Calculate the value of ΔG∘rxnΔGrxn∘ for the following reaction at 296 K. Ka = 2.9 × 10–8 and assume Ka does not change significantly with temperature. $$HClO(aq)+H2O(l) HClO−(aq)+H3O+(aq)

Answer:

The correct answer is -2878 kJ/mol.

Explanation:

The reaction that takes place at the time of the oxidation of glucose is,  

C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)

The standard free energy change for the oxidation of glucose can be determined by using the formula,  

ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)

The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.  

Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)

ΔG°rxn = -2878 kJ/mol

Answer:

1. REDOX

2. None of the above

3. Precipitation

4. Preicipitation

5. Acid base neutralization

Explanation:

Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.

Option 4 and 3.

3) CaO (s) + CO₂ (g) →  CaCO₃(s)

4) KOH (aq)  + AgCl (aq)  →  KCl (aq)  + AgOH(s)

Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.

5) Ba(OH)₂ (aq)  +  2HNO₂(aq)  →  Ba(NO₂)₂ (aq) + 2H₂O(l)

Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.

1) Ba(ClO₃)₂ (s)  → BaCl₂ (s) +  3O₂(g)

Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).

2.  2NaCl(aq)  +  K₂S(aq)  Na₂S (aq)  + 2KCl (aq)

None of the above

Answer: The density of the object is 1.10 g/ml

Explanation: Density of object = ?

Mass of object = 26.5 g

Volume of object = volume of water displaced = 24.1 ml

Putting values in above equation, we get:

[tex]Density: \frac{26.5g}{24.1ml} = 1.10g/ml[/tex]

Thus density of the object is 1.10 g/ml

Answer:

t = 2 seconds

Explanation:

It is given that,

Mass of a boy, m = 50 kg

Initial speed of boy, u = 0

Final speed of boy, v = 8 m/s

Force exerting by another boy, F = 200 N

Let t is the time of contact. The force acting on an object is given by :

F = ma

a is acceleration

So,

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times 8}{200}\\\\t=2\ s[/tex]

So, the contact time is 2 seconds.

Answer:

t=2 s

Explanation:

Answer:

0.89kg

Explanation:

Q=mL L=specific latent heat

Q=energy required in J

m=mass in Kg

Q=mL

m=Q/L

m=2000000J/2.25 x 10^6 J kg-1

m=0.89kg

Answer:

The molecular formula of the compound is [tex]C_{7}H_{6}O_{2}[/tex].

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

[tex]m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]

[tex]m_{C} = 84.067\,g[/tex]

Hydrogen

[tex]m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]

[tex]m_{H} = 6.057\,g[/tex]

Oxygen

[tex]m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]

[tex]m_{O} = 31.995\,g[/tex]

Now, the number of moles ([tex]n[/tex]), measured in moles, of each element are calculated by the following expression:

[tex]n = \frac{m}{M}[/tex]

Where:

[tex]m[/tex] - Mass of the element, measured in grams.

[tex]M[/tex]- Molar mass of the element, measured in grams per mol.

Carbon ([tex]m_{C} = 84.067\,g[/tex], [tex]M_{C} = 12.011\,\frac{g}{mol}[/tex])

[tex]n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }[/tex]

[tex]n = 7[/tex]

Hydrogen ([tex]m_{H} = 6.057\,g[/tex], [tex]M_{H} = 1.008\,\frac{g}{mol}[/tex])

[tex]n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }[/tex]

[tex]n = 6[/tex]

Oxygen ([tex]m_{O} = 31.995\,g[/tex], [tex]M_{O} = 15.999\,\frac{g}{mol}[/tex])

[tex]n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }[/tex]

[tex]n = 2[/tex]

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

[tex]C_{7}H_{6}O_{2}[/tex]

Answer:

Option C . CO2(g) + H2O(g)

Explanation:

When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

Thus, the product of the unbalanced combustion reaction is:

CO2(g) + H2O(g)

Thus, we can balance the equation as follow:

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)

There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:

C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)

Thus, the equation is balanced.

Answer:

See figure 1

Explanation:

In this case, we can start with the linear structure of D-Idose. Then, if we have a "D" configuration the "OH" in the last chiral (carbon 5) will be in the right. This carbon will attack carbon 1 and we will produce a cyclic structure with 6 members (pyranose). Additionally, we have to keep in mind that we want the "beta" structure. So, the "OH" on carbon 1 must point up (red arrow). Finally, we will have a cyclic structure with 6 atoms and the "OH" on carbon 1 pointing up.

See figure 1

I hope it helps!

Answer:

The answer to your question is given below.

Explanation:

To draw the structure of 2–methyl–1–butanamine, we following must be observed:

1. The functional group of the compound is amine –NH2.

2. The functional group is located at carbon 1.

3. The longest continuous carbon chain is carbon 4 i.e butane. Since the functional group is amine, the –e at the end of the butane is replaced with

–amine, making the name to be butanamine.

4. Methyl, CH3 is located at carbon 2.

5. Combine the above to get the structure of 2–methyl–1–butaamine.

Please see attached photo for the structure of 2–methyl–1–butanamine

Answer:

1. Na + O2 → Na2O (Balanced)

2. 4Al + 3O2 → 2(Al2O3) (Balanced)

3. H2 + i2 → 2HI (Balanced)

4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)

5. 2Ca +O2 → 2CaO (Balanced)

Answer:

The final temperature of sulfur dioxide gas is 215.43 C

Explanation:

Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.

Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:

[tex]\frac{P}{T}=k[/tex]

Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:

[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]

The reference temperature is the absolute temperature (in degrees Kelvin)

In this case:

Replacing:

[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]

Solving:

[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]

[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]

T2=488.58 K

Being 273.15 K= 0 C, then 488.58 K= 215.43 C

The final temperature of sulfur dioxide gas is 215.43 C

Answer:

58.45g is the answer

Explanation:

took the test

The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.

For a steady flow of charge through a conductor, the current can be determined with the following equation:

[tex]{\displaystyle I={Q \over t}[/tex]

where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.

Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.

Given, the amount of electric current flowing through the solution:

I = 45.2 A

The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec

The charge flowing through the solution, Q = I × t

Q = 45.2 × 7200

Q = 325440 C

The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol

We know Cr³⁺ + 3e⁻ →  Cr (s)

3 moles of electrons deposit of chromium  = 1 mol

3.37 mol of electrons deposit of chromium  = 3.37/3 = 1.12 mol

The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g

Learn more about electric current, here:

https://brainly.com/question/2264542

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Answer:

Option (b) Hydrocyanic acid, 4.9×10^-10

Explanation:

Data obtained from the question include:

Ka of Hydrofluoric acid = 3.5×10^-4

Ka of Hydrocyanic acid = 4.9×10^-10

Ka of Nitrous acid = 4.6×10^-4

To know which acid is least acidic, we shall determine the the pKa value for each acid.

This is illustrated below:

For Hydrofluoric acid

Ka = 3.5×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 3.5×10^-4

pKa = 3.5

For Hydrocyanic acid

Ka = 4.9×10^-10

pKa =..?

pKa = –Log Ka

pKa = –Log 4.9×10^-10

pKa = 9.3

For Nitrous acid

Ka = 4.6×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 4.6×10^-4

pKa = 3.3

Summary:

Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa

Hydrofluoric acid >> 3.5×10^-4 >> 3.5

Hydrocyanic acid >> 4.9×10^-10 > 9.3

Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3

NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.

From the above calculations, Hydrocyanic acid has the highest pKa value.

Therefore, Hydrocyanic acid is the least acidic compound

Answer:

See explanation

Explanation:

The calculated concentration of the sodium hydroxide is;

Number of moles= mass/molar mass = 1g/40gmol-1 = 0.025 moles

Concentration= number of moles/volume= 0.025×1000/250 = 0.1 M

This calculated concentration will be different from the molarity of NaOH obtained by standardization with acid. The result will not be more accurate if a different acid is used for the standardization this is because sodium hydroxide is deliquescent and absorbs moisture thereby leading to inaccuracy in the calculated molarity.

Any substance that must be used as a primary standard must not absorb moisture, it must be stable and it must be a substance in its pure form.

The given question is incomplete. The complete question is :

Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

a) [tex]O^{2-}[/tex]

b)  [tex]F^{-}[/tex]

c)  [tex]N^{3-}[/tex]

d)  [tex]S^{2-}[/tex]

Answer: b)  [tex]F^{-}[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here potassium is having an oxidation state of +1 called as  cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to  give neutral ionic compound.

Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

Answer:

41L

Explanation:

Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.

A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.

Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:

0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂

If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:

1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =

Answer:

495nm

Explanation:

The energy of a photon could be obtained by using:

E = hc / λ

Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.

The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:

242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.

Replacing in the equation:

E = hc / λ

4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ

λ = 4.946x10⁻⁷m

Is maximum wavelength  of light that could break a Cl-Cl bond.

Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:

4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

Answer:

Total momentum of both player after collision =93  Kg m/s

Explanation:

According to law of conservation of momentum

For an isolated system of bodies , momentum of bodies before and after collision remains same.

momentum is given by mass* velocity

_________________________________________

Here the isolated system of bodies are

two football players.

Momentum of player before collision

Momentum of player 1 = 105*8.6 = 903 Kg m/s

Momentum of player 2 = 90*-9 = -810 Kg m/s

Total momentum of both player before collision = 903 + (-810) = 93 Kg m/s

as by conservation of

Total momentum of both player before collision = Total momentum of both player after collision

Total momentum of both player after collision =93  Kg m/s

Answer:A is the Answer

Explanation:

Answer:

[tex]4.36~g~XY[/tex]

Explanation:

In this case, we can start with the reaction:

[tex]2X + Y_2~->~2XY[/tex]

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

[tex]3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X[/tex]

[tex]4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2[/tex]

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

[tex]\frac{0.04~mol~X}{1}=~0.04[/tex]

[tex]\frac{0.0875~mol~Y_2}{2}=0.04375[/tex]

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

[tex]0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY[/tex]

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol [tex]Y_2[/tex] = 48 g [tex]Y_2[/tex] (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

[tex]0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY[/tex]

I hope it helps!

Answer:

See figure 1

Explanation:

In this case, we have to start with the ionization reaction of HBr to produce the hydronium ion ([tex]H^+[/tex]) and the bromide ion ([tex]Br^-[/tex]). Then the double bond in the alkene can attack the hydronium ion to produce a carbocation. The most stable carbocation would be the tertiary one, therefore we have to put the positive charge in the tertiary carbon. Then, the bromide attacks the carbocation to produce the final halide.

See figure 1

I hope it helps!

Answer:

(c) is the correct answer

CH4(g)+2O2(g)⟶CO2(g)+2H2O(l)

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Answer:

Figure 1

Explanation:

In this case, we have to ester with a "malonic synthesis" in which we have to add a strong (ethoxide) to produce an enolate ion that would form a new C-C bond with an alkyl halide (ethyl bromide and bromo pentane). Then a "nucleophilic acyl substitution reaction" takes place to add thiourea, in this step two ethanol groups are eliminated to form a cyclic structure. Finally, an "elimination reaction" happen by the addition of sodium hydroxide generating a double bond and a negative charge in the sulfur atom that is neutralized with the positive charge of sodium.

See figure 1 for the total mechanism

I hope it helps!

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

Answer:

K = 2.7x10⁻⁵ at 25ºC

Explanation:

A way to write Arrhenius equation is:

ln K = - Ea/R × (1/T) + lnA

If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:

Y = -13815X +35.817

R² = 0.9927

(Taking the last k point as 0.0386) (ln 0.0386), 0.1386 has no sense)

Your slope is -13815

-13815K = - Ea/R

-13815K×8.314J/molK = 114858J/mol = Ea

And your intercept =

lnA = 35.817

A = 3.59x10¹⁵

Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):

Y = -13815X +35.817

Y = -13815(1/298.15K) +35.817

Y = -10.5187

lnK = -10.5187

Answer:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

The reactant that is reduced is Q.

Explanation:

The complete equation for the reaction is such that:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently,  [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].

From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.

Answer:

22.4 L H2

Explanation:

There is a better explanation https://brainly.com/question/9562878

Answer:

The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.

Explanation:

1s^2

2s^2

2p^6

3s^2

3p^6

4s^2

3d^10

4p^5

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

The volume of the glucose solution to be prepared = 500 [tex]cm^3[/tex]

Molarity of the glucose solution to be prepared = 1 M

i. Molar mass of glucose ([tex]C_1_2H_6O_6[/tex]) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii. mole = molarity x volume. Hence;

amount (in moles) of the glucose solution to be prepared

                 = 1 x 500/1000 = 0.5 mole