An oxidation-reduction reaction in which 3 electrons are Transferred has ∆G° = +18.55 kJ at 25°C? What is the value of E°?

Answer:

Option A. 1 0n

Explanation:

Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.

The missing part of the transmutation equation as it has been shown is 1/o n. Option A

Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.

The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.

We have the equation as;

238/92 U + 12/6 C  ----> 244/98 Cf + 6 1/0 n

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Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]

heat= 7.026 kJ

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Answer:

Pb + 2H2O --> PbO2 + 2H2

Explanation:

Products:

Solid metal; PbO2

Hydrogen; H

Reactants:

Metal; Pb

Steam; H2O

Reactants --> Products

Pb + H2O --> PbO2 + H2

Upon balancing we have;

Pb + 2H2O --> PbO2 + 2H2

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:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

Answer:

0.9718M

Explanation:

Rate constant, k =  0.255 M-1s-1

time, t = 8.00 s

Initial concentration, [A]o = 1.33 M

Final concentration, [A] = ?

These quantities are represented by the equation;

1 / [A] = 1 / [A]o + kt

1 / [A] = 1 /1.33 + (0.255 * 8)

1 / [A] = 0.7519 + 2.04

[A] = 1 / 2.7919 = 0.3582 M

How much of NO2 decomposed is obtained from the change in concentration;

Change in concentration = Initial - Final

Change = 1.33 - 0.3582 = 0.9718M

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

Answer: 2. Dalton Model

Explanation:

John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made  a similar statement in the past ( all matter are made up of small indivisible particle called atom).

As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

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

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

Explanation:

free your mind drink water and go outside take fresh air you will get answers

Answer:

Explanation:

This is a bit of a trick question.

Usually an exothermic reaction is written as

A + B - heat = C + D

The meaning of this equation is that when the bonds of the reactants break, heat has to be given away to the environment. On the left, exothermic means that heat has to be given.

The wording on this question means that heat is a product

A + B = C + D + heat.

In other words heat is given up to the environment. So this reaction is exothermic.

Answer:

W = 278.64375 Joules

Explanation:

The information given in this problem are;

Initial volume = 0L

Final volume = 2.50L

ΔV = 2.50 - 0 = 2.50 L

External pressure, P =  1.10 atm

Work = ?

These parameters are related by the equation;

w = - P ΔV

W = - (1.10 )(2.50)

W = 2.75 L atm

Upon conversion to joules;

1 liter atmosphere is equal to 101.325 joule

W = 278.64375 Joules

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.

Explanation:

Answer:

Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.

Explanation:

Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.

Small amounts of calcium hydroxide salt,  [tex]Ca(OH)_{2}_(s)[/tex]  is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.

I hope this explanation is helpful.

Answer:

There are no forces holding the gas particles together.

Explanation:

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  = 0.10 M

The chemical equation for this reaction is :

[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  [tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]

The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

Answer:

Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.

Explanation:

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:

[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:

[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]  (1)

Also, we know that:

[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex]    (2)

From equation (2) we have:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex]   (3)

By entering (3) into (1):

[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]

[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]

[tex] [NaHCO_{3}] = 0.103 M [/tex]  

Hence, the [Na_{2}CO_{3}] is:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]  

Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:

[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]

[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]

Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.      

I hope it helps you!

A 1.89 pints of blood would contain 873 grams of platelets.

To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:

                                  [tex]1.89 \times 473.2 = 894.3 mL[/tex]

                                         [tex]\frac{894.3}{1000}= 0.84L[/tex]

Now, just calculate the amount of platelets present in 0.84L:

                                    [tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]

                                       x = 873 grams

So, a 1.89 pints of blood would contain 873 grams of platelets.

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Answer:

Animals whose certain organs closely match those of humans or have similar genetic makeup are used in vaccine tests because the results can closely resemble those same results on humans.

Explanation:

Answer:

they use them to test the effectiveness of the vaccine.

Explanation:

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

Answer:

One carbon atom and the nitrogen atom have nonzero formal charges.

Explanation:

The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.

The formal charge an carbon and nitrogen is calculated as follows;

No. of valence electron on atom - [non bonded electrons + no. of bonds]

Therefore, for carbon and nitrogen, we have;

formal charge on carbon = 4 - (0 + 4) = 0

formal charge on nitrogen = 5 - (2 + 3) = 0

Hence carbon and nitrogen both possess zero formal charges.

Answer:Cobb

Explanation:What y'all

Answer:

True

Explanation:

The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:

Percentage by mass = mass of substance in solution/mass of solution x 100

In this case;

mass of KBr = 4.34 grams

mass of water = 17.4 grams

mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74

Percentage by mass of KBr = 4.34/21.74 x 100

                                              = 19.96 %

19.96 is approximately 20%.

Hence, the statement is true.

Answer:The concentration of the unknown HNO3 solution = 0.01809 M

Explanation:

For the acid-base reaction,  HNO3 + NaOH-----> NaN03 + H20

we have that

C1 V1 = C2 V2

Where ,

C1 = concentration of HNO3=?

V1 = volume of HNO3 = 25.00 mL,

V2 = volume of NaOH = 22.62 mL,  

C2 = concentration of NaOH = 0.02000 M

Therefore ,

25.00 mL x C1 = 22.62 mL x 0.02000 M    

 = (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M

The concentration of the unknown HNO3 solution = 0.01809 M

Answer:

1. the siren has a lower pitch to Jane

2. the siren has a higher pitch to John

3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter

Explanation:

The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.

This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.