When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of potassium bromide are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for potassium bromide in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.

Answer: The atomic number and mass number of the new element formed is 89 and 228 respectively.

Explanation:

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.  The released beta particle is also known as electron.

In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay of Ra-228 follows:

[tex]_{88}^{228}\tyextrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]

Hence, the atomic number and mass number of the new element formed is 89 and 228 respectively.

Answer:

Check the explanation

Explanation:

Lorenz force exerted on electron by magnetic field is

F=e*v*H, where e=1.602E-19 C, H=3.64E-3 T, v is speed of electron;

? meanwhile Lorenz force is centripetal force F=m*v^2/r, where mass of electron m=9.11E-31 kg, r is radius of the path of electron;

? therefore F=F; e*v*H = m*v^2/r; eH=m*(v/r), hence

v/r = eH/m =w is angular speed of electron, hence

T=2pi/w =2pi*m/(eH) is period of rotation of electron;

e- and e+ starting at the same point are moving in the same circular path and in opposite directions, and should meet in T/2 time;

T/2 = pi*m/(eH) = pi*9.11E-31 /(1.602E-19 *3.48E-3) =5.13E-9 s;

Answer: a) [tex]Cl_2[/tex] is the limiting reagent

b)  0.27 g of [tex]AlCl_3[/tex] will be produced.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]

[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

According to stoichiometry :

3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]Al[/tex]

Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.

b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]AlCl_3[/tex]

Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]

Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.

Answer:A and B

Explanation:a stats final stages of life depend on its mass and stars die when they run out of fuel

Answer:

49.544 g.

Explanation:

The balanced equation of reaction is given below;

Hg(ClO3)2 (aq) + Na2S (aq) --------> 2 NaClO3 (aq) + HgS (s).

So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.

Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles

(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.

Number of moles = 118.08/ 367.5.

Number of moles = 0.321 moles.

(2). For Na2S, the molar mass = 78.05 g /mol.

The number of moles = 16.643 / 78.05.

The number of moles= 0.213.

Therefore, the limiting reagent = Na2S.

This means that the excess reagent is Hg(ClO3).

==> In the balanced equation of reaction above, the solid precipitate = HgS.

Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.

This problem involves calculating the mass of a precipitate from a doubly replacement reaction. Sodium sulfide is the limiting reactant and by using stoichiometry, the moles of sodium sulfide will amount to the same moles of HgS, the precipitate. Multiplying the moles of HgS by its molar mass gives the mass of the precipitate.

This question requires a two-step process that starts with writing balanced chemical equations for the reaction and then using stoichiometry to calculate the mass of the precipitate. The balanced reaction is 2Na2S + Hg2(ClO3)2 -> 2NaClO3 + Hg2S2. Using stoichiometry and details given in the question, you can calculate the molar mass of the substances and find the limiting reactant.

Mercury(II) chlorate (Hg2(ClO3)2) has a molar mass of about 511.63 g/mol. Sodium sulfide (Na2S) has a molar mass of about 78.04 g/mol. Using stoichiometry, it's clear that sodium sulfide is the limiting reactant.

Using the moles of sodium sulfide calculated, you find that the reaction will produce the same number of moles of HgS as a precipitate. By multiplying this number by the molar mass of HgS (232.66 g/mol), you obtain the mass of the precipitate that will be formed. Deduce the exact values yourself for a better understanding of this chemical reaction involving stoichiometry and solubility product.

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Answer: starting on the left:

1. 0.1M HCI

2. 0.001M HCI

3. 0.00001M HCI

4. distilled water

5. 0.00001 NaOH

6. 0.001M NaOH

7. 0.1M NaOH

Explanation:

The solutions arranged from most acidic to most basic would be: 0.1 M HCl, 0.001 M HCl, 0.00001 M HCl, Distilled Water, 0.00001 M NaOH, 0.001 M NaOH, and finally 0.1 M NaOH.

The pH of a solution depends on the concentration of hydronium ions (H+) and hydroxide ions (OH-). In this case, a lower concentration means less acidity or basicity. The solutions arranged in order from most acidic to most basic would be:

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The intrinsic chemical properties of a reactant can affect the rate of reaction, as seen in the example of food spoiling more quickly on a kitchen counter than in a refrigerator due to temperature differences.

The option that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction is Food spoils more quickly on a kitchen counter than in a refrigerator. This is because the temperature difference between the kitchen counter and the refrigerator affects the rate of chemical reactions. Chemical reactions typically occur faster at higher temperatures, so food spoils more quickly when exposed to the higher temperature of the kitchen counter compared to the lower temperature inside the refrigerator.

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Answer:

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Explanation:

[tex]\Delta T_f=T-T_f[/tex]

[tex]\Delta T_f=K_f\times m[/tex]

[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]

where,

[tex]T[/tex] = Freezing point of solvent

[tex]T_f[/tex] = Freezing point of solution

[tex]\Delta T_f[/tex] =depression in freezing point

[tex]K_f[/tex] = freezing point constant

m = molality

we have :

Mass of ethylene glycol = 59.0 g

Molar mass of ethylene glycol = 62.1 g/mol

Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)

[tex]K_f[/tex] =1.86°C/m ,

[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]

[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]

[tex]\Delta T_f=3.26^oC[/tex]

Freezing point of pure water = T =  0°C

Freezing point of solution = [tex]T_f[/tex]

[tex]\Delta T_f=T-T_f[/tex]

[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Answer:

Entropy change is zero

Explanation:

∆S=S2-S1=0

We're S is entropy

Answer:

Check Explanation

Explanation:

a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below

2Al + 3Ag₂S → Al₂S₃ + 6Ag

Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.

b) Names of the products from this reaction

Ag is called Silver metal. Free Silver metal.

Al₂S₃ is called Aluminium Sulfide.

c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.

It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!

d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!

e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.

In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.

Al → Al³⁺

0 → +3 (Oxidation)

The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.

In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.

Ag⁺ → Ag

+1 → 0 (Reduction)

The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.

The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.

Hope this Helps!!!

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

Answer:

the answer 37

Explanation:

Answer:

P = 2.25 atm

Explanation:

Given data:

Mass of dry ice = 10 g

Volume of container = 2.5 L

Temperature = 25°C (25+273=298 K)

Pressure inside container = ?

Solution:

First of all we will calculate the number of moles of dry ice.

Number of moles = mass / molar mass

Number of moles = 10 g/ 44 g/mol

Number of moles = 0.23 mol

Now we will determine the pressure.

PV = nRT

P = nRT/V

P = 0.23 mol × 0.0821 atm.L/mol.K  × 298 K / 2.5 L

P = 5.63 atm  / 2.5

P = 2.25 atm

Answer:

c. 3.00 M HCl

Explanation:

From dilution formula

C1V1 = C2V2

C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml

Substitute and Simplify

C1×10= 1.5×20

C1= 3.00M

Answer:

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)

PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)              

The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which  means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.

Assuming all PbCrO₄ formed in the 175 ml solution remained ionized,  then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs, Qsp must be > than Ksp ...

=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 => This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...

          PbCrO₄(s) ⇄  Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³

C(i)           ---               0.00M      0.0336M

ΔC           ---                  +x                +x          

C(eq)       ---                    x           0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³

∴[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.

Answer:

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)

PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)              

The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which  means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.

By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,  

Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs,

Qsp must be > than Ksp ...

=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 =>

This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.

i.e .........

PbCrO₄(s) ⇄  Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³

C(i)           ---       0.00M      0.0336M

ΔC           ---       +x                +x          

C(eq)       ---        x          

0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³

Therefore;

[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

The balanced redox reaction is

5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)

The unbalanced redox reaction is :

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

The above equation can be balanced  by ensuring the atom of the elements

on the left hand side is equal to those on the right hand side.

We have 3 atoms of Mn on the left side of the equation and 1 atom on the

right hand side.This will be balanced by putting 3 in front of MnO₄− as shown

below:

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 3 atoms of O on the left hand side and 12 atoms of O on the right

hand side. This is  balanced by putting 5 in front of AsO₂− and 2 in front of

H₂O as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 4 atoms of H on the left hand side and 1 atom of H on the right

hand side.We can balance by putting 4 in front of H⁺ as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)

We have 5 atoms of As on the left hand side and 1 atom of As on the right

hand side. We can balance by putting 5 in front of As as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)

The equation is therefore now balanced as the number of atoms of the

element on the left hand side are equal with those on the right hand side.

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Answer: The monomer of nylon-6 is caprolactam.

Explanation:

From the question, we are asked to draw the cyclic monomer from the structure of Nylon-6.

Two repeating monomer units can be seen in the polymeric structure given in the problem.

From this we can see and conclude that The monomer of nylon-6 is caprolactam.

attached is a diagrammatic representation of the Structure (Nylon-6 monomer).

cheers i hope this helps!!!!

Final answer:

Nylon-6 is synthesized from the cyclic monomer epsilon-caprolactam through a process called ring-opening polymerization, unlike other nylons that are produced by condensation polymerization. Epsilon-caprolactam features a 6-carbon ring with an amide linkage, making it unique.

Explanation:

Nylon-6 is a unique type of polymer that is derived from a cyclic monomer known as epsilon-caprolactam. Unlike the Nylon 6,6 polymer, which is made from condensation polymerization of hexanedioic acid and 1,6-diaminohexane, Nylon-6 utilizes a process called ring-opening polymerization. This process involves breaking open the ring structure of the epsilon-caprolactam to form linear chains, which then link together head-to-tail to form the polymer. This is a key difference as it does not release a small molecule (like water or hydrogen chloride) during the polymerization process, making it a growth polymerization.

Epsilon-caprolactam, the monomer for Nylon-6, is a 6-carbon cyclic amide. To deduce and draw its cyclic monomer, one should recognize that the repeating unit in Nylon-6 consists of 6 carbons in a chain with an amide linkage connecting the ends. Thus, depicting epsilon-caprolactam involves drawing a ring structure containing 5 methylene (-CH2-) groups and one carbonyl group (=O) adjacent to an NH group, forming a lactam.

Answer:

13.279

Explanation:

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72=13.279

sorry it's not the best explanation but since you're on a time crunch i hope this helps

A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. The pH of the solution that the chemist prepared, to the nearest thousandth is 13.279.

pH is a numerical indicator of how acidic or basic aqueous or some other liquid solutions are. The phrase, which is frequently used during chemistry, biology, or agronomy, converts hydrogen ion concentrations.

The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral.

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72

   =13.279

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Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

          A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600g of water. From the given information, the final temperature of the reaction is 22.71° C

From the given parameters:

From the listed parameters, the first step is to determine the number of moles of the sample by using the relation:

[tex]\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

[tex]\mathbf{number \ of \ moles = \dfrac{0.500 \ g}{213 \ g/mol}}[/tex]

number of moles of C7H5N2O6 = 0.00235 moles

However, the heat of combustion is the amount or quantity of heat energy released when one mole of a sample is burned.

Mathematically;

[tex]\mathbf{\Delta H_c =\dfrac{q_{rxn} }{ n}}[/tex]

where;

[tex]\mathbf{ q_{rxn}}[/tex] = quantity of heat released

n = number of moles

Making  [tex]\mathbf{ q_{rxn}}[/tex] the subject of the formula:

[tex]\mathbf{ q_{rxn}= \Delta H_{rxn} \times n }[/tex]

[tex]\mathbf{ q_{rxn}=-3374 \ kJ/moles \times 0.00235 \ moles }[/tex]

[tex]\mathbf{ q_{rxn}=-7.9289 \ kJ }[/tex]

[tex]\mathbf{ q_{rxn}=-7928.9 \ J }[/tex]

The heat released [tex]\mathbf{ q_{rxn}}[/tex] = heat absorbed by bomb calorimeter + heat absorbed by water

∴

[tex]\mathbf{ q_{rxn}= -(m\times c_{bc} \Delta T + c_{water} \times \Delta T)}[/tex]

[tex]\mathbf{-7928.9 \ J = -( 420 J/^0C \times \Delta T + 600 \ g \times 4.18 \ J/g ^0 C\times \Delta T)}[/tex]

[tex]\mathbf{7928.9 = (2928 \Delta T )}[/tex]

[tex]\mathbf{ \Delta T = \dfrac{7928.9}{2928} }[/tex]

[tex]\mathbf{ \Delta T =2 .71 ^0 \ C}[/tex]

Since ΔT = [tex]\mathbf{T_2 - T_1}[/tex]

2.71° C = T₂ - 20° C

T₂ = 20° C + 2.71° C

T₂ = 22.71° C

Therefore, we can conclude that the final temperature of the reaction is:

22.71° C

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Answer:

The answer is 4.8659

Explanation:

Given:

[C₆H₅COOH]=0.11M

[C₆H₅COO]=2*0.2=0.4M

Ka=6.3x10⁻⁵

First, calculate the pKa:

[tex]pKa=-logKa=-log(6.3x10^{-5} )=4.2007[/tex]

The pH is:

[tex]pH=pKa+log\frac{C6H5COO]}{[C6H5COOH]} =4.2007+log\frac{0.4}{0.11} =4.7614[/tex]

Like the volume is 5L, the volume of C₆H₅COO is x, then, the volume of C₆H₅COOH is 5-x

[tex]4.7614=4.2007+log\frac{0.4x}{0.11*(5-x)}[/tex]

[tex]0.5607=log\frac{0.4x}{0.11*(5-x)}[/tex]

Solving for x:

x=2.49L=2490mL of C₆H₅COO

2510mL of C₆H₅COOH

The milimoles of C₆H₅COOH and C₆H₅COO is:

nC₆H₅COOH=(0.11*2510)-50=226.1mmol

nC₆H₅COO=(0.4*2490)+50=1046mmol

The pH is:

[tex]pH=4.2007+log\frac{1046}{226.1} =4.8659[/tex]

Answer:

4.86

Explanation:

The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.

Please kindly check attachment for the step by step solution of the given problem.

Answer:

The weight of aspirin lost is  [tex]W =0.03267(V_A +V_B) N[/tex]

The percent yield is lowered by [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]  

Explanation:

From the question we are told that

       Since the solubility of aspirin in water is 1 g per 300 mL it implies that after crystallization the solution would contain 1 g for every 300mL of water  at 25°

     Let assume that the volume of the solution is [tex]V_A[/tex]

          The the aspirin lost after filtration would be

                       [tex]= \frac{1}{300} * V_A[/tex]

                       [tex]=\frac{V_A}{300}g[/tex]

Let assume that you used water of volume [tex]V_B[/tex] to wash the crystallized aspirin then the lost during washing would be

                   [tex]= \frac{1}{300} * V_B[/tex]

                   [tex]= \frac{V_B}{300}[/tex]

So the total loss is

                 [tex]= \frac{V_A}{300} + \frac{V_B}{300}[/tex]

               [tex]\frac{V_A +V_B}{300}g[/tex]

So the weight of aspirin lost denoted by W is

                [tex]W = \frac{V_A +V_B}{300} *9.8[/tex]

                     [tex]W =0.03267(V_A +V_B) N[/tex]

Let denote How much was your percent yield lowered by this loss by A

    So

             [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]  

Answer:

The solution is basic

Explanation:

Low pH means that a solution is acidi while low pOH means that a solution is basic.

Answer:

pH = 11.7

Explanation:

pOH= -log [OH]=-log[0.05]

=2.3

pOH+pH= 14

pH= 14-2.3= 11.7

Answer:

85.0 kJ

Explanation:

Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

To melt 160 g of octane and bring it's temperature to 99.2°C

(from literature)

Heat of fusion of Octane = 20.740 kJ/mol

Melting point of octane = -57°C

Boiling point of Octane = 125.6 °C

Molar mass of octane = 114.23 g/mol

Heat capacity of octane = 255.68 J/K.mol

So, it is evident that Octane is still a liquid at 99.2°C.

So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C

First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.

Number of moles = (mass)/(Molar mass)

Number of moles of octane = (160/114.23) = 1.401 moles

Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ

Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C

= nCΔT

n = 1.401 moles

C = 255.68 J/K.mol

ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)

Heat required to raise the temperature of already melted octane from -57°C to 99.2°C

= (1.401×255.68×156.2)

= 55,952.04 J = 55.952 kJ

Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C

= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ

Hope this Helps!!!

The heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process - melting and then heating. You use the latent heat of fusion and specific heat capacity of octane to calculate the total heat required.

The amount of heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process – melting, and heating. The process should be looked at in separate steps, similar to how the enthalpy of combustion is used to calculate the heat produced in the combustion of 1.00 L of isooctane in the reference provided. Note that different substances have different heat capacities and heat of fusion, so the associated values (likely given in the problem or textbook) for octane specifically would need to be used to get the final answer.

First, you would need to know the latent heat of fusion, which is the amount of energy required to change one kilogram of the substance from solid to liquid without changing its temperature. Then, you would multiply the mass of the octane by the latent heat of fusion to find out how much energy is required for it to go from solid to liquid.

Once the octane is in liquid form, we then need to heat it to the desired temperature. To do this, you use the specific heat capacity formula – the amount of energy required to raise one kilogram of the substance by one degree Celsius. Multiply the mass of the material by the specific heat capacity of the octane and the change in temperature (final temperature – initial temperature).

The added heat required for both of these processes would give you your final answer.

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Answer:

The empirical formula of the compound is [tex]CrBr_2[/tex].

Explanation:

Mass of chromium = 19.13 g

Mass metal bromide formed = 77.92 g

Mass of bromine in metal bromide = 77.92 g - 19.13 g = 58.79 g

Moles of chromium metal :

[tex]=\frac{19.13 g}{52 g/mol}=0.3679 mol[/tex]

Moles of bromine:

[tex]=\frac{58.79 g}{80 g/mol}=0.7349 mol[/tex]

For empirical formula divide the smallest number of moles of element from the all the moles of elements:

chromium : [tex]\frac{0.3679 mol}{0.3679}=1[/tex]

bromine : [tex]\frac{0.7349 mol}{0.3679}=2[/tex]

The empirical formula of the compound is [tex]CrBr_2[/tex].

To find the empirical formula, the moles of each element involved must be determined. When the sample weights are converted to moles, the ratio of chromium (Cr) to bromine (Br) in the bromide compound is found to be 1:2, making the empirical formula CrBr2.

To determine the empirical formula of the compound, we first need to identify the moles of each element. The chromium sample weighs 19.13g. Using the atomic weight of chromium, which is 52.00g/mol, the number of moles of chromium is 19.13g / 52.00g/mol = 0.368 mol.

Considering the entire mass of the metal bromide, which is 77.92 g, the mass of bromine in the compound must be 77.92 g - 19.13 g = 58.79 g. As the atomic weight of bromine is approximately 79.90g/mol, we can find that the moles of bromine are 58.79 g / 79.90 g/mol = 0.736 mol.

The ratio of Cr to Br in the compound is 0.368 : 0.736, which is approximately 1 : 2 when we divide both numbers by the smallest to obtain the simplest whole number ratio. Therefore, the empirical formula of the compound is CrBr2.

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Answer:

0.2 mol/kJ

Explanation:

Methane is stored at a temperature of 111.4K, [tex]T_c[/tex]. The heat source to vapourization of methane is ambient air which is at 300 K. [tex]T_H[/tex]

Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.

Calculate the vaporization rate from the given data by relation shown below:

[tex]Vaporization rate = \frac{Q_c}{[\frac{\delta H_n^{lv}}{W}]} ..........(1)[/tex]

here,

[tex]Q_c[/tex] is the heat at temperature [tex]T_c, \delta H_n^{lv}[/tex] is the phase transition enthalpy of methane and W is the work

Calculate [tex]Q_c[/tex] from the equation shown below:

[tex]Q_c = Q_n (1 - \eta_{HE}) .............(2)[/tex]

where Q_n is the  heat at temperature of [tex]T_n[/tex] and [tex]\eta_{HE}[/tex] is the efficiency of heat engine

calculate [tex]Q_H[/tex] from the relation shown below:

[tex]Q_H = \frac{W}{\eta_{HE}} ..........(3)[/tex]

calculate the heat engine efficiency from the given carnot engine efficiency as shown below

[tex]\eta_{HE} = 0.6 \times \eta_{carnot} ............(4)[/tex]

here, [tex]\eta_{carnot}[/tex] is the carnot engine efficiency

[tex]\eta{carnot} = 1 - \frac{T_c}{T_H}[/tex]

substituting the values of temperature, we have

[tex]= 1 - \frac{111.4K}{300K}\\= 0.629\\[/tex]

substitute values of [tex]\eta_{carnot}[/tex] in equation 4, we get

[tex]\eta_{HE} =0.6 \times \eta_{carnot}\\ = 0.6 \times 0.629\\ = 0.3774\\\\[/tex]

check the attached file for additional solution

In a neutral atom, there are equal numbers of electrons and protons, which gives the atom its neutral charge. Neutrons do not factor into the atom's charge, and allotropes are different forms of an element, not part of its atomic structure.

In a neutral atom, there are equal numbers of electrons and protons. The number of protons in an atom's nucleus determines the atomic number, and the atom's charge is neutral because the positive charge of the protons is balanced by the negative charge of the electrons. Neutrons, on the other hand, do not carry an electronic charge and are not necessarily equal in number to the protons or electrons. Concerning allotropes, they are different forms of the same element and it's not directly related to this question about atomic structure.

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In neutral atoms, there are equal numbers of electrons and protons. Each element, when neutral, has a characteristic number of electrons that matches its atomic number (the number of protons). Neutrons are electrically neutral and their number does not affect the atom's charge.

In all neutral atoms, there are equal numbers of electrons and protons. This is because the number of protons, which carry a positive charge, and the number of electrons, which carry a negative charge, balance each other in a neutral atom, resulting in an overall charge of zero.

Each element, when electrically neutral, has a characteristic number of electrons equal to its atomic number, which is the number of protons it contains. For example, in a carbon-12 atom (which is neutral), there are six protons and six electrons, yielding a net charge of zero.

It's important to note that while neutrons also exist within an atom, they are electrically neutral. Therefore, the number of neutrons does not affect the atom's charge and doesn't necessarily match the number of protons or electrons.

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Final answer:

To find the molar mass of the diprotic acid, divide the mass of the sample by the number of moles. The molar mass of the acid is 307.03 g/mol.

Explanation:

To find the molar mass of the diprotic acid, we can use the information from the titration. First, we need to determine the number of moles of NaOH used in the titration. The volume of NaOH solution used is 35.8 mL, which is equal to 0.0358 L. The molarity of the NaOH solution is 0.160 M.

Therefore, the number of moles of NaOH used is 0.0358 L x 0.160 mol/L = 0.005728 mol.

Since the diprotic acid is diprotic, it can donate two moles of H+ ions per mole of acid. Therefore, the number of moles of the diprotic acid is half of the number of moles of NaOH used, which is 0.005728 mol / 2 = 0.002864 mol.

To calculate the molar mass of the acid, we need to divide the mass of the sample by the number of moles. The mass of the acid sample is 0.881 g. Therefore, the molar mass of the acid is 0.881 g / 0.002864 mol = 307.03 g/mol.

Final answer:

To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.

Explanation:

Empirical Formula Calculation

To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.

To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.