2.61 kilograms of water in a container have a pressure of 200 kPa and temperature of 200°C . What is the volume of this container? m (Round to three decimal places)

Answer:

31.9178 °C is the final temperature of the water

Explanation:

[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ[/tex]

Mass of benzene burned = 6.200 g

Moles of benzene burned = [tex]\frac{6.200 g}{78 g/mol}=0.0794 mol[/tex]

According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.

Then 0.0794 mol of benzene on combustion will give:

[tex]\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q[/tex]

Mass of water in which Q heat is added = m = 5691 g

Initial temperature = [tex]T_i=21^oC[/tex]

Final temperature = [tex]T_f[/tex]

Specific heat of water = c = 4.18 J/g°C

Change in temperature of water = [tex]T_f-T_i[/tex]

[tex]Q=mc\Delta t=mc(T_f-T_i)[/tex]

[tex]259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)[/tex]

[tex]T_f=31.91 ^oC[/tex]

31.9178 °C is the final temperature of the water

The final temperature of the water : 31.916 °C

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

From reaction:

2C₆H₆ (l) + 15O₂ (g) ⟶12CO₂ (g) + 6H₂O (l) +6542 kJ, heat released by +6542 kJ to burn 2 moles of C₆H₆

If there are 6,200 g of C₆H₆ then the number of moles:

mol = mass: molar mass C₆H₆

mol = 6.2: 78

mol C6H6 = 0.0795

so the heat released in combustion 0.0795 mol C6H6:

[tex]\rm Q=heat=\dfrac{0.0795}{2}\times 6542\:kJ\\\\Q=260.0445\:kJ[/tex]

the heat produced from the burning is added to 5691 g of water at 21 ∘ C

So :

Q = m . c . ∆T  (specific heat of water = 4,186 joules / gram ° C)

260044.5 = 5691 . 4.186.∆T

[tex]\rm \Delta T=\dfrac{260044.5}{5691\times 4.186}\\\\\Delta T=10.916\\\\\Delta T=T(final)-Ti(initial)\\\\10.916=T-21\\\\T=31.916\:C[/tex]

the difference between temperature and heat  

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Specific heat  

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relationships among temperature, heat, and thermal energy.  

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When heat is added to a substance  

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Answer:

The answer is lithium(LI)

Answer:

Roughly C100 H140 N3 O

Explanation:

Gilsonite is a bituminous product that resembles shiny black obsidian.

It contains more than 100 elements.

Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.

Its empirical formula is roughly C100 H140 N3 O.

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

Now we have to determine the oxidation state of the elements in the compound.

(a) [tex]H_2SO_4[/tex]

Let the oxidation state of 'S' be, 'x'

[tex]2(+1)+x+4(-2)=0\\\\x=+6[/tex]

Hence, the oxidation state of 'S' is, (+6)

(b) [tex]Ca(OH)_2[/tex]

Let the oxidation state of 'Ca' be, 'x'

[tex]x+2(-2+1)=0\\\\x=+2[/tex]

Hence, the oxidation state of 'Ca' is, (+2)

(c) [tex]BrOH[/tex]

Let the oxidation state of 'Br' be, 'x'

[tex]x+(-2)+1=0\\\\x=+1[/tex]

Hence, the oxidation state of 'Br' is, (+1)

(d) [tex]ClNO_2[/tex]

Let the oxidation state of 'N' be, 'x'

[tex]-1+x+2(-2)=0\\\\x=+5[/tex]

Hence, the oxidation state of 'N' is, (+5)

(e) [tex]TiCl_4[/tex]

Let the oxidation state of 'Ti' be, 'x'

[tex]x+4(-1)=0\\\\x=+4[/tex]

Hence, the oxidation state of 'Ti' is, (+4)

(f) [tex]NaH[/tex]

Let the oxidation state of 'Na' be, 'x'

[tex]x+(-1)=0\\\\x=+1[/tex]

Hence, the oxidation state of 'Na' is, (+1)

Under standard conditions :

E(cell) = E(cathode) - E(anode)

Note : cathode has the larger numeric value and anode has the smaller. Therefore

E(cell) = +1.36V - ( -3.04V)

= 1.36 + 3.04

= +4.40V

Answer:

(a) forward direction

(b) forward direction

(c) backward direction.

Explanation:

Given , the chemical reaction in equilibrium is,

SO₂(g)  + Cl₂(g)  ⇄  SO₂Cl₂ (g)

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equlibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.

(a)

For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.

Hence, reaction will go in forward direction.

(b)

Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂  , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.

Hence, reaction will go in forward direction.

(c)

Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.

Hence, reaction will go in backward direction.

Answer:

so the reaction rate increases by a factor 6.

Explanation:

For the given equation the reaction is first order with respect to both ester and sodium hydroxide

So we can say that the rate law is

[tex]Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}][/tex]

now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration

The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.

The new rate law is

[tex]Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}][/tex]

the final rate = 6 X initial rate

so the reaction rate increases by a factor 6.

The reaction rate of ethyl acetate with sodium hydroxide would increase by a factor of 6 if the concentration of ethyl acetate is increased by half and the concentration of NaOH is quadrupled, as it is first order in both reactants.

The reaction of ethyl acetate with sodium hydroxide is:

This reaction is first order concerning both CH₃COOC₂H₅ and NaOH. The rate law for this reaction can be written as:

If the concentration of CH₃COOC₂H₅ is increased by half, its new concentration becomes 1.5 times its initial concentration. If the concentration of NaOH is quadrupled, its new concentration becomes 4 times its initial concentration. Therefore, the rate of the reaction increases by a factor of:

So, the reaction rate would increase by a factor of 6.

Answer:

Choice D) The nebula is moving away from the observer.

Explanation:

Is the emission here a result of electron transition or thermal radiation?

The red-pink emission here is as a line rather than a continuous spectrum. In other words, the red-pink line observed is a result of electron transition. The energy difference will be constant. That should be the same case on the earth as it is in space at the nebula.

Also, this energy difference does not depend on the temperature of the hydrogen. Only that at higher temperature, low-energy radiations will be less prominent. The wavelength will still be 656 nm when the light was emitted from the nebula.

The wavelength observed on the earth is longer than the wavelength emitted. The Doppler's effect is likely to be responsible. As the star moves away from the earth, the distance that light from the star needs to travel keep increasing. Consider two consecutive peaks from the star. When compared with the first peak, the second peak will need to travel a few more kilometers and will need a few more fractions of a second to get to the earth. It would appear to an observer on the earth that the frequency of the light is lower than it actually is. Accordingly, the wavelength will appear to be longer than it was when emitted from the star.

Conversely, the wavelength will appear shorter if the source is moving toward to observer. For this star, the wavelength appears to be longer than it really is. In other words, the star is moving away from the earth.

The ratio between the speed at which the star moves away from the earth and the speed of the light can be found using the equation: (Source: AstronomyOnline)

[tex]\displaystyle \frac{v}{c} = \frac{\Delta \lambda}{\lambda_0} \approx 0.009[/tex].

The gas cloud is moving away from Earth at about 1% the speed of light due to the Doppler effect.So,option D is correct.

The gas cloud is moving away from Earth at about 1% the speed of light. This can be inferred from the observed shift in the wavelength of the hydrogen emission line from 656.3 nm to 656.6 nm.

The shift in wavelength is due to the Doppler effect, indicating the motion of the source relative to the observer.

Answer : The heat of this reaction of AgCI formed will be, 66.88 KJ

Explanation :

First we have to calculate the heat of the reaction.

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = amount of heat = ?

[tex]c[/tex] = specific heat capacity = [tex]4.18J/g.K[/tex]

m = mass of substance = 120 g

[tex]T_{final}[/tex] = final temperature = [tex]22^oC=273+22=295K[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.8^oC=273+22.8=295.8K[/tex]

Now put all the given values in the above formula, we get:

[tex]q=120g\times 4.18J/g.K\times (295.8-295)K[/tex]

[tex]q=401.28J[/tex]

Now we have to calculate the number of moles of [tex]AgNO_3[/tex] and [tex]HCl[/tex].

[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }AgNO_3=0.20mole/L\times 0.03L=0.006mole[/tex]

[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }HCl=0.10mole/L\times 0.1L=0.01mole[/tex]

Now we have to calculate the limiting reactant.

The balanced chemical reaction will be,

[tex]AgNO_3+HCl\rightarrow AgCl+HNO_3[/tex]

As, 1 mole of [tex]AgNO_3[/tex] react with 1 mole of HCl

So, 0.006 mole of [tex]AgNO_3[/tex] react with 0.006 mole of HCl

From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgCl.

The given balanced reaction is,

[tex]Ag^++Cl^-\rightarrow AgCl[/tex]

From this we conclude that,

1 mole of [tex]Ag^+[/tex] react with 1 mole [tex]Cl^-[/tex] to produce 1 mole of [tex]AgCl[/tex]

0.006 mole of [tex]Ag^+[/tex] react with 0.006 mole [tex]Cl^-[/tex] to produce 0.006 mole of [tex]AgCl[/tex]

Now we have to calculate the heat of this reaction of AgCI formed.

As, 0.006 mole of AgCl produced the heat = 401.28 J

So, 1 mole of AgCl produced the heat = [tex]\frac{401.28}{0.006}=66880J=66.88KJ[/tex]

Therefore, the heat of this reaction of AgCI formed will be, 66.88 KJ

Answer:

-3.7555 M/s is the rate of change of B.

Explanation:

3A + 5B  → 4C

Given that rate of the reaction ,R= 0.7511 M/s

Rate of the reaction is defined as change in concentration of any one of the reactant or product with respect to time.

[tex]R=\frac{-1}{3}\frac{dA}{dt}=\frac{-1}{5}\frac{dB}{dt}=\frac{1}{4}\frac{dC}{dt}[/tex]

[tex]R=0.7511 M/s=\frac{-1}{5}\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt}=5\times 0.7511 M/s=-3.7555 M/s[/tex]

The negative sign indicates the concentration of reactant B is decreasing with progress in time. This mean reactant B is disappearing.

-3.7555 M/s is the rate of change of B.

Answer:

For a: The mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]

For b: The volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]

For c: The edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]

For d: The radius of a copper atom 127.82 pm.

Explanation:

We know that:

Mass of copper atom = 63.55 g/mol

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

If, [tex]6.022\times 10^{23}[/tex] number of atoms occupies 63.55 grams.

So, 1 atom will occupy = [tex]\frac{63.55g}{6.022\times 10^{23}atom}\times 1 atom=1.0553\times 10^{-22}g[/tex]

Hence, the mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]

Copper crystallizes with a face-centered cubic lattice. This means that 4 number of copper atoms are present in 1 units cell.

Mass of 4 atoms of copper atom = [tex]1.0553\times 10^{-22}g/atom \times 4atoms=4.2212\times 10^{-22}g[/tex]

We are given:

Density of copper = [tex]8.93g/cm^3[/tex]

To find the volume of copper, we use the equation:

[tex]\text{Density of copper}=\frac{\text{Mass of copper}}{\text{Volume of copper}}[/tex]

Putting values in above equation, we get:

[tex]8.93g/cm^3=\frac{4.2212\times 10^{-22}}{\text{Volume of copper}}\\\\\text{Volume of copper}=4.726\times 10^{-23}cm^3[/tex]

Hence, the volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]

Edge length of the unit cell is taken as 'a'

Volume of cube = [tex]a^3[/tex]

Putting the value of volume of unit in above equation, we get:

[tex]\sqrt[3]{4.726\times 10^{-23}}cm^3=3.615\times 10^{-8}cm[/tex]

Hence, the edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]

The relation of radius and edge length for a face-centered lattice follows:

[tex]a=r\sqrt{8}[/tex]

Putting values in above equation, we get:

[tex]3.615\times 10^{-8}=r\sqrt{8}\\\\r=1.2782\times 10^{-8}cm[/tex]

Converting cm to pm, we get:

[tex]1cm=10^{10}pm[/tex]

So, [tex]1.2782\times 10^{-8}cm=127.82pm[/tex]

Hence, the radius of a copper atom 127.82 pm.

Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane  to be stable it  must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane  does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.

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Answer:

False

Explanation:

Glucose is a monosachharide carbohydrate, with the molecular formula C₆H₁₂O₆.

Glucose molecule can exist in two forms-

1. Open chain form

2. cyclic form  

The open chain form of the glucose is an unbranched 6 carbon atom  chain. The carbon 1 of the molecule is an aldehyde group and the rest of the five carbon atoms have one hydroxyl group each.

The cyclic form of the glucose can be-

a. Pyranose: The pyranose form is a 6-membered cyclic ring, which consists of 5 carbon atoms and 1 oxygen atom in the ring.

b. Furanose: The furanose form is a 5- membered cyclic ring, which consists of 4 carbon atoms and 1 oxygen atom in the ring.

In an aqueous solution, 99% glucose molecule exists in the cyclic pyranose form as it is energetically more stable.

Therefore, in aqueous solution, the glucose molecule does not prefer the open-chain conformation.

Therefore, the statement is false.

Hey there!:

Total volume of wine = 750ml

volume ℅ of ethanol = 12 %

volume of ethanol = (12ml/100ml)*750ml = 90ml

Density of Ethanol = 0.789 g/ml

Mass of Ethanol = 0.789 g/ml × 90ml = 71.01 g

Molar mass of ethanol = 46 g/mol  Nº of mole of ethanol = Mass/molar mass

=>  71.01 g /46(g/mol)= 1.5437 moles

Hope this helps!

1.541 moles of ethanol are present in a 750 mL bottle of wine.

Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]

The substance per unit volume is called Density. SI unit of density is kg/m.

It is expressed as:

Density = [tex]\frac{\text{Mass}}{\text{Volume}}[/tex]

Volume of ethanol = 12%

                               = [tex]\frac{12}{100}[/tex]

                               = 0.12

Volume of ethanol = 0.12 × 750

                               = 90

Density of ethanol = [tex]\frac{\text{Mass of ethanol}}{\text{Volume of ethanol}}[/tex]

0.789 g/mL = [tex]\frac{\text{Mass of ethanol}}{90}[/tex]

Mass of ethanol = 0.789 × 90

                           = 71.01 g

Now put the value in above formula we get

Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]

                             = [tex]\frac{71.01\ g}{46.06\ \text{g/mol}}[/tex]

                             = 1.541 mol

Thus from the above conclusion we can say that 1.541 moles of ethanol are present in a 750 mL bottle of wine.

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Answer:

-4.15*[tex]10^{-5}mol/(L.s)[/tex]

Explanation:

Average rate            = [tex]\frac{final concentration -initial concentration}{change in time} = \frac{0.1076mol/L-0.1103mol/L}{65s}=-4.15.10^{-5}mol/(L.s)[/tex]

The rate is negative because it is a decomposition and our focus is the reactant which is depleting.

Answer : The partial pressure of helium in this mixture is, 12.4 atm.

Explanation : Given,

Partial pressure of oxygen = 0.10 atm

Total partial pressure = 12.5 atm

Now we have to calculate the partial pressure of helium.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{He}+p_{O_2}[/tex]

where,

[tex]P_T[/tex] = total partial pressure = 12.5 atm

[tex]P_{O_2}[/tex] = partial pressure of oxygen = 0.10 atm

[tex]P_{He}[/tex] = partial pressure of hydrogen = ?

Now put all the given values is expression, we get the partial pressure of the helium gas.

[tex]12.5atm=p_{He}+0.10atm[/tex]

[tex]p_{He}=12.4atm[/tex]

Therefore, the partial pressure of helium in this mixture is, 12.4 atm.

The partial pressure of helium in a deep-sea diving mixture of heliox, given a total pressure of 12.5 atm and an oxygen partial pressure of 0.10 atm, is 12.4 atm.

The subject of this question pertains to the principles of gas laws and partial pressures in a gas mixture, specifically applicable to scuba diving environments where deep-sea divers use a unique mixture of gases like heliox (mixture of helium and oxygen).

When the total pressure is 12.5 atm and the oxygen has a partial pressure of 0.10 atm, the partial pressure of helium in the mixture can be found via subtraction. Hence, the partial pressure of helium in this mixture can be found by subtracting the partial pressure of oxygen from the total pressure. That is, 12.5 atm - 0.10 atm = 12.4 atm. In other words, helium, as part of this heliox mixture, offers a partial pressure of 12.4 atm.

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Answer:

the percentage by mass of Nickel(II) iodide = 23.58%

Explanation:

% by mass of solute = (mass of solute/mass of solution) x 100%

% by mass of NiI2 = (mass of NiI2/mass of solution) x 100%

% by mass of NiI2 = (5.47 grams/23.2 grams) x 100% = 23.58% m/m

Try the suggested option; answer is marked with red colour (18.4953 °C).

All the details are in the attached picture.

The equilibrium temperature of the system is required.

The equilibrium temperature of the mixture is [tex]18.48^{\circ}\text{C}[/tex].

[tex]m_i[/tex] = Mass of ice = 25 kg

[tex]m_s[/tex] = Mass of steam = 4 kg

[tex]T_i[/tex] = Temperature of ice = [tex]0^{\circ}\text{C}[/tex]

[tex]T_s[/tex] = Temperature of steam = [tex]100^{\circ}\text{C}[/tex]

[tex]L_f[/tex] = Latent heat of fusion = [tex]3.34\times 10^5\ \text{J/kg}[/tex]

[tex]L_v[/tex] = Latent heat of vaporization = [tex]2.23\times 10^6\ \text{J/kg}[/tex]

[tex]c_w[/tex] = Specific heat of water = [tex]4180\ \text{J/kg}^{\circ}\text{C}[/tex]

The heat balance of the system will be

[tex]m_iL_f+m_ic_w(T-T_i)=m_sL_v+m_sc_w(T_s-T)\\\Rightarrow m_iL_f+m_ic_wT-m_ic_wT_i=m_sL_v+m_sc_wT_s-m_sc_wT\\\Rightarrow T=\dfrac{m_ic_wT_i+m_sL_v+m_sc_wT_s-m_iL_f}{m_ic_w+m_sc_w}\\\Rightarrow T=\dfrac{25\times 4180\times 0+4\times 2.23\times 10^6+4\times 4180\times 100-25\times 3.34\times 10^5}{25\times 4180+4\times 4180}\\\Rightarrow T=18.49^{\circ}\text{C}[/tex]

The equilibrium temperature of the mixture is [tex]18.49^{\circ}\text{C}[/tex].

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hey there!:

Electron density on Cl atom depends on electronegativity difference between Cl and other bonded atom. If the electrnegativity difference is more then Cl has greater electron density, that menas if the bonded atom has less electronegativity then bonded electrons are more attracted by Cl and it has greater electron density.  

Among the five atoms which are bonded to Cl atom H has low electronegativity. So in H-Cl two bonded electrons are closer to Cl atom as it has greater electronegativity than H. This results more electron density on Cl atom.

Hence in H-Cl bond Cl atom have the highest electron density

Hope this helps!

The bond in which the chlorine atom has the highest electron density is H − C l.

The polarity of a bond depends on the magnitude of electronegativity difference between the atoms in the bond. The greater the electronegativity difference between the atoms in a bond the more the polarity of the bond.

The magnitude of electron density on each atom in a bond depends on its electronegativity. The more electronegative an atom is, the more it is able to accommodate larger electron density. Looking at the options, hydrogen is far less electronegative than chlorine so a large magnitude of electron density resides on the chlorine atom in HCl.

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Answer : The final concentration of the seawater is, 2.909 mole/L

Explanation :

Formula used for osmotic pressure :

[tex]\pi=CRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure  = 70.0 bar = 70 atm

R = solution constant  = 0.0821 Latm/moleK

T= temperature of solution = [tex]20^oC=273+20=293K[/tex]

C = final concentration of seawater = ?

Now put all the given values in the above formula, we get the concentration of seawater.

[tex]70atm=C\times 0.0821Latm/moleK\times 293K[/tex]

[tex]C=2.909mole/L[/tex]

Therefore, the final concentration of the seawater is, 2.909 mole/L

To determine the final concentration of seawater when reverse osmosis stops at a pressure of 70.0 bar and 20 °C, additional data such as the initial osmotic pressure of the seawater is required. Without this data, the final concentration cannot be calculated.

In reverse osmosis, water purification occurs by forcing water from a more concentrated solution to a less concentrated one by applying pressure greater than the osmotic pressure. When a pressure of 70.0 bar is applied to seawater at 20 °C, the process will continue until the osmotic pressure of the sea water is equal to the applied pressure. Since the question asks for the final concentration at which the reverse osmosis stops, we would need information about the initial osmotic pressure of seawater to calculate the final concentration. Typically, however, this value can vary, and since the necessary data to perform the calculation is not provided, we cannot accurately provide the final concentration of the seawater.

Reverse osmosis systems, such as those used in desalination plants, continuously introduce seawater under pressure and collect pure water, hence the process carries on indefinitely and the actual concentration in the plants would be constantly changing based on the amount of seawater processed and pure water extracted.

Answer:

The molecular formula of mono sodium glutamate is [tex]C_5H_8O_4N_1Na_1[/tex]

Explanation:

Molar mass of sodium glutamate,M = 169 g/mol

let the molecular formula be [tex]C_aH_bO_cN_dNa_e[/tex]

Percentage of carbon in the M.S.G. =35.52 %

[tex]35.51\%=\frac{a\times 12 g/mol}{169 g/mol}[/tex]

a = 5

Percentage of Hydrogen in the M.S.G. = 4.77 %

[tex]4.77\%=\frac{b\times 1 g/mol}{169 g/mol}[/tex]

b = 8

Percentage of oxygen in the M.S.G. =37.85 %

[tex]8.29\%=\frac{c\times 16 g/mol}{169 g/mol}[/tex]

c = 3.99 ≈ 4

Percentage of nitrogen in the M.S.G. = 8.29 %

[tex]4.77\%=\frac{d\times 14 g/mol}{169 g/mol}[/tex]

d = 1

Percentage of sodium in the M.S.G. =13.60 %

[tex]13.60\%=\frac{c\times 23g/mol}{169 g/mol}[/tex]

e = 0.99 ≈ 1

The molecular formula be :[tex]C_aH_bO_cN_dNa_e=C_5H_8O_4N_1Na_1[/tex]

The molecular formula of Monosodium glutamate is C2H4O4N2Na2

To determine the molecular formula of monosodium glutamate based on its elemental composition, we'll first find the empirical formula, and then calculate the molecular formula.

Find the moles of each element:

Carbon (C): 35.51%

Hydrogen (H): 4.77%

Oxygen (O): 37.85%

Nitrogen (N): 8.29%

Sodium (Na): 13.60%

Calculate the moles of each element using their molar masses:

Moles of C = (35.51/100) * 169 g / (12.01 g/mol) ≈ 5.97 moles

Moles of H = (4.77/100) * 169 g / (1.01 g/mol) ≈ 7.90 moles

Moles of O = (37.85/100) * 169 g / (16.00 g/mol) ≈ 8.37 moles

Moles of N = (8.29/100) * 169 g / (14.01 g/mol) ≈ 9.99 moles

Moles of Na = (13.60/100) * 169 g / (22.99 g/mol) ≈ 9.43 moles

Find the smallest whole number ratio of moles.

Divide all moles by the smallest number of moles (approximately 5.97).

Empirical Formula:

C1H1.32O1.4N1.68Na1.58

Round the subscripts to whole numbers (since you can't have fractions of atoms):

CH2O2N2Na2

The empirical formula of Monosodium glutamate is CH2O2N2Na2.

To find the molecular formula, you need to determine the molar mass of the empirical formula and compare it to the given molar mass (169 g/mol).

The empirical formula mass is 85 g/mol (approximately), which is half of the molar mass.

Therefore, the molecular formula is twice the empirical formula:

Molecular Formula: C2H4O4N2Na2

Learn more about Monosodium glutamate here:

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Answer:

The correct option is : b) noncompetitive

Explanation:

There are three main types of inhibition:

1. Competitive: In this inhibition, the inhibitor molecule competes with the substrate to bind on the active site of the enzyme.

2. Uncompetitive: In this inhibition, the inhibitor molecule binds to the enzyme- substrate activated complex and thus, does not compete with the substrate to bind on the active site of the enzyme.

3. Non-competitive: In this inhibition, the inhibitor molecule can bind to both the enzyme molecule or to the enzyme-substrate activated complex.

Therefore, In non- competitive inhibition, the inhibitor molecule binds to the enzyme regardless of whether the substrate molecule is bound to the enzyme active site or not.

Answer : The percent yield is, 32.79 %

Explanation :  

First we have to calculate the moles of [tex]CH_3CH_2OH[/tex].

[tex]\text{Moles of }CH_3CH_2OH=\frac{\text{Mass of }CH_3CH_2OH}{\text{Molar mass of }CH_3CH_2OH}=\frac{65.2g}{46.07g/mole}=1.415mole[/tex]

Now we have to calculate the moles of [tex]CH_3CH_2OCH_2CH_3[/tex]

The balanced chemical reaction will be,

[tex]2CH_3CH_2OH(l)\rightarrow CH_3CH_2OCH_2CH_3(l)+H_2O(l)[/tex]

From the balanced reaction, we conclude that

As, 2 moles of [tex]CH_3CH_2OH[/tex] react to give 1 mole of [tex]CH_3CH_2OCH_2CH_3[/tex]

So, 1.415 moles of [tex]CH_3CH_2OH[/tex] react to give [tex]\frac{1.415}{2}=0.7075[/tex] mole of [tex]CH_3CH_2OCH_2CH_3[/tex]

Now we have to calculate the mass of [tex]CH_3CH_2OCH_2CH_3[/tex]

[tex]\text{Mass of ether}=\text{Moles of ether}\times \text{Molar mass of ether}[/tex]

[tex]\text{Mass of }ether=(0.7075mole)\times (74.12g/mole)=52.44g[/tex]

The theoretical yield of ether, [tex]CH_3CH_2OCH_2CH_3[/tex]  = 52.44 g

Now we have to calculate the percent yield of [tex]CH_3CH_2OCH_2CH_3[/tex]

[tex]\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}\times 100=\frac{17.2g}{52.44g}\times 100=32.79\%[/tex]

Therefore, the percent yield is, 32.79 %

Answer:The reaction would shift to the left on adding Br2(g)

             The reaction would shift to right on adding BrNO(g)

Explanation:

The concept can be explained on the basis of LeChateliers principle.

LeChateliers principle explain the effect on equilibrium when a reaction system at equilibrium is subjected to any disturbance or change in conditions then the reaction shifts in such a way so as to reduce the effect of that change or disturbance and again establish the equilibrium.

For example if we add more reactants in a given reaction at equilibrium, the reaction would shift in such a way so that it can reduce the effect of increasing the concentration of reactants and hence the reaction would favor that direction in which it can reduce the concentration of reactants.So when we increase the concentration of reactants  the reaction would move towards the formation of more products and so the concentration of reactants would be less in the reaction.

Likewise if we increase the concentration of products so the reaction would shift in such a way so that it can oppose the increased concentration of products so the reaction moves towards more formation of reactants that is the products decompose to form reactants and reaction moves backwards.

In this reaction:

2BrNO(g)⇄2NO(g)+Br₂(g)

When we add more amount of Br₂(g) the reaction would proceed in such a ways so that oppose the increased concentration of Br₂(g).Hence the reaction would move towards left that is backwards.

When we add more amount of BrNO(g)the reaction would proceed in such a ways so that oppose the increased concentration of BrNO(g).Hence the reaction would move towards right that is forward.

Answer:

50 Joule

Explanation:

Diatomic gas

Q = Heat given = 70 J

n = number of moles

Cp = specific heat at constant pressure

ΔT = Change in temperature

R = Gas constant  

Change in energy

ΔE = Q-w

⇒ΔE = n(Cp)ΔT-nRΔT

As it is a diatomic gas Cp = (7/2)R

Putting the value of Cp in the above equation we get

Q = (7/2)RΔT

ΔE = (5/2)RΔT

Dividing the equations we get

ΔE/Q = 5/7

⇒ΔE = (5/7)Q

⇒ΔE = (5/7)×70

⇒ΔE = 50 J

∴ The internal energy change is 50 Joule

To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.

To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:

2SO2 + O2 → 2SO3

Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:

Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.

Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.

Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.

Next, we compare the theoretical yield to the actual yield to determine the percent yield:

Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%

The percent yield of SO3 in this experiment is 26.32%.

The percent yield of SO3 in this experiment is 26.32%.

To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.

To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:

2SO2 + O2 → 2SO3

Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:

Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.

Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.

Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.

Next, we compare the theoretical yield to the actual yield to determine the percent yield:

Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%

The added KI does not have any impact  

The reaction invovles Titration of vitaminc ( Ascorbic acid)

ascorbic acid + I₂ → 2 I⁻  +  dehydroascorbic acid

the excess iodine is free reacts with the starch  indicator, forming the blue-black starch-iodine complex.  

This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.

Answer B

Hope this helps!

Final answer:

Adding excess potassium iodide (KI) to a vitamin C sample for titration with N-bromosuccinimide (NBS) results in increased production of iodine, which requires more sodium thiosulfate (Na₂S₂O₃) for back titration. Therefore, the amount of NBS needed to titrate the sample will increase.

Explanation:

When a student mistakenly adds a potassium iodide (KI) solution twice while preparing a vitamin C sample for titration with N-bromosuccinimide (NBS), this will lead to the addition of excess KI in the solution. The presence of additional KI will reduce the titrand more, producing a stoichiometric amount of iodine (I₂). This increased amount of I₂ will then be determined by back titration using sodium thiosulfate (Na₂S₂O₃) as a reducing titrant. Adding excess KI does not change the amount of vitamin C in the sample, but it does result in an increased production of I₂ which needs to be titrated with sodium thiosulfate. Therefore, the larger quantity of KI will increase the amount of NBS needed to titrate the sample since there is more I₂ produced that needs to be accounted for during the titration process.

Answer:So we can say that the rate of reaction increases by factor of 6.

Explanation:

The rate law for any given reaction  

A+B⇄C+D

Rate law for the above reaction is:

R=K[A]ᵃ[B]ᵇ

a and b are the order of reaction and it is an experimentally determined quantity.

K is the rate constant and it is constant for a given reaction

[A] and [B] are the concentrations of the reactants.

R is the rate of reaction

For the given reaction :

H₂O₂(aq.)+I₂(aq.)⇆OH⁻(aq.)+HIO(aq.)

Also it is given for the reaction that order with respect to H₂O₂ is 1 and order with respect to I₂ is also 1

The rate law can be written as :

R=K[H₂O₂]¹[I₂]¹

k=rate constant

When we increase the  concentration of H₂O₂ by half which meansthat new concentration of H₂O₂ will be= 3/2[H₂O₂].

When we increase the  concentration of I₂ by 4 which means that new concentration of I₂ will be= 4[I₂]

Putting the values of our new concentration in the rate law:

R=K3/2[H₂O₂]¹4[I₂]¹

R=6K[H₂O₂]¹[I₂]¹=New rate

So as we can see that the new rate of the reaction using the new concentration is 6 times the older rate of reaction.

So we can say that the rate of reaction increases by factor of 6.

Answer : The Lewis structures for the two molecules are shown below.

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 52.2 g

Mass of H = 13.1 g

Mass of O = 34.7 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{52.2g}{12g/mole}=4.35moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.1g}{1g/mole}=13.1moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{34.7g}{16g/mole}=2.17moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{4.35}{2.17}=2.00\approx 2[/tex]

For H = [tex]\frac{13.1}{2.17}=6.03\approx 6[/tex]

For O = [tex]\frac{2.17}{2.17}=1[/tex]

The ratio of C : H : O = 2 : 6 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_2H_6O_1[/tex]  = [tex]C_2H_6O[/tex]

The empirical formula weight = 12(2) + 6(1) + 1(16) = 46 gram/eq

Now we have to calculate the value of 'n'.

Formula used :

[tex]n=\frac{\text{Molecular formula weight}}{\text{Empirical formula weight}}=\frac{45g/mole}{46g/eq}=0.9\approx 1[/tex]

Molecular formula = [tex](C_2H_6O)_n=(C_2H_6O)_1=C_2H_6O[/tex]

So, there are two possibilities for the arrangements of atoms. That means, it will be an ethanol [tex](H_3C-CH_2-OH)[/tex] or dimethyl ether [tex](H_3C-O-CH_3)[/tex].

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

As we know that carbon has '4' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in [tex]C_2H_6O[/tex] = 2(4) + 6(1) + 6 = 20

According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons.

Thus, the Lewis structures for the two molecules are shown below.