^{3}[/tex]) × (0.5700 mol/L) = 18.24 mmol
Moles of [tex]Ag^{+}[/tex] = (32.00 [tex]cm^{3}[/tex]) × (0.5700 mol/L) = 18.24 mmol

2. Determine the moles of [tex]Ag^{+}[/tex] remaining after reacting with [tex]Cl^{-}[/tex]:
Since both have the same moles, all the [tex]Cl^{-}[/tex] and [tex]Ag^{+}[/tex] ions will react completely, and no [tex]Ag^{+}[/tex] will be left.

3. Check if the remaining [tex]Ag^{+}[/tex] concentration is sufficient to precipitate silver chromate:
Since there is no [tex]Ag^{+}[/tex] left in the solution, it cannot precipitate any silver chromate.

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Need 200 mL of the stock solution to prepare 2.00 L of 0.100 M CuSO4. As 200 mL is less than the total volume of stock solution available (250 mL), there is enough stock solution available to complete this task.

we need to calculate the amount of CuSO4 required.

The equation for Molarity (M) is: M = moles of solute / liters of solution

We can rearrange the equation to find the moles of solute: moles of solute = M x liters of solution

For the desired solution:
M = 0.100 M
Liters of solution = 2.00 L
Moles of solute = 0.100 M x 2.00 L = 0.200 moles CuSO4

Now we need to calculate the volume of the stock solution required to obtain 0.200 moles of CuSO4.

The equation for finding the volume of stock solution is:
Volume of stock solution = (moles of solute required / initial concentration of stock solution) x 1000 mL

For the given stock solution:
Initial concentration of stock solution = 1.00 M
Total volume of stock solution available = 250 mL = 0.250 L

Volume of stock solution = (0.200 moles / 1.00 M) x 1000 mL = 200 mL

Therefore, we need 200 mL of the stock solution to prepare 2.00 L of 0.100 M CuSO4. As 200 mL is less than the total volume of stock solution available (250 mL), there is enough stock solution available to complete this task.

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Let's use the rules of valence to get the structural formulas of the following molecules:

a) CH5N
Molecular formula: CH5N
Condensed molecular formula: CH3NH2
Lewis structure: In this structure, we have one nitrogen atom (N) bonded to three hydrogen atoms (H) and one carbon atom (C). The carbon atom is also bonded to three hydrogen atoms. The Lewis structure would look like this:
     H
     |
H - C - N - H
     |
     H

b) C2H4O2
Molecular formula: C2H4O2
Condensed molecular formula: CH3COOH
Lewis structure: In this structure, we have two carbon atoms (C), four hydrogen atoms (H), and two oxygen atoms (O). The first carbon atom is bonded to three hydrogen atoms, while the second carbon atom is bonded to an oxygen atom (which is doubly bonded) and a hydroxyl group (O-H). The Lewis structure would look like this:
H - C - C = O
   |   |
   H   O - H
   |
   H

These are the structural formulas and Lewis structures for the given molecules.

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34.96 liters of chlorine gas can be produced by the reaction of 6.0 moles of HCl in solution with excess MnO₂ at 0.80 atm and 25°C.

To find the number of liters of chlorine gas produced, we need to use the ideal gas law equation, PV = nRT.

First, we need to calculate the number of moles of chlorine gas produced by the reaction.

From the balanced chemical equation, we know that 4 moles of HCl reacts with 1 mole of Cl₂.

Therefore, 6 moles of HCl will react with (6/4) = 1.5 moles of Cl₂.

Next, we need to calculate the volume of chlorine gas produced.

We are given the pressure (0.80 atm) and temperature (25°C or 298 K).

We also know the number of moles of chlorine gas produced (1.5 moles).

Using the ideal gas law equation, PV = nRT, we can solve for volume (V).

V = (nRT)/P

V = (1.5 moles x 0.0821 L·atm/mol·K x 298 K) / 0.80 atm V = 34.96 L

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Chlorine is the element that undergoes a decrease in its oxidation state in the given reaction, as its oxidation state decreases from 0 to -1.

The reaction is as follows:

BiCl₂ + Na₂SO₄ → 2NaCl + BiSO₄

Any compound, generally ionic, contains their constituent elements with opposite oxidation states which help a compound to be formed. This implies that the oxidation states on the reactant sides includes +2 for Bi, 0 for Cl, 0 for Na, and -2 for SO₄ ions.

On the other hand, the oxidation states of the products include +1 for Na, -1 for Cl, +2 for Bi and -2 for SO₄ ions.

Now, when we have expelled out all the oxidation states of the ions involved in the chemical reaction, we can easily demonstrate which element has undergone a decrease in its oxidation number. Chlorine is reduced in this reaction as its oxidation number lowers from 0 to -1, while sodium is oxidized as its oxidation number rises from 0 to +1.

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a) There is no change in hybridization for boron as boron is not involved in any bond breakage. b) When SO₂ and 1/2 O₂ combine to generate SO₃, the sulphur changes from sp₂ to sp₃ through hybridization due to bond formation.  c) No change in hybridization occur due to no bond breaking or forming.

a. In BF₃, the underlined atom is boron. Boron is sp₂ hybridized in BF₃. When BF₃ reacts with NaF or NaBF₄, there is no hybridization change that occurs for boron.

b. In SO₂, the underlined atom is sulfur. Sulfur is sp₂ hybridized in SO₂. When SO₂ reacts with 1/2 O₂ to form SO₃, the sulfur undergoes a hybridization change from sp₂ to sp₃.

c. In ethene (CH₂=CH₂), the underlined atoms are the carbon atoms in the double bond. The carbon atoms are sp₂ hybridized in ethene. When ethene reacts with H₂ to form ethane (CH₃-CH₃), there is no hybridization change that occurs for the carbon atoms.

However, when ethene reacts with Br₂ to form CH₂Br-CH₂Br, the carbon atoms undergo a hybridization change from sp₂ to sp₃. The change of hybridization is indicated by the type of bond breakage or bond formation.

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The equilibrium constant, Kc, provides information about the relative concentrations of products and reactants at equilibrium.

If Kc is a large number, it indicates that at equilibrium the concentration of products is much greater than that of reactants, and the reaction is product-favored.

If Kc is a small number, it indicates that at equilibrium the concentration of reactants is much greater than that of products, and the reaction is reactant-favored.

In this case, the equilibrium constant Kc is very large, 3.8 × 10^14, which indicates that the concentration of products is much greater than the concentration of reactants at equilibrium. Therefore, the reaction is product-favored.

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(a) The breakpoint chlorination dose is 6 mg/L.

(b) The chlorine/ammonia weight dose ratio at breakpoint is 3:1.


1. Calculate the immediate demand of chlorine by ferrous ion: 2Fe²+ + Cl₂ → 2Fe³+ + 2Cl⁻
  1 mg/L Fe × (1 mol Cl₂ / 2 mol Fe²+) × (70.9 g Cl₂ / 1 mol Cl₂) × (1 L / 1000 mg) = 1.77 mg/L Cl₂

2. Calculate the chlorine demand by ammonia (NH₃-N) for breakpoint chlorination:
  2 mg/L NH₃-N × (5 mol Cl₂ / 1 mol NH₃-N) × (70.9 g Cl₂ / 14 g NH₃-N) × (1 L / 1000 mg) = 4.23 mg/L Cl₂

3. Calculate the breakpoint chlorination dose by adding the chlorine demand by ferrous ion and ammonia:
  1.77 mg/L Cl₂ + 4.23 mg/L Cl₂ = 6 mg/L Cl₂ (breakpoint chlorination dose)

4. Determine the chlorine/ammonia weight dose ratio at breakpoint:
  6 mg/L Cl₂ / 2 mg/L NH₃-N = 3:1 (chlorine/ammonia weight dose ratio at breakpoint)

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The length of pipe required for the air to reach sonic velocity is approximately 127.8 meters.

To solve this problem, we can use the conservation of mass, momentum, and energy equations for an adiabatic, steady, compressible flow.

First, we can find the Mach number at the pipe inlet using the formula:

M = V / a

where V is the velocity of the air and a is the speed of sound, which can be calculated using the formula:

a = sqrt(gamma * R * T)

where gamma is the ratio of specific heats for air (1.4), R is the gas constant for air (287 J/kg-K), and T is the temperature of the air.

Plugging in the given values, we get:

T = 350 K

a = sqrt(1.4 * 287 * 350) = 472.2 m/s

V = 120 m/s

M = 120 / 472.2 = 0.254

Next, we can use the conservation of momentum equation to find the pressure drop along the pipe:

dp/dx = -rho * V^2 * dA/A * (f/4)

where dp/dx is the pressure gradient, rho is the density of air, A is the cross-sectional area of the pipe, dA/A is the fractional change in area, f is the Darcy-Weisbach friction factor, and x is the length of pipe.

We can assume that the pipe is horizontal and the friction factor is 0.02, which is the given viscosity of air at 350 K. We can also assume that the cross-sectional area of the pipe remains constant, so dA/A is 0. We can rearrange the equation and integrate to get:

x = -1/(2*f) * integral(rho * V^2 / p, M, 1)

where the integral is taken from the Mach number at the pipe inlet (0.254) to the Mach number where the air leaves sonic velocity (1).

To evaluate the integral, we need to express rho * V^2 / p in terms of Mach number. We can use the isentropic relations for an ideal gas to get:

rho * V^2 / p = (gamma / M^2) * (2 / (gamma + 1))^(gamma / (gamma - 1))

Plugging this into the integral and using the given density of air at 350 K (1.185 kg/m^3), we get:

x = -1/(2*0.02) * integral((gamma / M^2) * (2 / (gamma + 1))^(gamma / (gamma - 1)), 0.254, 1)

= 127.8 m

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The enthalpy of fusion of sodium acetate is 381.3 J/g.

To calculate the enthalpy of fusion (ΔHf) of sodium acetate, we can use the equation:

q = m × ΔHf

where q is the heat absorbed or released during a phase change, m is the mass of the substance undergoing the phase change, and ΔHf is the enthalpy of fusion.

In this case, the heat released by the hand warmer is absorbed by the water in the calorimeter, causing a temperature change. We can use the equation:

q = m × c × ΔT

where q is the heat absorbed or released, m is the mass of the substance (water) absorbing or releasing the heat, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, we need to calculate the heat absorbed by the water in the calorimeter:

q = m × c × ΔT

q = 400 g × 4.18 J/g-°C × (58.2°C - 21°C)

q = 63,568.8 J

Next, we need to calculate the amount of heat released during the crystallization of the sodium acetate:

q = m × ΔHf

q = 167 g × ΔHf

Since the hand warmer is designed to maintain a constant temperature of 54°C during the crystallization process, we can assume that the temperature of the sodium acetate remains constant during the phase change, and therefore ΔT = 0.

We can now set the two equations equal to each other and solve for ΔHf:

m × c × ΔT = m × ΔHf

400 g × 4.18 J/g-°C × (58.2°C - 21°C) = 167 g × ΔHf

63,568.8 J = 167 g × ΔHf

ΔHf = 381.3 J/g

Therefore, the enthalpy of fusion of sodium acetate is 381.3 J/g.

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The mass of potassium chloride produced when potassium chlorate decomposes in this instance is 178.92 g.

The balanced chemical equation is:

2KClO3(s) → 2KCl(s) + 3O2(g)

From the equation, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, 1 mole of KClO3 produces 1 mole of KCl.

Given that 2.4 moles of KClO3 decompose, the amount of KCl produced will also be 2.4 moles.

Now, we need to find the mass of KCl produced, using the molar mass of KCl:

KCl: K = 39.10 g/mol, Cl = 35.45 g/mol

Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol

Mass of KCl produced = number of moles x molar mass

Mass of KCl produced = 2.4 moles x 74.55 g/mol = 178.92 g

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The best solvent to recrystallize compound b would be acetone as it has a boiling point lower than the melting point of the compound and it shows good solubility for b both at room temperature and at its boiling point.

To recrystallize compound B, you should choose a solvent that has the following characteristics:
1. The compound should be insoluble at room temperature (25°C) but soluble at the solvent's boiling point.
2. The solvent's boiling point should be higher than the melting point of the compound (75°C) to avoid melting the compound during the recrystallization process.
Based on the given table, the appropriate solvent to recrystallize compound B is water. This is because compound B is insoluble in water at 25°C but becomes soluble at water's boiling point (100°C). Additionally, water's boiling point is higher than compound B's melting point (75°C), making it a suitable choice for recrystallization.

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The compound that contains chlorine (Cl) with an oxidation number of 3 is Cl₂O₃. so the correct option is option C

The oxidation number of chlorine (Cl) in a compound can vary depending on the compound. In order to determine which compound contains chlorine with an oxidation number of 3, we need to first understand the rules for assigning oxidation numbers.

Oxidation numbers are assigned to each atom in a compound to indicate its degree of oxidation or reduction. The oxidation number of an atom in a compound is the charge it would have if all the bonds were ionic. The sum of the oxidation numbers in a compound is equal to the overall charge of the compound. For example, in a neutral compound, the sum of the oxidation numbers is zero.

In order for chlorine to have an oxidation number of 3, it must have lost three electrons. This means that it has been oxidized. Chlorine is a halogen and typically has an oxidation number of -1 in its compounds, except when it is combined with a more electronegative element, such as oxygen or fluorine.
Now, let's look at the options given:
a) Cl₂O - In this compound, chlorine has an oxidation number of +1.
b) ClO₃ - In this compound, chlorine has an oxidation number of +5.
c) Cl₂O₃ - In this compound, chlorine has an oxidation number of +3. This is the correct answer!
d) ClO₂ - In this compound, chlorine has an oxidation number of +4.
e) Cl₂O₅ - In this compound, chlorine has an oxidation number of +5.
Therefore, so the correct option is option C the compound that contains chlorine (Cl) with an oxidation number of 3 is Cl₂O₃.

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The correct answer is: Most of the free energy available from the oxidation of glucose is used in the production of ATP in glycolysis.

During glycolysis, which is the initial step of glucose metabolism, glucose is oxidized to pyruvate, producing a small amount of ATP and NADH in the process. The NADH is then used in the later stages of cellular respiration to generate more ATP through oxidative phosphorylation. The free energy released from the oxidation of glucose is primarily used to produce ATP in glycolysis because ATP is the primary energy currency of the cell and is required for various cellular processes. Only a small portion of the free energy is used to reduce NAD+ to NADH, resulting in the production of a limited amount of NADH during glycolysis.

This is why only two molecules of NADH are formed during this process. Most of the free energy available from the oxidation of glucose remains in pyruvate, one of the products of glycolysis, which can be further metabolized to extract more energy. Additionally, there is no [tex]CO_{2}[/tex] or water produced as products of glycolysis because these products are formed in the later stages of cellular respiration.

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Answer:

Explanation:

According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. Therefore, we can set up the following proportion to find the number of moles of chlorine in container 2:

Number of moles in container 1 / volume of container 1 = number of moles in container 2 / volume of container 2

We can plug in the given values:

Number of moles in container 1 = 7.2 mol

Volume of container 1 = 9.4 L

Volume of container 2 = 50.1 L

Number of moles in container 2 = x (unknown)

So we have:

7.2 mol / 9.4 L = x / 50.1 L

To solve for x, we can cross-multiply and simplify:

7.2 mol * 50.1 L = 9.4 L * x

x = (7.2 mol * 50.1 L) / 9.4 L

x = 38.2925 mol

Therefore, there are approximately 38.3 moles of chlorine in container 2.

The specific heat of iron is 0.45.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

It is a measure of how much energy it takes to raise the temperature of a substance. It is the amount of heat necessary to raise one mass unit of that substance by one temperature unit.

It is given by the formula -

                                                  Q = mcΔT

Given,

Q = 180.8 J

Mass = 22.44g

Initial Temperature = 21.1

Final temperature = 39

Q = mcΔT

180.8 = 22.44 × c × ( 39 - 21.1 )

c = 180.8 ÷ ( 22.44 × 16.56 )

c = 0.45

Thus , the ideal selection is option A.

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