For a particular reaction, ΔH∘=−28.4 kJ and ΔS∘=−87.9 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer:2.62 L

Explanation:

A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.

An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.

By using ideal gas equation,

P₁ V ₁ ÷ T = P₂V₂ ÷ T

1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50

V₂ = 1 × 2.62 × 50 ÷ 25 × 2

V₂ = 2.62 liters.

Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

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Answer:

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Explanation:

Given that,

Anode : Zn electrode in a solution of 0.050 M Zn(NO₃)₂

Cathode : Pt electrode with 0.500 atm H₂(g) in 0.010 M HNO₃

Anode :

[tex]Zn(s)\Rightarrow Zn^{2+}(aq) + 2e^{-}[/tex]

Cathode :

[tex]2H+(aq)+2e^{-}\Rightarrow H_{2}(g)[/tex]

We need to write the overall balanced cell reaction

Using anode and cathode

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Hence, This is required answer.

Answer:

3.11

Explanation:

Any buffer system can be described with the reaction:

[tex]HA~->~H^+~+~A^-[/tex]

Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]

With all this in mind, we can write the reaction for our buffer system:

-) Nitrous acid: [tex]HNO_2[/tex]

-) Sodium nitrate: [tex]NaNO_2[/tex]

[tex]HNO_2~->~H^+~+~NO_2^-[/tex]

In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).

We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:

-) moles of [tex]HNO_2[/tex]:

[tex]mol=0.150~M*0.045~L=0.00657[/tex]

-) moles of [tex]NO_2^-[/tex]:

[tex]mol=0.175~M*0.020~L=0.0035[/tex]

The total volume would be:

0.020 L + 0.045 L = 0.065 L

With this in mind, we can calculate the molarity of each compound:

-) Concentration of [tex]HNO_2[/tex]

[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]

-) Concentration of [tex]NO_2^-[/tex]

[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]

The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:

[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]

The final pH value would be 3.11

I hope it helps!

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.

Nitrous acid is a weak acid and nitrate ion is its conjugate base.

It is a solution used to resist abrupt changes in pH when acids or bases are added.

We do so by multiplying the molar concentration by the volume in liters.

HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol

NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol

The total volume will be the sum of the volumes of each solution.

V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L

HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M

NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M

We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.

pH = pKa + log [NaNO₂]/[HNO₂]

pH = 3.16 + log 0.0538/0.104 = 2.87

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

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Answer: C) [tex]CO^{2-}_{3}[/tex]

Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.

To determine molecular geometry:

1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;

2) Count the number of electron pairs (count multiple bonds as 1 pair);

3) Arrange electron pairs to minimise repulsion;

4) Position the atoms to minimise the lone pair;

5) Name the molecular geometry from the atom position;

Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.

The Lewis structure of each molecule is shown in the attachment.

Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

Explanation:

After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the  last step when balancing a redox reaction under basic condition.

The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.

Answer:

2, hydrogen

Explanation:

i think

Answer:

Answer is not hydrogen

Explanation:

did the test and got it wrong

Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]

Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:

[tex]\Delta H=\Delta E+\Delta(PV)[/tex]

So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.

Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.

Analysing each reaction above:

Reaction A has the same number of moles in reagents and products, so  enthalpy change and internal energy change will be equal;

Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.

Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

Answer:

4.16x10⁻³m

Explanation:

Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.

As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.

Moles glucose:

There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.

Molar mass glucose:

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

72.06 + 12.10 + 96 = 180.16g/mol

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Molality of the solution:

4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =

The molarity of the solution is 4.16x10⁻³m

We know that the molarity refers to the ratio that arise between the moles of a solute.

Since there are 100 g of solvent so here the kg should be 0.1.

Likewise there is 75 mg so it should be 0.075g

Now the Molar mass glucose should be

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

So,

= 72.06 + 12.10 + 96

= 180.16g/mol

Now

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Now finally

Molality of the solution:

= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent

=4.16x10⁻³m

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Answer:

12-crown-4

Explanation:

We must recall that any structural moiety in organic chemistry having the R-O-R unit is an ether. If the oxygen form a ring in which they are sandwiched in between carbon atoms, the compound is known as a crown ether. The name emanates from the close resemblance of the compound to an actual crown.

If we want to name the crown ether, we first count the number of carbon atoms present and the number of oxygen atoms present. The correct name is now, total number of carbon + oxygen atoms -crown- number of oxygen atoms, in this case; 12-crown-4, hence the answer.

Answer:

12-crown-4

Explanation:

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Answer:

[tex]m=0.127g[/tex]

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]

Best regards.

Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

Answer:

the chemical equation of ethanol as fuel is

C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)

the preparation process of ethanol from cassava is

cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

Answer:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]

The equilibrium expression is:

[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:

[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]

Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Regards.

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

The dissociation of calcium fluoride has been given by:

[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]

The solubility constant, ksp has been given as:

[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]

From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.

The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.

The concentration of Ca in the solution has been resulted as x + 0.01 M.

The solubility product can be given as:

[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

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Answer:

The products are KNO3 + PbCl2.....

Espero que te sirva.

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

Answer:

Yes, same piece can be used.

Explanation:

The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.

The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.

Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.

The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.

Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.

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Answer:

whats the ph  ofpoh=9.78

Explanation:

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

Answer:

-A nucleus with a neutron:proton ratio of 1.49

-The nucleus of Sb-123

-A nucleus with a mass of 187 and an atomic number of 75

Answer:

Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

Explanation:

The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).

∆G is given by;

∆G= ∆H -T∆S

Where;

∆H= change in enthalpy of the system

T= absolute temperature of the system

∆S= change in entropy

Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]

Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.

Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:

    0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to: