Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known in popular culture as truth serum. It is synthesized like other barbiturates, but uses thiourea, (H2N)2C=S, in place of urea. The mechanism involves the following steps:
1. Ethoxide ion deprotonates malonic ester, forming enolate anion 1;
2. Enolate anion 1 acts as a nucleophile in an SN2 reaction with ethyl bromide, forming alkylated intermediate 2;
3. Ethoxide ion deprotonates alkylated intermediate 2, forming enolate anion 3;
4. Enolate anion 3 acts as a nucleophile in an SN2 reaction with 2-bromopentane, forming alkylated intermediate 4;
5. Alkylated intermediate 4 reacts with thiourea to form tetrahedral intermediate 5;
6. Tetrahedral intermediate 5 collapses, expelling ethoxide ion and forming intermediate 6;
7. Intermediate 6 reacts with sodium hydroxide to form sodium pentothal.
Draw the mechanism out on a separate sheet of paper and then draw the structure of enolate anion 1. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

Answer:

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K →  23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K  . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

Answer and Explanation:

Given the following chemical equation:

Cl₂(g) + F₂(g) ⇒ 2ClF(g)

The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:

On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).

On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).

Answer:

the amount of grams of dipotassium succinate trihydrate  K₂C₄H₄O₄·3H₂O = 41.798g

Explanation:

Given that:

The volume of  [tex]K_2C_4H_4O_4.3H_2O[/tex] = 660.mL

The molarity of succinic acid = 0.0577 M

The pH of the solution = 5.869

The pKa₁ = 4.207

The pKa₂ = 5.636

The pKa₂ is required to be used for the determination of given that succinic acid is dissociated twice.

Using Henderson-HasselBalch Equation,

[tex]PH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]

we know that :

The volume of  [tex]K_2C_4H_4O_4[/tex] = 660.mL

The molarity of succinic acid = 0.0577 M

number of moles = Molarity [tex]\times[/tex] Volume (liters)

number of moles =  0.0577 [tex]\times[/tex]  0.660 L

number of moles =  0.038082 mol

The balanced chemical reaction for this equation can be expressed as follows:

[tex]K_2C_4H_4O_4+C_4H_4O_4^{2-} \longleftrightarrow 2KC_4H_4O_4^-[/tex]

Here, the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]

number of moles of [tex]2KC_4H_4O_4^-[/tex] = 2 × 0.038082 mol

number of moles of [tex]2KC_4H_4O_4^-[/tex] = 0.076164 mol

From the Henderson-HasselBalch Equation,

[tex]pH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]

[tex]pH =pKa_2 + log\dfrac{[K_2C_4H_4O_4]}{[KC_4H_4O_4^-]}[/tex]

[tex]pH =pKa_2 + log\dfrac{n_{K_2C_4H_4O_4}}{n_{KC_4H_4O_4^-}}[/tex]

[tex]5.869 =5.636+ log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]

[tex]0.233= log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]

[tex]10^{0.233}= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]

[tex]1.710015315= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]

[tex]1.710015315 \times 0.076164= n_{K_2C_4H_4O_4[/tex]

[tex]n_{K_2C_4H_4O_4}= 0.1302416[/tex]

SInce; the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]

[tex]n_{C_4H_4O_4^{2-}} =[/tex]( 0.1302416 + 0.038082) mol

[tex]n_{C_4H_4O_4^{2-}} =[/tex] 0.1683236 mol

The grams of   dipotassium succinate trihydrate = numbers of moles of dipotassium succinate trihydrate × Molar mass

The grams of   dipotassium succinate trihydrate = 0.1683236 mol × 248.32 g/mol

The grams of   dipotassium succinate trihydrate = 41.798g

Thus , the amount of grams of dipotassium succinate trihydrate  K₂C₄H₄O₄·3H₂O = 41.798g

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

Note the following:

Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.

Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

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Answer:

D. [NO₂]²/[N₂O₄]

Explanation:

The equilibrium constant expression for a reaction is products over reactants. Since NO₂ has a coefficient of 2, it will become an exponent.

So, it would be:

[NO₂]²/[N₂O₄]

Hope that helps.

The correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide (N₂O₄) and nitrogen dioxide (NO₂) is option d. [NO₂]²/[N₂O₄].

In the balanced equation N₂O₄(g) ⇌ 2NO₂(g), the stoichiometric coefficients indicate that two moles of NO₂ are produced for every one mole of N₂O₄ consumed.

The equilibrium constant expression is derived from the molar concentrations of the species involved in the equilibrium. In this case, the expression is written as [NO₂]²/[N₂O₄], where [NO₂] represents the molar concentration of NO₂ and [N₂O₄] represents the molar concentration of N₂O₄.

By squaring the concentration of NO₂ and dividing it by the concentration of N₂O₄, the equilibrium-constant expression correctly accounts for the stoichiometry of the reaction and the relative concentrations of the species involved.

Hence, the correct option is option d.

Learn more about equilibrium here:

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#SPJ 4

Answer:

3-methylpentan-3-ol

Explanation:

In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand  how this molecule is produced we have to check the mechanism:

The mercury(II) acetate ([tex]Hg(OAC)_2[/tex]) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation [tex]HgOAc^+[/tex] and the anion [tex]AcO^-[/tex]. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the [tex]NaBH_4[/tex] would reduce the compound to produce the final alcohol.

See figure 1

I hope it helps!

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

Answer:

The purpose of adding an aqueous solution of sodium bicarbonate is either to extract a certain compound or to remove/neutralize acidic compounds present in the reaction mixture.

Explanation:

Upon adding sodium bicarbonate, carbon dioxide is released (gaseous state at room temperature), which helps build up pressure that is able to push out the unwanted gas/liquid.

Answer:

C. Two hexagon rings with carbons at each corner have alternating double bonds and share one side. H is single bonded to all the C's except the ones on the shared side

Explanation:

The structure of aromatic hydrocarbon contain benzene (C6H6) which is a cyclic hydrocarbon.

Alternating single and double bond at each corner of two hexagon rings is an aromatic hydrocarbon as the alternating single and double bond form a benzene ring and H is single bonded to all the carbon's (C) except on the shared side.

Hence, the correct answer is "C"

Answer:

if you look up "propyne hydrocarbon" online, the model of D should pop up!

H - C = C - C - H

Middle C after "=" should have H on both sides (top and bottom)

Explanation:

Explanation:

first find the the number of moles of of zinc .

as the number of moles of zinc and ZnCl2 is same we can calculate the mass of ZnCl2.

Answer:

1. False

2. True

3. True

4. True

5. True

6. True

7. True

8. False

9. True

Explanation:

Density is a physical property since its measurement does not involve any chemical process.

Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.

Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.

The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.

When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.

Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.

The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.

10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.

If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.

Answer:

(d) 2

Explanation:

Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer

Answer:

[tex]M=0.727M[/tex]

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

[tex]n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-[/tex]

So the total mole of chloride ions:

[tex]N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-[/tex]

And the total volume by adding the volume of each solution in L:

[tex]V=0.500L+0.425L=0.925L[/tex]

Finally, the molarity turns out:

[tex]M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M[/tex]

Best regards.

Answer:

The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Explanation:

The half reactions are:

2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)

Sn(s) ⟶ Sn²⁺(aq) + 2 e-  (oxidation)

In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:

anode reaction∥ cathode reaction

where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.  

In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).

The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:

Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Answer:

B. 3

Explanation:

To decrease the temperature of your coffee from 85°C to 55°C your system need to absorb energy. This energy will be absorbed from the addition of some ice.

How many energy must be absorbed? You can use:

Q = C×m×ΔT

Where Q is heat (Energy), C is specific heat of your solution (4.184J/g°C), m is its mass (mass of 355mL of coffee = 355g) and ΔT is change in temperature (85°C-55°C = 30°C)

Replacing, your ice needs to absorb:

Q = C×m×ΔT

Q = 4.184J/g°C×355g×30°C

Q = 44559.6J

The energy that is taken from an ice cube to change its temperature from -15°C to 55°C is:

Energy from -15°C to 0°C (C of ice = 2.95J/g°C):

Q = C×m×ΔT

Q = 2.95J/g°C×24g×15°C

Q = 1062J

Now the energy taken to pass the ice from solid to liquid is:

Q = ΔHf×m

Q = 334J/g×24g

Q = 8016J

And the energy to increase the temperature of 0°C to 55°C of 24g of ice:

Q = 4.184J/g°C×24g×55°C

Q = 5522.9J

And the total energy that 1 ice cube needs is:

Q = 1062J + 8016J + 5522.9J

Q = 14600.9J

But you need 44559.6J to decrease the temperature of your coffee, that is:

44600J / 14600.9J = 3.05

≈ 3 ice cubes to decrease the temperature of the coffee.

Right solution:

Answer:

3

Explanation:

I’m not positive but I think it’s correct

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

And mass of AgCl = 65.08 - 50.24

The masses of the compounds in the precipitate can be found my knowing

the number of moles of chloride ion contributed by each compound.

Reasons:

The given parameter are;

Volume of KCl solution added = 0.242 L

Concentration of KCl solution = 1.92 M KCl

The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions

Precipitates formed = AgCl and PbCl₂

The mass of the precipitate = 65.08 g

Required:

The mass of PbCl₂ and AgCl in the precipitate

Solution;

Number of moles of chloride ions in a mole of PbCl₂ = 2 moles

Number of moles of chloride ions in a mole of AgCl = 1 mole

Let X represent the mass of PbCl₂ in the precipitate, we have;

The mass of AgCl in the precipitate = 65.08 g - X

[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]

Number of moles of chloride ions from PbCl₂ is therefore;

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]

The number of moles of chloride ions from one mole of KCl = 1 mole

Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;

0.242 L × 1.92 moles/L = 0.46464 moles

Number of moles of chloride ions from KCl = 0.46464 moles

[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]

Which gives;

[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]

Therefore;

[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]

The mass of PbCl₂ in the precipitate, X ≈ 49.24 g

The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g

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Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

Answer:

[tex]m_{air}=0.947g[/tex]

[tex]n_{O_2} =0.00686molO_2[/tex]

Explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]

Best regards.

Answer: The net ionic equation is [tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

[tex](NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)[/tex]

The equation can be written in terms of ions as:

[tex]2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)[/tex]

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The ions which are present on both the sides of the equation are ammonium and chlorate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is :

[tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]

Answer:

The value of Kp at this temperature is 9.0*10³

Explanation:

The equilibrium constant Kp describes the relationship that exists between the partial pressures of the reactants and products, while Kc represents the relationship that exists between the concentrations of the reactants and products that participate in the reaction.

The general relationship between the constants Kp and Kc results:

Kp=Kc*[tex](R*T)^{moles of product - moles of reagent}[/tex]

In this case:

Replacing:

Kp=2.2*10⁵*[tex](0.0821*298)^{-1}[/tex]

Solving:

Kp≅9.0*10³

The value of Kp at this temperature is 9.0*10³

Answer:

1-cyclopentylhexan-2-one

Explanation:

1-cyclopentylhexan-2-one

Answer:

[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]

Explanation:

There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the Al be Component 1 and the H₂O be Component 2.

Data:  

For the Al:

[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

For the water:

[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

Calculations

(a) The relative temperature changes

[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]

(b) Final temperature

[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]

[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]

Check:

[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]

The second term has only two significant figures because ΔT₂ has only two.

It agrees to two significant figures

Answer:

d. 2500

Explanation:

In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:

v = k [A]ᵃ [B]ᵇ [C]ⁿ

If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:

v = k [A]ᵃ [B]ᵇ [C]⁰

v = k [A]ᵃ [B]ᵇ×1

Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:

v = k [A]ᵃ [B]ᵇ

v = k [A]ᵃ [B]²

Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:

v = k [A]ᵃ [B]²

v = k [A] [B]²

Relpacing this equation in any experiment (Experiment 5, for example):

20.0 = k [0.200] [0.200]²

2500 = k

That means right answer is:

Answer:

O-H stretch signal at 3300 cm-1

Explanation:

In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).

In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).

I hope it helps!

Answer:

Compound are defined as the containing two or more different element .

(1) Ionic compound and (2) Covalent compound.

Explanation:

Covalent compound :- covalent compound are the sharing of electrons two or more atom.

Covalent compound are physical that lower points and compared to ionic .

Covalent compound that contain bond are carbon monoxide (co), and methane .

Covalent compound are share the pair of electrons.

Covalent compound are bonding a hydrogen atoms electron.

Ionic compound a large electrostatic actions between atoms.

Ionic compound are higher melting points and covalent compound.

Ionic compound are bonding a nonmetal electron.

Ionic electron can be donate and received ionic bond.

Ionic compound bonding kl.

Answer:

100.2g of CuI₂ you must add

Explanation:

Molality, m, is defined as the ratio between moles of solute and kg of solvent.

In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: ‎317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.

As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:

0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂

Using molar mass, grams of CuI₂ in the solution are:

0.316moles CuI₂ ₓ (317.35g / mol) =

Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.