At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction
2 NOCl(g) reverse reaction arrow 2 NO(g)+ Cl2(g)

Calculate the concentrations of all species at equilibrium if

2.0 mol NO and 1.0 mol of Cl2 are placed in a 1.0 L flask

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = [tex]\frac{90 g}{47.87 g/mol}=1.8800 mol[/tex]

Total number of atoms of titanium ,[tex]a_t=1.8800 mol\times N_A[/tex]

Mass of aluminum = 6 g

Moles of aluminium = [tex]\frac{6 g}{26.98 g/mol}=0.2223 mol[/tex]

Total number of atoms of aluminium,[tex]a_a=0.2223 mol\times N_A[/tex]

Mass of vanadium  = 4 g

Moles of vanadium= [tex]\frac{4 g}{50.94 g/mol}=0.0785 mol[/tex]

Total number of atoms of vanadium[tex]a_v=0.0785 mol\times N_A[/tex]

Total number of atoms in an alloy = [tex]a_t+a_a+a_v[/tex]

Atomic percentage:

[tex]Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100[/tex]

Atomic percentage of titanium:

:[tex]\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%[/tex]

Atomic percentage of Aluminium:

:[tex]\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%[/tex]

Atomic percentage of vanadium

:[tex]\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%[/tex]

Answer:

Explanation:

Butane is an alkane and has no multiple bond in it.

The isomers of butane are

a) n-butane

b) t-butane

The structures are shown in the figure.

The most stable radical will be tertiary butyl radical.

The structures are shown in figure.

Answer:

On the attached document.

Explanation:

Hello,

At first, if we're looking for the tertiary arrangement we must consider butane's branched structure which is shown on the attached document whose name is tert-butane. Afterwards, one must remove one hydrogen to let the tert-butyl radical, which is stood for a dot above the tertiary carbon, to be formed.

Best regards.

Answer:

Explanation:

Mixing Ammonia gas into a solution of Copper(II) Sulfate will give Ammonium Sulfate and a precipitate of Copper(II) Hydroxide (Cu(OH)₂). The Ksp of Cu(OH)₂ is published => 2.2 x 10⁻²². Such gives a solubility* of the Cu(OH)₂ to be ~1.77 x 10⁻⁷M =>  [Cu⁺²] ~1.77 x 10⁻⁷M and [OH⁻] = 2(1.77 x 10⁻⁷)M = 3.53 x 10⁻⁷M.  The reaction of Ammonium Hydroxide and Copper(II) Sulfate will generate 1 x 10⁻⁴ mole Cu(OH)₂ as a precipitate but only 1.77 x 10⁻⁷ mole of the hydroxide will remain in  1 Liter of solution b/c of extreme limited solubility.

*Solubility of 1:2 ionization ratio salts = CubeRt(Ksp/4).

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion=[tex]\frac{Volume}{Time}[/tex]

Effusion rate of oxygen gas after time t = [tex]E=\frac{4.64 mL}{t}[/tex]

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = [tex]E'=\frac{3.12 mL}{t}[/tex]

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

[tex]\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]

[tex]\frac{E}{E'}=\sqrt{\frac{M'}{M}}[/tex]

[tex]\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}[/tex]

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

r ∝ 1/√M

where,

Under identical conditions and after the same amount of time (t), 4.64 mL of O₂ and 3.12 mL of an unknown gas X effuse. The ratio of their rates of effusion is:

[tex]\frac{rO_2}{rX} = \frac{\frac{vO_2}{t} }{\frac{vX}{t} } = \frac{vO_2}{vX} = \frac{4.64 mL}{3.12 mL} = 1.48[/tex]

We can calculate the molar mass of the unknown gas by applying Graham's law.

[tex]\frac{rO_2}{rX} = 1.48 = \sqrt{\frac{M(X)}{M(O_2)} } = \sqrt{\frac{M(X)}{32.00 g/mol} }\\\\M(X) = 70.1 g/mol[/tex]

Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.

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Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture  = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

[tex]\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100[/tex]

Percent recovery of compound A:

[tex]\frac{83 mg}{119 mg}\times 100=69.74\%[/tex]

Mass of compound B in a mixture  = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

[tex]\frac{79 mg}{97 mg}\times 100=81.44\%[/tex]

Answer: 161.8 torr

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]

where, x = mole fraction

[tex]p^0[/tex] = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2[/tex][tex]p_{total}=x_Ap_A^0+x_BP_B^0[/tex]

[tex]x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{5.50}{5.50+8.50}=0.39[/tex],

[tex]x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{8.50}{5.50+8.50}=0.61[/tex],

[tex]p_{A}^0=264torr[/tex]

[tex]p_{B}^0=96.5torr[/tex]

[tex]p_{total}=0.39\times 264+0.61\times 96.5=161.8torr[/tex]

The total vapor pressure above the solution is 161.8 torr.

An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.

A chemist has a sample with a mass of 894.0 g. When it absorbs 4.90 kJ of heat its temperature increases from -5.8 °C to 17.5 °C. The chemist can calculate the specific heat capacity of the substance using the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

[tex]Q = c \times m \times \Delta T\\\\c = \frac{Q}{m \times \Delta T} = \frac{4.90 kJ}{894.0 g \times (17.5 \° C - (-5.8 \° C))} = 2.35 \times 10^{-4} kJ/g\° C[/tex]

An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.

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The specific heat capacity of the substance is approximately 0.234 J/g°C. This was found by applying the formula for specific heat capacity (q = mcΔT) and doing the necessary calculations with the given values.

To determine the specific heat capacity of the substance, we can use the formula q = mcΔT, where q is the heat energy absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. Given that q equals 4.90 kJ (or 4900 J to match the unit of specific heat capacity), m equals 894.0g and ΔT is 17.5°C - (-5.8°C) = 23.3°C, we can rearrange the formula to solve for c: c = q / (mΔT).

Substituting the given values, we find c = 4900 J / (894.0g * 23.3°C), yielding a specific heat capacity of approximately 0.234 J/g°C to three significant figures.

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Answer: NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

We are given:

Moles of NO = 0.857 mol

Moles of oxygen = 0.498 mol

By stoichiometry of the reaction:

If 2 moles of NO reacts with 1 mole of oxygen gas.

So, 0.857 moles of NO will react with = [tex]\frac{1}{2}\times 0.857=0.4285mol[/tex] of [tex]O_2[/tex]

As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, NO is considered as the limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

If 2 moles of NO produces 2 moles of nitrogen dioxide gas.

So, 0.857 moles of NO will produce = [tex]\frac{2}{2}\times 0.857=0.857mol[/tex] of [tex]NO_2[/tex]

Hence, NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.

Answer:

The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.

Explanation:

4% acetic acid by mass means that 4 gram of acetic acid in 100 g solution.

Given that density of the vinegar is same is that of water =  1 g/mL

Mass of the vinegar solution = 100 g

Volume of the vinegar solution = V

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

[tex]1 g/mL=\frac{100 g}{V}[/tex]

V = 100 mL = 0.1 L

Moles of acetic acid =[tex]\frac{4 g}{60.052 g/mol}=0.0666 mol[/tex]

[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}[/tex]

[tex]M=\frac{0.0666 mol}{0.1 L}=0.6660 mol/L[/tex]

The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.

Answer: The atomic weight of the second isotope is 64.81 amu.

Explanation:

Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]     .....(1)

We are given:

Let the mass of isotope 2 be 'x'

Mass of isotope 1 = 62.9 amu

Percentage abundance of isotope 1 = 69.1 %

Fractional abundance of isotope 1 = 0.691

Mass of isotope 2 = 'x'

Percentage abundance of isotope 2 = 30.9%

Fractional abundance of isotope 2 = 0.309

Average atomic mass of copper = 63.5 amu

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)][/tex]

[tex]x=64.81amu[/tex]

Hence, the atomic weight of second isotope will be 64.81 amu.

By using the weighted average formula and the given details, we can determine the atomic weight of the second naturally occurring isotope of copper approximately to be 64.93 amu.

The atomic weight of an isotope is calculated by adding the products of the abundance percentages and the atomic weights of each isotope. For copper (Cu), one of its isotopes is Copper-63, which makes up 69.1% of naturally occurring copper, and it has an atomic weight of 62.9 amu. Therefore, we can calculate the atomic weight of the second isotope using the given average atomic weight of copper (63.5 amu) as follows:

Average atomic weight = (abundance of Copper-63 * atomic weight of Copper-63) + (abundance of Copper-65 * atomic weight of Copper-65)

63.5 = (0.691*62.9) + (0.309 * x)

After we calculate the first part, we subtract it from 63.5 to solve for the atomic weight (x) of the second isotope, known as Copper-65.

Therefore, the atomic weight of the second isotope of copper is approximately 64.93 amu.

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The concentration of S in the 10.0 mL solution is calculated by considering the initial 8.48 mM concentration and the dilution factor due to the increase in volume to 10.0 mL. The final concentration of S is found to be 0.848 mM.

To determine the concentration of S in the 10.0 mL solution, you must take into account the initial concentration of S and how it changes with dilution. Initially, 1.00 mL of 8.48 mM (millimolar) S was added to the solution. The dilution can be calculated using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume after dilution.

The initial volume of S (V1) is 1.00 mL and the final volume of the diluted solution (V2) is 10.0 mL. Therefore, using the initial concentration of S (C1) as 8.48 mM, we can solve for the final concentration (C2) as follows:

C1V1 = C2V2
(8.48 mM)(1.00 mL) = (C2)(10.0 mL)

Solving for C2 gives us:

C2 = (8.48 mM)(1.00 mL) / (10.0 mL)
C2 = 0.848 mM

The final concentration of S in the 10.0 mL solution is 0.848 mM (millimolar).

Answer: The mass of calcium hydroxide that are present in the solution is 0.3256 grams.

Explanation:

To calculate the moles of a solute, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Volume of hydrochloric acid = 42.7mL = 0.0427 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.208 moles/ L

Putting values in above equation, we get:

[tex]0.208mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0427L}\\\\\text{Moles of hydrochloric acid}=0.0088mol[/tex]

For the given chemical reaction:

[tex]Ca(OH)_2(aq.)+2HCl\rightarrow CaCl_2(aq.)+2H_2O(l)[/tex]

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.0088 moles of hydrochloric acid will react with = [tex]\frac{1}{2}\times 0.0088=0.0044mol[/tex] of calcium hydroxide.

To calculate the mass of calcium hydroxide, we use the equation:

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

Moles of calcium hydroxide = 0.0044 moles

Molar mass of calcium hydroxide = 74 g/mol

Putting values in above equation, we get:

[tex]0.0044mol=\frac{\text{Mass of calcium hydroxide}}{74g/mol}\\\\\text{Mass of calcium hydroxide}=0.3256g[/tex]

Hence, the mass of calcium hydroxide that are present in the solution is 0.3256 grams.

0.325 grams of Ca(OH)₂ must be in the solution of HCl to neutralize a solution.

Molarity of any solution is defined as:

M = n/V, where

n = no. of moles

V = volume

Given chemical reaction is:

Ca(OH)₂(aq) + 2HCl → CaCl₂(aq) + 2H₂O(l)

From the stoichiometry of the reaction, it is clear that:
2 moles of HCl = react with 1 mole of Ca(OH)₂

1 mole of HCl = react with 1/2 mole of Ca(OH)₂

Moles of HCl will be calculated by using the molarity formula and given data as:

Given concentration of HCl = 0.208M

Given volume of HCl = 42.7mL = 0.0427L

Moles of HCl = 0.208 × 0.0427 = 0.0088 moles

0.0088 moles of HCl = react with 0.0088 × 1/2 = 0.0044moles of Ca(OH)₂

Mass of Ca(OH)₂ will be calculated by using the formula:

n = W/M, where

W = required mass

M = molar mass = 72g/mole

W = 0.0044mole × 72g/mole = 0.325 grams

Hence, 0.325 grams of Ca(OH)₂ will be in the solution.

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Answer:

[tex]E_{red}^{0}[/tex] for X is -1.20 V

Explanation:

Oxidation: [tex]2\times[/tex][[tex]X^{2-}(aq.)-2e^{-}\rightarrow X(s)[/tex]]

reduction: [tex]2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)[/tex]

---------------------------------------------------------------------------------------------------

overall:[tex]2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)[/tex]

So, [tex]E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})[/tex]

or, [tex]0.63=-0.57-E_{red}^{0}(X\mid X^{2-})[/tex]

or, [tex]E_{red}^{0}(X\mid X^{2-})= -1.20[/tex]

So, [tex]E_{red}^{0}[/tex] for X is -1.20 V

Answer: The standard reduction potential of X is -1.20 V

Explanation:

For the given chemical equation:

[tex]2X^{2-}(aq.)+2SO_3^{2-}+3H_2O(l)\rightarrow 2X(s)+S_2O_3^{2-}(aq.)+6OH^-[/tex]

The half reaction follows:

Oxidation half reaction:  [tex]X^{2-}(aq.)\rightarrow X(s)+2e^-[/tex]     ( × 2 )

Reduction half reaction:  [tex]2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)E^o=-0.57V[/tex]

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

We are given:

[tex]E^o_{cell}=0.63V[/tex]

Putting values in above equation, we get:

[tex]0.63=-0.57-E^o_{anode}\\\\E^o_{anode}=-0.57-0.63=-1.20V[/tex]

Hence, the standard reduction potential of X is -1.20 V

Answer : The entropy change of the system is, 19.5 J/K

Solution :

First we have to calculate the moles of benzene.

[tex]\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=\frac{17.5g}{78.11g/mole}=0.224moles[/tex]

Now we have to calculate the entropy change of the system.

Formula used :

[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = entropy change of the system = ?

[tex]\Delta H[/tex] = enthalpy of vaporization = 30.7 kJ/mole

n = number of moles of benzene  = 0.224 mole

[tex]T_b[/tex] = normal boiling point of benzene = [tex]80.1^oC=273+80.1=353.1K[/tex]

Now put all the given values in the above formula, we get the entropy change of the system.

[tex]\Delta S=\frac{0.224mole\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K[/tex]

Therefore, the entropy change of the system is, 19.5 J/K

A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.

When a sample of magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity (C) of 3024 J/°C, the temperature increases by 1.126°C (ΔT). We can calculate the heat absorbed by the calorimeter (Qc) using the following expression.

[tex]Qc = C \times \Delta T = \frac{3024J}{\° C} \times 1.126 \° C \times \frac{1kJ}{1000J} = 3.405 kJ[/tex]

According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the reaction (Qr) is zero.

[tex]Qc + Qr = 0\\\\Qr = -Qc = -3.405 kJ[/tex]

3.405 kJ are released by the combustion of 0.1375 g of Mg. The heat released per gram of Mg is:

[tex]\frac{-3.405kJ}{0.1375g} = -24.76 kJ/g[/tex]

Finally, we will convert -24.67 kJ/g to kJ/mol using the molar mass of Mg (24.31 g/mol).

[tex]\frac{-24.76kJ}{g} \times \frac{24.31g}{mol} = -601.9 kJ/mol[/tex]

A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.

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hey there!:

A) Knowing theatre the protease is showing the highest activity at pH 4-6, implies that the amino acid that amino acid that it is acting in is an amino acid with a basic side chain. Therefore, the residues can be any one of the three basic amino acids being histidine, arginine or lysine , having basic side chains at neutral pH.

b) The mechanism of reaction of cysteine proteases is as follows:

 First step in the reaction is the deprotonation of a thiol in the cysteine proteases's active site by an adjacent amino acid with a basic side chain, which might be a histidine residue. This is followed by a nucleophilic attack by the anionic sulfur of the deprotonated cysteine on the substrate carbonyl carbon.

Here, a part of the substrate is released with an amine terminus, restoring the His into a deprotonated form, thus forming a thioester intermediate, forming a link between the carboxy-terminal of the substrate and cysteine, resulting in thiol formation. Thus the name thiol proteases. The thioester bond is then hydrolyzed into a carboxylic acid moiety while again forming the free enzyme.

C) cysteine proteases have a pka of 8-9 but when they are deprotonated by a His residue, their pka would come down to 6-8, which would be their optimal pH for functioning. This is because there is a deprotonation of the thiol group , later restoring the HIS deprotonated form and then formation of a thioester bond. This thioester bond when hydrolysed will a carboxylate moeity , which is responsible for bringing the pH down towards a more acidic side.  

d) at the optimal pH , the fraction of deprotonated cysteine and protonated B will be equal which will change with the change in pH.

Hope this helps!

Answer:

The correct answer is: Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction.

Explanation:

Dynamic equilibrium is a chemical equilibrium between froward reaction and backward or reverse reaction where rate of reaction going forwards is equal to the rate of reaction going backward (reverse).  

Some other properties of dynamic equilibrium are:

Answer:

B: At dynamic equilibrium, the rate of the forward reaction is higher than the rate of the reverse reaction.

Answer:

The main difference between suspension and emulsion polymerization is that suspension polymerization requires a dispersing medium, monomer(s), stabilizing agents and initiators whereas emulsion polymerization requires water, monomer and a surfactant.

Explanation:

Answer:

Molar concentration of [tex]CuSO_4[/tex] solution = 0.0056 moles / liter

Explanation:

Looking at the chemical reaction we realize that the precipitate, formed by adding the iron powder, is cooper.

Then for finding the number of moles of precipitated copper:

[tex] number.of.moles= \frac{mass (g)}{molecular.mass (g/mole)}[/tex]

[tex]number.of.moles.of.solid.copper = \frac{0.089}{63.5} =1.4*10^{3}[/tex]

From the chemical reaction we deduce that [tex]1.4*10^{3}[/tex] moles of Cu equals to [tex]1.4*10^{3}[/tex] moles of [tex]CuSO_4[/tex] in the initial solution.

So molar concentration is defined as:

[tex]molar.concentration = \frac{number.of.moles}{solution.volume (liters)}[/tex]

[tex]molar.concentration.of.CuSO_4.solution= \frac{1.4*10^{3}}{0.250}= 5.6*10^{3} moles/liter = 0.0056 moles/liter[/tex]

Answer:

[tex]M=5.6x10^{-3}M[/tex]

Explanation:

Hello,

In this case, we first must consider the given already-balanced chemical reaction to realize that 89 mg of copper were recovered, moreover we can relate such mass with the employed moles of copper (II) sulfate via this reaction's stoichiometry as follows:

[tex]n_{CuSO_4}=89mg Cu*\frac{1gCu}{1000mgCu} *\frac{1molCu}{63.546gCu} *\frac{1molCuSO_4}{1molCu} \\n_{CuSO_4}=1.4x10^{-3}molCuSO_4[/tex]

Now, if we state the molarity (mol/L) as the required concentration, we apply its mathematical definition as shown below:

[tex]M=\frac{n_{CuSO_4}}{V_{sln}} =\frac{1.4x10^{-3}molCuSO_4}{0.250L} \\M=5.6x10^{-3}M[/tex]

Best regards.

Answer:

2C₈H₁₈(g) + 25O₂(g)→16CO₂(g) + 18H₂O

Explanation:

To balance an equation, the moles of one element on one side of the equation should be the same as those on the other side of the equation. This is because (as a law of thermodynamics), in a chemical reaction, the matter is not destroyed nor created - atoms are only rearranged.

Answer: The coefficients are 2, 17, 8 and 18

Explanation:

A balanced chemical equation always follow law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.  

For the given chemical reaction, the balanced equation follows:

[tex]2C_8H_{18}(g)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]

By Stoichiometry of the reaction:

2 moles of octane reacts with 25 moles of oxygen gas to produce 16 moles of carbon dioxide gas and 18 moles of water vapor.

Hence, the coefficients are 2, 25, 16 and 18

Answer: Option (1) is the correct answer.

Explanation:

Melting point is defined as the point at which a solid state of a substance starts to melt and it converts into liquid state.

Also, it is known that more is the number of linear carbon atoms attached to each other, more is the melting point of a substance. Whereas more is the branching present in a compound least will be its melting point.

For example, [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has only 5 carbon atoms attached to carboxylic acid group. Whereas [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has 9 carbon atoms attached to carboxylic acid group.

Therefore, compound [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] will have low melting point.

Thus, we can conclude that out of the given options, fatty acid [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has the lowest melting point.

Answer:

Mole fraction of alcohols in liquid phase [tex]x_1=0.2727\& x_2=0.7273[/tex].

Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex].

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the  n-propyl alcohol =[tex]p^{o}_1=21.0 Torr[/tex]

Partial vapor pressure of the isopropyl alcohol =[tex]p^{o}_2=45.2 Torr[/tex]

[tex]p=x_1\times p^{o}_1+x_2\times p^{o}_2[/tex]  (Raoult's Law)

[tex]p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2[/tex]

[tex]38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr[/tex]

[tex]x_1=0.2727[/tex]

[tex]x_2=1-0.2727=0.7273[/tex]

[tex]x_1\& x_2[/tex] is mole fraction in liquid phase.

Mole fraction of components in vapor phase [tex]y_1\& y_2[/tex]

[tex]p_1=y_1\times p[/tex] (Dalton's law of partial pressure)

[tex]y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}[/tex]

[tex]y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468[/tex]

[tex]y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}[/tex]

[tex]y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516[/tex]

Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex]

Answer:

[tex]\boxed{3}[/tex]

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

[tex]\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )[/tex]

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

[tex]\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}[/tex]

Data:

[tex]n_{f} = 2[/tex]

λ = 657 nm

Calculation:  

[tex]\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}[/tex]

Using Bohr's model and the energy-wavelength relationship for photons, we can explore the transition of an electron in a hydrogen atom from a higher energy level to a lower one. Given the emitted photon's wavelength is 657 nm, we can calculate the energy difference that represents this transition. Substituting these values into the formula, we get an approximation for the initial principal quantum number ni.

The question pertains to the transition of an electron in a hydrogen atom from a higher principle quantum state (ni) to a lower principle quantum state (n=2). The wavelength of the emitted photon during this transition is 657 nm. The energy levels of atoms are quantized, and this is especially evident in the hydrogen atom. According to Bohr's model, the energy of each level is given by the equation E=-13.6/n² eV, where n is the principal quantum number. The difference in energy ∆E between two levels ni and nf (final), given that energy has to be conserved during the transition, is given to the photon.

 Given this information, and the relationship between energy and wavelength of a photon (E=hc/λ where h is Planck's constant, c is the speed of light, and λ is the wavelength), we can determine the starting principal quantum number, ni. Emission of photons means energy is released from the atom, and hence ∆E = Ef - Ei, will be negative; i.e., the energy of the final state is lesser than the initial state. Substituting the values, we get ni as the integral part of the solution.

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Answer: Both the salts are basic salts.

Explanation:

Salts are formed when an acid reacts with a base during a neutralization reaction.

When a strong acid and a weak base reacts, it leads to the formation of acidic salt.

When a strong base and weak acid reacts, it leads to the formation of basic salt.

When a strong acid and strong base or weak acid and weak base reacts, it leads to the formation of neutral salts.

[tex]Ba_3(PO_4)_2[/tex] is a salt which is formed by the combination of [tex]H_3PO_4\text{ and }Ba(OH)_2[/tex]. Phosphoric acid is a weak acid and barium hydroxide is a strong base. So, the given salt is a basic salt.

[tex]NaCN[/tex] is a salt which is formed by the combination of [tex]HCN\text{ and }NaOH[/tex]. Hydrogen cyanide is a weak acid and sodium hydroxide is a strong base. So, the given salt is a basic salt.

Hence, both the salts are basic salts.

Ba3(PO4)2 forms a basic solution due to the phosphate ions derived from a weak acid, while NaCN also yields a basic solution because the cyanide ion acts as a weak base.

To determine whether an aqueous solution of a salt is acidic, basic, or neutral, one can consider the hydrolysis of the ions formed when the salt dissolves in water. The acid or base strength of the parent acid and base from which the salt is derived plays a key role in this behavior.

Regarding Ba3(PO4)2, barium ions (Ba2+) are from a strong base (barium hydroxide, Ba(OH)2), and phosphate ions (PO43-) are from a weak acid (phosphoric acid, H3PO4). The Ba2+ ion does not hydrolyze significantly because it is from a strong base, whereas the PO43- ions will hydrolyze to produce hydroxide ions (OH-), making the solution basic.

For NaCN, the sodium ion (Na+) does not affect the pH since it is from a strong base (sodium hydroxide, NaOH). However, the cyanide ion (CN-) is a weak base since it comes from a weak acid (hydrocyanic acid, HCN). Therefore, the CN- ion will accept protons from water to produce hydroxide ions (OH-), resulting in a basic solution.

Answer:

(a) backward direction

(b) forward direction

(c) no change

Explanation:

(a)

On increasing the concentration of NO , the reaction will go in backward direction , as, according to le chatelier's principle ,

In a chemical equilibrium , when the concentration , pressure or temperature is changed , the equilibrium of the reaction is disturbed , hence, in order to gain attain the equilibrium the reaction moves in order to counteract the changes.

So, in this case, the reaction goes back , to reduce the increased concentration of NO.

(b)

On increasing the concentration of NO2 , the reaction goes in forward direction ,

according to le chatelier's principle ,

the reaction moves in forward direction , to reduce the increased concentration of NO2.

(c)

No change ,

As, changing the volume , does not effect change in the concentration,

Hence,

No change , as the system is allowed to expand to 5 times its initial value.

Answer:Kindly find the structure of A and B in attachment.

Explanation:

Lindlars catalyst is basically a palladium metal based catalyst which reduces specifically the alkynes to just alkenes and only reduce them in cis manner.

The palladium catalyst is poisoned by lead or quinoline so that the reduction reaction can be stopped at a point when alkynes are just reduced to alkenes because palladium in presence of hydrogen reduces the alkyne completely to alkane.

So lindlars catalys is very specific in its action and even in terms of stereochemistry as it only reduces alkynes to alkenes and the stereochemistry of reduced alkyne is cis.

The general formula of lindlars catalyst is : 5%Pd-CaCO3,Pb(OCOCH3)2 and quinoline.

A chiral alkyne would have all 4 different substituents present at the carbon next to triple bond.

since the reduction of chiral alkyne A would be done using Lindlars catalyst hence the reduced product formed would have Cis -stereochemistry.

The structures of A&B are drawn in attachments. Kindly find in attachment.

The chiral alkyne is 2-hexyne and when it's reduced using H2 and Lindlar catalyst, it forms cis-2-hexene which has the R configuration at its stereogenic center.

The molecule A is a chiral alkyne with the molecular formula C6H10. This indicates that this alkyne molecule contains a chiral center. The most likely structure is 2-hexyne (CH3CH2C≡CCH2CH3), which features a triple bond between the 2nd and 3rd carbons.

When reduced with H2 and Lindlar catalyst, the molecule A is converted into molecule B. The Lindlar catalyst is used specifically to reduce alkynes to cis-alkenes. Thus, molecule B would be cis-2-hexene (CH3CH2CH=CHCH2CH3) as this is the alkene version with a chiral center.

It is specified that molecule B has the R configuration at its stereogenic center. The R or S configuration can be determined by the Cahn-Ingold-Prelog (CIP) priority rules. In the case of cis-2-hexene, assuming the hydrogen atom is bonded to the chiral carbon, a clockwise (R) configuration can be achieved.

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Answer : The fugacity in the solution is, 16 bar.

Explanation : Given,

Fugacity of a pure component = 40 bar

Mole fraction of component  = 0.4

Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.

Now we have to calculate the fugacity in the solution.

Formula used :

[tex]f_i=X_i\times f_i^o[/tex]

where,

[tex]f_i[/tex] = fugacity in the solution

[tex]f_i^o[/tex] = fugacity of a pure component

[tex]X_1[/tex] = mole fraction of component

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

[tex]f_i=0.4\times 40\text{ bar}[/tex]

[tex]f_i=16\text{ bar}[/tex]

Therefore, the fugacity in the solution is, 16 bar.

Answer:

[tex]\boxed{\math{_{31}^{71}\text{Ga}^{3+}}}\boxed{\math{_{35}^{80}\text{Br}^{-}}} \boxed{\math{_{90}^{232}\text{Th}^{4+}}} \boxed{\math{_{38}^{87}\text{Sr}^{2+}}}[/tex]

Explanation:

(a)

If the ion has 28 electrons and a 3+ charge, it normally has 31 electrons. It must also have 31 protons. Atom 31 is gallium.

[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{31}^{71}\textbf{Ga}^{3+}}}[/tex]

(b)

If the ion has 35 protons, it normally has 35 electrons. It has 36 electrons, so it has a negative charge.  Atom 35 is bromine.

A = p + n = 35 + 45 = 80, so the isotopic mass number is 80

\text{The symbol for the ion is }\boxed{\mathbf{_{35}^{80}\textbf{Br}^{-}}}

(c)

If the ion has 86 electrons and a charge of 4+, it normally has 90 electrons. It must also have 90 protons. Atom 90 is thorium.

A = p + n = 90 + 142 = 232, so the isotopic mass number is 232.

[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{90}^{232}\textbf{Th}^{4+}}}[/tex]

(d)

If the ion has 38 protons, it is strontium.

[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{38}^{87}\textbf{Sr}^{2+}}}[/tex]

The symbol of ion will be:[tex]^{71}_{31}Ga^{+3}[/tex]

The symbol of ion will be:[tex]^{80}_{35}Br^{-1}[/tex]

The symbol of ion will be:[tex]^{232}_{90}Th^{+4}[/tex]

The symbol of ion will be:  [tex]^{87}_{38}Sr^{+2}[/tex]

Explanation:

The symbol of an ion of an element is written as:

[tex]^A_ZX^C[/tex]

Where:

A = Mass number of the atom

Z = Atomic number of the atom

X = Symbol the atom

C = Charge on the atom

Given:

a) The ion with a 3+ charge, 28 electrons, and a mass number of 71.

b) The ion with 36 electrons, 35 protons, and 45 neutrons

c) The ion with 86 electrons, 142 neutrons, and a 4+ charge

d) The ion with a 2+ charge, atomic number 38, and mass number 87

To find:

The symbols of given ions

Solution:

a)

Charge on the ion = C = 3+

Number of electrons in the ion = 28

Number of electrons in ion's parent atom = 28 +3 = 31

Number of electrons in parent atom = Number of proton in ion/ paarent atom

= 31 = 31

Atomic number = Number of protons

Atomic number of the ion = Z = 31

Mass number of ion = A = 71

The element with atomic number 31 is gallium with the chemical symbol Ga, so the symbol of ion will be:

[tex]^{71}_{31}Ga^{+3}[/tex]

b)

Charge on the ion = C = ?

Number of electrons in the ion = 36

Number of protons in the ion = 35

Number of neutrons = 45

Number of electrons in parent atom = Number of protons in ion/ parent atom

= 35 = 35

Atomic number = Number of protons

Atomic number of the ion = Z = 35

Number of electrons in ion's parent atom = Number of protons in the ion = 35

Charge on the ion = C = 36 - 35 = 1-

Mass number of ion = Number of protons + Mass of neutrons

A = 35 + 45 = 80

The element with atomic number 35 is bromine with the chemical symbol Br, so the symbol of ion will be:

[tex]^{80}_{35}Br^{-1}[/tex]

c)

Charge on the ion = C = 4+

Number of electrons in the ion = 86

Number of electrons in ion's parent atom = 86+4 = 90

Number of electrons in parent atom = Number of protons in ion/ parent atom

= 90= 90

Atomic number = Number of protons

Atomic number of the ion = Z = 90

Number of neutrons = 142

Mass number of ion = Number of protons + Mass of neutrons

A = 90+ 142 = 232

The element with atomic number 90 is thorium with the chemical symbol Th, so the symbol of ion will be:

[tex]^{232}_{90}Th^{+4}[/tex]

d)

Charge on the ion = C = 2+

Atomic number of the ion = Z = 38

Mass number of ion = A = 87

The element with atomic number 38 is strontium with chemical symbol Sr, so the symbol of ion will be:

[tex]^{87}_{38}Sr^{+2}[/tex]

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Cyclopentadiene is unusually acidic for a hydrocarbon because its conjugate base, the cyclopentadienyl anion, is aromatic and thus highly stable, unlike an unstable diradical.

Cyclopentadiene is unusually acidic for a hydrocarbon due to its ability to form a stable cyclopentadienyl anion, which is aromatic after deprotonation. When a hydrogen atom is removed from cyclopentadiene, the remaining electrons are delocalized around the ring structure, satisfying Huckel's rule (4n+2 π electrons, where n is a non-negative integer), and thus making the anion particularly stable. The conjugate base of cyclopentadiene is not an unstable diradical; instead, it is stabilized by the delocalization of electrons which is a characteristic feature of aromatic compounds like benzene. For comparison, in cyclohexane and other cycloalkanes, carbon atoms are sp³ hybridized and the rings feature different conformations that alleviate strain, such as the chair conformation in cyclohexane, which has minimal ring strain and no aromatic stabilization.