A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder? (1 LLaTeX: \cdot⋅atm = 101.3 J) Group of answer choices

Answer:

The correct option is: D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

Explanation:

The standard enthalpy change of a given chemical reaction ([tex]\Delta H^{\circ }_{r}[/tex]) is equal to the difference of the sum of formation enthalpy ([tex]\Delta H^{\circ }_{f}[/tex]) of products and the sum of formation enthalpy of reactants.

[tex]\Delta H^{\circ }_{r} = \sum v. \Delta H^{\circ }_{f}(products) - \sum v. \Delta H^{\circ }_{f}(reactants)[/tex]

The standard enthalpy of formation is zero for elements that exist in their standard states.

The standard enthalpy change of a reaction is equal to the standard enthalpy of formation of methanol (l) [[tex]\Delta H^{\circ }_{r} = \Delta H^{\circ }_{f}(CH_{3}OH, l)[/tex]], only if:

1. The standard enthalpy of all the other reactants and products should be zero, i.e. all the other reactants and products should be in their standard states.

2. The stoichiometric coefficient of methanol (CH₃OH, l) should be 1.

Among the given options, only the option D satisfies both the conditions.

D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

The standard enthalpy change for this chemical reaction:

[tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [\Delta H^{\circ }_{f}(C, s, graphite) + \Delta H^{\circ }_{f}(H_{2}, g) + \Delta H^{\circ }_{f}(O_{2}, g)\right ][/tex]

The standard enthalpy of formation:

[tex]\Delta H^{\circ }_{f}(C, s, graphite) = 0 kJ/mol[/tex]

[tex]\Delta H^{\circ }_{f}(H_{2}, g) = 0 kJ/mol[/tex]

[tex]\Delta H^{\circ }_{f}(O_{2}, g) = 0 kJ/mol[/tex]

⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [0 kJ/mol + 0 kJ/mol + 0 kJ/mol\right ] [/tex]

⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ][/tex]

Therefore, the correct option is (D).

Answer:

The correct answer is 1 the total number of proton plus neutrons in the products and reactants must be same.

Explanation:

During a radioactive reaction the nucleus of the radioactive atom undergo disintegration or breakdown to form new nucleus.

   Radioactive disintegration deals with the emission of α particle or β particle from a radioactive nucleus resulting in the formation of new nucleus.Although new nucleus is formed the total number of proton plus neutrons in product and reactant must be same.

A  rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

A nuclear equation for a radioactive reaction, shows the reactants and product, where the proton number and atomic mass are conserved, unlike in chemical equations.

In a radioactive decay, there is emission of α particle or β particle from a radioactive nucleus to form a new nucleus.

Therefore, a rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

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Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

Explanation:

Fe3+ + SCN- -----> [FeSCN]2+

a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]  

[Ca2+]= 0.050 M   [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

Answer:

There is 182 mL of 3.00 M H2SO4 needed

Explanation:

Step 1: Data given

Mass of BaO2 = 92.5 grams

Molarity of 3.00 M H2SO4

Step 2: The balanced equation

BaO2 ( s ) + H2SO4(aq ) ⟶ BaSO4(s) + H2O2(aq )

For 1 mole of BaO2  we need 1 mole of H2SO4 to produce 1 mole of BaSO4 and 1 mole H2SO2

Step 3: Calculate moles of  BaO2

Moles BaO2 = mass BaO2 / Molar mass BaO2

Moles BaO2 = 92.5 / 169.33 g/mol

Moles BaO2 = 0.546 moles

Step 4: Calculate moles of H2SO4

For 1 mole BaO2 we need 1 mole of H2SO4

For 0.546 moles of BaO2 we need 0.546 moles of H2SO4

Step 5: Calculate the volume of H2SO4 needed

Volume = moles / molarity

Volume = 0.546 mol/ 3.00 mol/L

Volume = 0.182 L = 182 mL

There is 182 mL of 3.00 M H2SO4 needed

Answer:

CCl₄, because it is the heaviest compound.

Explanation:

When a liquid is in a closed container, evaporation occurs, in what the molecules with the highest kinetic energy can scape of the liquid. The vapor that was formed does pressure on the liquid that remains, and when both phases stay in equilibrium the pressure is called vapor pressure.

We can notice that the vapor pressure is a measure of the volatility of a liquid. Substances with higher vapor pressure are more volatile. The volatility, however, depends on the nature of the forces in the compound and on the molar mass of it.

For the substances given, they are all covalent compounds and have dipole-induced-dipole-induced bonds between the molecules (because they are nonpolar). So, the lowest vapor pressure is in the heaviest compound, which is the most substituted: CCl₄.

Final answer:

Carbon tetrachloride (CCl4) has the lowest vapor pressure at room temperature due to its highest molecular weight and strongest intermolecular forces among the products of the reaction between methane and chlorine.

Explanation:

Methane, CH4, reacts with chlorine, Cl2, to produce chlorinated hydrocarbons such as methy chloride (CH3Cl), methylene chloride (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4). Looking at the structures of these molecules, we can see that carbon tetrachloride (CCl4) has the highest number of chlorine atoms, which are highly electronegative and contribute to more significant intermolecular forces like London dispersion forces. Consequently, the compound with the highest molecular weight and strongest intermolecular forces will exhibit the lowest vapor pressure at room temperature, which, in this case, is carbon tetrachloride (CCl4).

a. 2257.7 kW  b. 6.057kw/K

Steam existing at 100KPa with t = 20C, we can obtain values of enthalpy and entropy from table

h1 = hf = 417.4 kJ/kg

s1 = 1.302 KJ/kg.K

h2 = hg = 2675.1 kJ/kg

s2 = 7.359 kJ/K

a. Electrical Power consumption of system is given by Wel = mass x Change in Enthalpy

Wel = 1 x (h2 - h1) = 1 x (2675.1 - 417.4) = 2257.7 kw

b . Entropy Generation is given by Change in Entropy  = Entropy generation - heat dissipated/temperature, since we have a well insulated system with no losses, q =0, hence

ΔS = Sgen + 0/T = Sgen

m(S2 - S1) = Sgen

Sgen = 1 x (7.359 - 1.302) = 6.057 kW/K

Answer:

[tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

Explanation:

Initial pressure of NOBr is "x" atm.

34% = 0.34

The equilibrium chemical reaction is a follows.

       [tex]2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)[/tex]

Initial          x                                      0          0

Change    -0.34x                             +0.34x    +0.17x

Final        (x-0.34x)                          +0.34x     +0.17x

Total pressure = sum of the pressure at equilibrium.

[tex]0.25atm=x-0.34x+0.34x+0.17x[/tex]

[tex]0.25atm= 1.17x[/tex]

[tex]x= \frac{0.25}{1.17}=0.213[/tex]

Partial pressure of NOBr=  x-0.34x =x (1-0.34)

= 0.213(0.66) = 0.14 atm.

Hence, partial pressure of [tex][]P_{NOBr}][/tex] is 0.14 atm.

[tex]P_{NO}=0.34x = 0.34 \times 0.213 = 0.072[/tex]

[tex]P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm[/tex]

From equation (1)

[tex]K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}[/tex]

[tex]= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}[/tex]

Therefore, [tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex]K_{p}= K_{c}(RT)^{\Delta n}[/tex]

Rearrange the equation is as follows.

[tex]K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}[/tex]

[tex]\Delta n[/tex] = number of moles of reactants - Number of moles of products

= 3-2 = 1

[tex]= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}[/tex]

Therefore,[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

The [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

[tex]2NOBr\rightleftharpoons 2NO+Br_2[/tex]

​When the initial pressure of [tex]NOBr[/tex] is p, then at equilibrium, therefore the dissociation percentage of [tex]NOBr \ \ is\ \ 34\%[/tex]

[tex]\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p\\\\[/tex]

Calculating the total pressure:

[tex]\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm\\\\[/tex]

[tex]\to p=0.214 \ atm\\\\[/tex]

The equilibrium partial pressures:

[tex]\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm\\\\[/tex]

Calculating the [tex]K_p[/tex]:

[tex]\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}\\\\[/tex]

          [tex]=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm[/tex]

Calculating the [tex]K_c[/tex]:

Using formula:

[tex]\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}[/tex]

            [tex]= (2+1) - 2\\\\= 3 - 2\\\\=1 \\[/tex]

The relationship between [tex]K_p[/tex] and [tex]K_c[/tex] is:

[tex]\to K_p = K_C(RT)^{\Delta ng}[/tex]

Hence, by putting the values of  [tex]K_p[/tex], gas constant [tex]R, T,[/tex] and [tex]\Delta ng[/tex], we can calculate the value of [tex]K_c[/tex].

[tex]\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1[/tex]

[tex]\to K_c = \frac{0.0096 }{0.0821\times 298}[/tex]

         [tex]= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}[/tex]

Therefore, the [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

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Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex]  [tex]\Delta H=-2220.1kJ/mol[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})][/tex]

[tex]\Delta H_{C_3H_8}=72.4kJ/mol[/tex]

Therefore, the heat of formation of propane is 72.4 kJ/mol

The heat of formation of propane can be found using Hess's Law and the given enthalpy formations of CO2 and H2O, along with the enthalpy combustion of propane. Calculations reveal a heat of formation for propane of -103.85 kJ/mol.

The heat of formation of propane can be calculated using Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the pathway between the initial and final states. In this case, we know the enthalpy of combustion for propane (C3H8) is -2220.1kJ/mol, and that the standard enthalpy of formation (ΔHf°) for CO2 and H2O are -393.5kJ/mol and -241.8kJ/mol, respectively.

Therefore, by rearranging the equation ΔH(combustion) = Σ(ΔHf° products) - Σ(ΔHf° reactants), we can solve for the ΔHf° of propane.

Inserting the given values into the equation, -2220.1 kJ/mol = [ 3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [x + 5(0)], where x represents the ΔHf° of propane, and '5(0)' is the contribution from the oxygen, which is zero because the change in enthalpy formation for an element in its stable form is zero. Thus, solving for x, you get a heat of formation for propane of -103.85 kJ/mol.

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Answer: The decreasing order of [tex]K_{sp}[/tex] is [tex]AgSCN>AgBr>AgCN[/tex]

Explanation:

The balanced equilibrium reaction for the ionization of silver bromide follows:

[tex]AgBr\rightleftharpoons Ag^{+}+Br^-[/tex]

                 s      s

The expression for solubility constant for this reaction will be:

[tex]K_{sp}=[Ag^{+}][Br^-][/tex]

We are given:

[tex]s=7.3\times 10^{-7}mol/L[/tex]

Putting values in above equation, we get:

[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}[/tex]

Solubility product of AgBr = [tex]5.33\times 10^{-13}[/tex]

The balanced equilibrium reaction for the ionization of silver cyanide follows:

[tex]AgCN\rightleftharpoons Ag^{+}+CN^-[/tex]

                    s      s

The expression for solubility constant for this reaction will be:

[tex]K_{sp}=[Ag^{+}][CN^-][/tex]

We are given:

[tex]s=7.7\times 10^{-9}mol/L[/tex]

Putting values in above equation, we get:

[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}[/tex]

Solubility product of AgCN = [tex]5.33\times 10^{-17}[/tex]

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

[tex]AgSCN\rightleftharpoons Ag^{+}+SCN^-[/tex]

                     s      s

The expression for solubility constant for this reaction will be:

[tex]K_{sp}=[Ag^{+}][SCN^-][/tex]

We are given:

[tex]s=1.0\times 10^{-6}mol/L[/tex]

Putting values in above equation, we get:

[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}[/tex]

Solubility product of AgSCN = [tex]1.0\times 10^{-12}[/tex]

The decreasing order of [tex]K_{sp}[/tex] follows:

[tex]AgSCN>AgBr>AgCN[/tex]

Answer:

Statements Y and Z.

Explanation:

The Van der Waals equation is the next one:

[tex] nRT = (P + \frac{an^{2}}{V^{2}})(V -nb) [/tex] (1)

The ideal gas law is the following:

[tex] nRT = PV [/tex] (2)

where n: is the moles of the gas, R: is the gas constant, T: is the temperature, P: is the measured pressure, V: is the volume of the container, and a and b: are measured constants for a specific gas.  

As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2), by the incorporation of the intermolecular interaction between the gases and the gases volume. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, taking into account the volume occuppied by the gas in the total volume, which implies a reduction of the total space available for the gas molecules.          

So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.            

Have a nice day!

Answer:

Explanation: AH, AG and AS for the reaction at 873 K

delta G =-RT ln Kp

G= -8.314 x 873 ln 0.9

= 764.7J/mol

AG = AH-TAS

AH = 764.7 - 8314(0.9)

Kp =pCO/pCo2

Substitute the values at different temp. Solve simultaneously

Since mole fraction =kp/pCo

Mf =1.5

Answer:

you looking beautiful

To find the molar mass of compound X, we use the formula moles of solute = (m)(mass of solvent), and then divide the mass of the solute by the moles of solute to calculate the molar mass. In this case, the molar mass of compound X is 5510 g/mol.

To find the molar mass of compound X, we need to use the formula:

Molality (m) = moles of solute / mass of solvent (in kg)

In this case, we know the molality (m) = 1.0 m, and the mass of the solvent (benzonitrile) = 100 g = 0.1 kg. We can rearrange the formula to solve for the moles of solute:

moles of solute = (m)(mass of solvent)

moles of solute = (1.0 m)(0.1 kg) = 0.1 mol

To calculate the molar mass, we divide the mass of the solute by the moles of solute:

Molar mass = mass of solute / moles of solute

Molar mass = 551 mg / 0.1 mol = 5510 g/mol

Therefore, the molar mass of compound X is 5510 g/mol.

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Answer:

The solution boils at 102.76 °C

Explanation:

Δt = m* Kb x i  

i = 1 with 1 particle in solution

m = molality

Kb = 0.512 °C/molal

Step 2: Calculate moles C2H6O2

molar mass of  C2H6O2 = 62.068 g/mol

Calculate number of moles:

Moles = Mass / molar mass

Moles = 1400 grams / 62.068 g/mol

Moles C2H6O2 = 22.56 moles

Step 3: Calculate molality

these are in 4192 g of water:

22.56 moles / 4.192 kg water

⇒ moles / kg of water = 5.38 moles / Kg = m olal

Step 4: Calculate  ΔT

ΔT= Kb * molal = 5.38 molal* 0.512 °C/m

ΔT = 2.76 °C

Boiling point = 100°C + 2.75 °C = 102.76 °C

the solution boils at 102.76 °C

The boiling point of the ethylene glycol solution is calculated by first finding the molality of the solution and then using this to find the boiling point elevation. The normal boiling point of water is thus raised by this amount, giving a final boiling point of the solution as 102.75 degrees Celsius.

To solve this problem, we need to calculate the molality of the ethylene glycol solution and then use this to determine the boiling point elevation of the solution.

Firstly, we need to convert the mass of ethylene glycol (C2H6O2) and water into moles. The molar mass of ethylene glycol is approximately 62.07 g/mol, so 1.4 kg (or 1400 g) of ethylene glycol is equivalent to approximately 22.54 moles. The mass of water is 4192 g, or roughly 4.192 kg.

The molality (m), is thus given by the formula m = moles of solute / kg of solvent, therefore the molality of the ethylene glycol solution is 22.54 moles / 4.192 kg = 5.38 m.

Next, we use the formula for boiling point elevation: ΔTb = Kb * m, where Kb is the ebullioscopic constant for water (given as 0.512 °C/m), and m is the molality calculated earlier. Therefore, the boiling point of the solution rises by ΔTb = 0.512 × 5.38 = 2.75 °C.

The normal boiling point of water is 100 °C, thus, the boiling point of the ethylene glycol solution is 100 °C + 2.75 °C = 102.75 °C.

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Answer:

The molar heat of solution of potassium hydroxide = 57.7 kJ /mol

Explanation:

Step 1: Data given

Mass of potassium hydroxide = 4.21 grams

Volume of water = 250 mL

Temperature rise = 4.14 °C

Density = 1g/mL

Specific heat = 4.184 J/g°C

Step 2: Calculate the heat absorbed by water

q = m*c*ΔT

⇒ with m = the mass of water = 250 grams

⇒ with c= the specific heat of the solution = 4.184 J/g°C

⇒ with ΔT = 4.14 °C

q = 250 * 4.184*4.14 = 4330.4 J

Step 3: Calculate moles of KOH

Moles KOH = Mass KOH / Molar mass KOH

Moles KOH = 4.21 grams / 56.106 g/mol

Moles KOH = 0.075 moles

Step 4: Calculate molar heat of solution

4330.4 J / 0.075 moles = 57738.7 j/mol = 57.7 kJ /mol

(Note that the enthalpy change for the reaction is negative because the reaction is exothermic)

The molar heat of solution of potassium hydroxide = 57.7 kJ /mol

Final answer:

To calculate the molar heat of solution of potassium hydroxide, use the formula q = m × C × ΔT and divide the heat absorbed by the moles of potassium hydroxide.

Explanation:

To calculate the molar heat of solution of potassium hydroxide, we first need to find the amount of heat absorbed by the solution. This can be done using the formula:

q = m × C × ΔT

Where q is the heat absorbed, m is the mass of the solution, C is the specific heat capacity, and ΔT is the change in temperature. Rearranging the formula, we get:

q = molar heat of solution × moles of potassium hydroxide

Now, we can solve for the molar heat of solution by dividing the heat absorbed by the moles of potassium hydroxide.

NO2Cl is a polar molecule because the electronegativity difference between atoms of the molecule, being the chloride tho most electronegative. This fact and the asymmetry of the molecule makes a dipole moment.

In order to don’t include formal charges in the drawing you can use the image number 1. Nevertheless is more usual to use a lewis structure like image number 2.

Answer:

A. Soaps react with ions in hard water to create a precipitate.

B. Soaps are both hydrophobic and hydrophilic.

D. Soaps should be weakly alkaline in solution.

Explanation:

A. Hard water contains magnesium and calcium minerals like calcium and magnesium carbonates, sulfates and bicarbonates. As soon as these minerals come in contact with soap their ions like Mg²⁺ & Ca²⁺ form precipitates.

B. Soap are both hydrophilic and hydrophobic. They reason why they exhibit both the properties is really important for their functionality. The hydrophobic part of soap makes interaction with oil/dust particles while the hydrophilic part makes interaction with water. When the cloth is rinsed the dirt/soap particles are removed from the dirty clothes thereby making them clean.

C. Soaps have alkaline pH i.e. more than 7 that is why they have bitter taste.

Statements A, B, D, and E correctly describe soap: they react with hard water ions to create precipitate, are hydrophilic and hydrophobic, they are weakly alkaline in solution, and they are a mixture of fatty acid salts and glycerol. Statement C is incorrect, as soaps do not work well in all pH conditions.

The following statements accurately describe soap:

Statement C. Soaps work in solutions of any pH, however, is not correct. Soap works best in slightly alkaline conditions.

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The theoretical yield of Li3N is 1.24 mol or 138.16 g. The percent yield of the reaction is 4.23%.

To find the theoretical yield of Li3N, we first need to determine the limiting reactant. This is done by comparing the moles of each reactant to the stoichiometric ratio of the balanced equation. The molar mass of Li is 6.94 g/mol and the molar mass of N2 is 28.02 g/mol. First, convert the given masses of Li and N2 to moles using their respective molar masses:

12.7 g Li X (1 mol Li / 6.94 g Li) = 1.83 mol Li

34.7 g N2 X (1 mol N2 / 28.02 g N2) = 1.24 mol N2

Then, divide the number of moles of each reactant by its stoichiometric coefficient to find the mole ratio:

1.83 mol Li / 6 = 0.305 mol Li3N

1.24 mol N2 / 1 = 1.24 mol Li3N

Since N2 gives a smaller value, it is the limiting reactant. Therefore, the theoretical yield of Li3N is 1.24 mol or 138.16 g.

To calculate the percent yield, divide the actual yield by the theoretical yield, then multiply by 100:

Percent yield = (Actual yield / Theoretical yield) X 100

Percent yield = (5.85 g / 138.16 g) X 100 = 4.23%

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Answer:

[tex]Cx = .376 \times 0.26 = 0.0978 wt% N[/tex]

Explanation:

Given details

concentration of surface is Cs 0.26 wt%

distance from surface is x 2.8 mm

time t = 3.6 h

original concentration is Co 0

we know that

[tex]\frac{Cx - Co}{Cs - Co} = 1 - erf(\frac{x}{2\sqrt{Dt}}[/tex]

[tex]\frac{Cx - 0}{0.26 - Co} = 1 - erf(\frac{2\times 10^{-3}}{2\sqrt{2.2\times 10^{-11} 5 hr (3600 /1 hr)}}[/tex]

[tex]\frac{Cx}{0.26} = 1 - erf(0.626)[/tex]

FROM TABLE OF ERROR FUNCTION WE HAVE

Z           erf(z)

0.60        0.6039

0.65         0.6420

so by interpolation technique we have

for z = 0.626 we have value of erf(0.626) =   0.623

[tex]\frac{Cx}{0.26} = 1 - 0.623[/tex]

[tex]Cx = .376 \times 0.26 = 0.0978 wt% N[/tex]

The concentration in weight percent from the surface is 0.0962%.

Given the following data:

Note: The original concentration of iron (Co) is 0.

To determine the concentration in weight percent:

In this exercise, we would apply Fick's second law and error function:

[tex]\frac{C_x - C_o}{C_s - C_o} =1-erf\frac{x}{2\sqrt{Dt} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{2.2 \times 10^{-10} \times 22680} })\\\\\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{4.99 \times 10^{-6}} })\\\\\frac{C_x }{0.26 } =1-erf(\frac{2.8 \times 10^{-3}}{2\times 0.0022 })\\\\\frac{C_x }{0.26} =1-erf(0.6364)\\\\\frac{C_x}{0.26 } =1-0.63\\\\\frac{C_x}{0.26 } =0.37\\\\C_x = 0.26 \times 0.37[/tex]

Cx = 0.0962%.

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Answer:

CH3CHCHCOOH : 11 sigma and 2 pi bonds

HCCCH2CHCHNH2: 12 sigma and 3 pi bonds

H2CCHCH2CCH: 10 sigma and 3 pi bonds

Explanation:

>In this carboxylic acid there are 11 sigma bonds between C-H, C-C ,C-O and  O-H atoms,while there are two pi bonds ,one is made by carbonyl carbon and other from alkene C=C carbons

> In this amine there are 12 sigma bonds made by C-C,C-H and N-H atoms,while 2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms

>In this organic molecule there are 10 sigma bonds made by C-C and C-H bonds ,while  2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms.

Answer:

The number of sigma and pi bonds in (a) 11 sigma, 2 pi, (b) 12 sigma, 3 pi, (c) 10 sigma and 3 pi bonds.

Explanation:

Sigma bonds are formed by the linear interactions of the two atoms. The pi bonds are covalent bonds formed at top and bottom of the sigma bonds.

(a) [tex]\rm CH_3CHCHCOOH[/tex]

The compound is bonded with 7 Hydrogen with sigma bond. There is 3 sigma bond in between the 4 carbon atoms. One carbon atom is attached with OH group of carboxylic acid with 1 sigma bond. So, total, 11 Sigma bonds are present in the structure.

The pi bonds are 2 in number, 1 in between C-C in backbone. The pi bond is in carboxylic group in between C-O.

(b) [tex]\rm HCCCH_2CHCHNH_2[/tex]

There are 12 sigma bonds in the backbone of given amine. The number of pi bonds are 3. There are two pi bonds in between the alkyene carbons and one pi bond is present in the alkene carbon-carbon bond.

(c) [tex]\rm H_2CCHCH_2CCH[/tex]

The structure comprises of 10 sigma bonds in the backbone. There is presence of 3 pi bonds in the given compound structure.

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Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5

- There are transferred 5 moles of e-

Explanation:

This is the reaction:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Let's think the oxidation numbers:

Fe2+ changed to Fe3+

It has increased the oxidation number → OXIDATION

Mn in MnO₄⁻ acts with +7 and it changed to Mn²⁺

It has decreased the oxidation number → REDUCTION

Let's make the half reactions:

Fe²⁺ → Fe³⁺  +  1e⁻    (it has lost 1 mol of e⁻)

MnO₄⁻ + 5e⁻ →  Mn²⁺  (it has gained 5 mol of e⁻)

Now we have to ballance the O. In acidic medium we complete with water as many oxygens we have, in the opposite side. We have 4 O in reactant side, so we fill up with 4 H2O in products side.

MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Now we have to ballance the H, so as we have 8 H in products side, we complete with 8H⁺ in reactants, this is the complete half reaction:

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Notice that have 1e⁻ in oxidation and 5e⁻ in reduction. We must multiply (x5) the half reaction of oxidation, so the electrons can be cancelled.

(Fe²⁺ → Fe³⁺  +  1e⁻ ) .5

5Fe²⁺ → 5Fe³⁺  +  5e⁻  

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

We sum both half reactions:

5Fe²⁺  + 8H⁺  + MnO₄⁻ + 5e⁻ →  5Fe³⁺  +  5e⁻   + Mn²⁺  + 4H₂O

The electrons are cancelled, so the ballanced reaction is this:

5Fe²⁺  + 8H⁺  + MnO₄⁻  →  5Fe³⁺  + Mn²⁺  + 4H₂O

The given redox reaction balances to 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l), with coefficients of 8 for H⁺ and 5 for Fe3+ in the balanced reaction. A total of 5 moles of electrons are transferred.

To balance the given redox reaction, you need to separate the two half-reactions, one for oxidation and one for reduction. In the case of Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq), the oxidation half-reaction is Fe2+(aq) → Fe3+(aq) + e⁻, and the reduction half-reaction is MnO4⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn2+(aq) + 4H2O(l).

After balancing the half-reactions, you can add them together to get 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l).  This reaction shows that the coefficients in front of H⁺ and Fe3+ in the balanced reaction are 8 and 5, respectively. Also, the stoichiometric coefficients in front of the electrons in the two half-reactions indicate that a total of 5 moles of electrons are transferred during the reaction.

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Answer:

See the figure

Explanation:

As the first step, the Cl leaves the molecule and the double bond is moved to obtain a tertiary carbocation (the most stable one). Then the molecule of water attacks the primary carbon of the double bond, therefore the double bond moves again to the initial place. Finally, a hydronium ion is produced to remove the positive charge of the oxygen.

Answer:

The IUPAC of 1-pentanone is not correct.

Explanation:

The structures of each of the given compounds with correct IUPAC names are shown in the figure.

In case of ketones there must be two alkyl groups attached to the carbonyl carbon. In case of aldehydes there must be atleast one hydrogen attached to carbonyl carbon.

Now if we are saying that 1-pentanone, it means we are actually naming the compound with wrong functional group. This is infact an aldehyde with the correct name as "pentanal".

Rest of the given IUPAC names are correct.

All solutions A through E, being strong acids at the same concentration, will have virtually the same pH value under the same conditions. None will have a significantly higher pH. In general, the solution with the highest pH would be the one derived from the weakest acid or strongest base.

The pH scale is used to determine the acidity or basicity of a solution. The solutions given are all acidic, being solutions of hydrobromic acid (HBr), hydroiodic acid (HI), hydrofluoric acid (HF), hydrochloric acid (HCl), and perchloric acid (HClO4), respectively. Each of these acids dissociates in water to a different extent, affecting the pH of the solution. However, common to all of these acids is that they are strong acids and will fully dissociate in water.

In this case, all solutions have the same molarity (concentration) of 0.10 M. Because all the acids are strong acids and will fully dissociate, the pH of these solutions will be virtually the same. Therefore, none of the solutions will have a significantly higher pH than the others if measured under the same conditions.

In general, to identify the solution with the highest pH (or lowest acidity), you would look for the weakest acid or the strongest base out of the options provided.

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Among the given solutions, C) 0.10 M HF(aq) has the highest pH because HF is a weak acid and only partially dissociates in water, resulting in a higher pH compared to the strong acids HCl, HI, HBr, and HClO₄.

To determine which solution has the highest pH, we need to understand that pH is a measure of the hydrogen ion concentration in a solution. A lower pH signifies a higher hydrogen ion concentration. Since all the given solutions are acids, we must determine which one produces the least amount of hydrogen ions (H⁺).

HCl, HI, HBr, and HClO₄ are strong acids, meaning they fully dissociate in water. This full dissociation results in a high concentration of H⁺ ions, giving them all a pH of 1 for a 0.10 M solution:

HCl: pH = 1.0

HI: pH = 1.0

HBr: pH = 1.0

HClO4: pH = 1.0

HF, however, is a weak acid and only partially dissociates in water. This partial dissociation results in fewer H+ ions compared to the strong acids listed. Consequently, the pH of a 0.10 M HF solution will be higher (less acidic) than the pH of the other solutions.

Given this information, the solution that has the highest pH is:

C) 0.10 M HF(aq)

Answer:

[tex]V_f=552 mL[/tex]

Explanation:

Initially:

Total pressure: 735 mmHg

Water vapor pressure: 21 mmHg

Hydrogen pressure: 714 mmHg

(This is because the total pressure is divided between both gases)

When water vapor is removed:

Total pressure: 735 mmHg

Hydrogen pressure: 735 mmHg

Assuming ideal gases:

Boyle-Mariotte's Law:

[tex]P_i*V_i = P_f * V_f[/tex]

[tex]V_f=\frac{P_i*V_i}{P_f}[/tex]

[tex]V_f=\frac{714 mmHg*568 mL}{735 mmHg}[/tex]

[tex]V_f=552 mL[/tex]

Answer:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Explanation:

In the anode takes place the oxidation, in which the reducing agent loses electrons.

In the cathode takes place the reduction, in which the oxidizing agent gains electrons.

The corresponding half-reactions are:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Answer:

V = 18.8 L

Explanation:

This question is solved by using the ideal gas law :

PV = nRT ∴  V= nRT/P

where n= 0.788 mmol = 0.788 mol (1 mmol is a thousandths of a mol)

R = R constant for ideal gases =  0.08206 Latm/Kmol

T = 17.0 ºC =  (17.0 + 273) K = 290 K

and V = volume in liters our unknown

Plugging our values and solving for V,

V = 0.788 mol x 0.08206 Latm/Kmol x 290 K / 1atm = 18.8 L

Applying the Ideal Gas Law with the given number of moles, temperature and pressure shows that the volume of dinitrogen difluoride gas collected is approximately 19.0 L.

The volume of a gas can be calculated using the Ideal Gas Law if the number of moles, temperature and pressure are known. In your question, we're given that 788mmol of dinitrogen difluoride gas is evolved at a temperature of 17.0 degree C and a pressure of 1 atm. Firstly, convert the temperature to Kelvins by adding 273.15 to the Celsius temperature, giving us 290.15K.

The Ideal Gas Law is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/ K·mol for this situation) and T is temperature. We can rearrange the equation to solve for volume: V=nRT/P. Therefore, substituting in the known values: V = 0.788 mol * 0.0821 L·atm/ K·mol * 290.15K / 1 atm, we find that the volume of dinitrogen difluoride gas collected is approximately 19.0 L, given correct to the number of significant digits.

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Answer:

1) 1.092x10⁻¹² M

2) 4.89x10⁻⁶ M

3) Ag₃PO₄

4) Yes, it is.

Explanation:

1) The dissolution reaction of AgI is:

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

The constant of equilibrium Kps is:

Kps = [Ag⁺]*[I⁻]

So, the minimum concentration of Ag⁺ is that one in equilibrium

8.30x10⁻¹⁷ = [Ag⁺]*7.60x10⁻⁵

[Ag+] = 1.092x10⁻¹² M

2) The dissolution reaction of Ag₃PO₄ is:

Ag₃PO₄(s) ⇄ 3Ag⁺(aq) + PO₄³⁻(aq)

The constant of equilibrium, Kps is:

Kps = [Ag⁺]³*[PO₄³⁻]

So, the minimum concentration of Ag⁺ is that one in equilibrium

8.90x10⁻¹⁷ = [Ag⁺]³*0.760

[Ag⁺]³ = 1.171x10⁻¹⁶

[Ag⁺] = 4.89x10⁻⁶ M

3) Because Ag₃PO₄ needs less Ag⁺ to precipitated, it will precipitate first.

4) To see if the separation will occur, let's verify is the ratio of the molar concentration of Ag⁺ for the substances is less then 0.1%:

(1.092x10⁻¹²)/(4.89x10⁻⁶) * 100% = 2.23x10⁻⁵ %

So, the separation is complete.

The Ksp is used to denote the extent of dissolution of compounds in water.

1) The equilibrium for the dissolution of AgI is set up as follows:

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

Hence;

Ksp = [Ag⁺] [I⁻]

At equilibrium, the minimum concentration of Ag⁺ is

8.30x10⁻¹⁷ = [Ag⁺]*7.60x10⁻⁵

[Ag+] = 1.092x10⁻¹² M

2) The equilibrium for the dissolution of Ag₃PO₄ is set up as follows:

Ag₃PO₄(s) ⇄ 3Ag⁺(aq) + PO₄³⁻(aq)

Hence:

Ksp = [Ag⁺]³*[PO₄³⁻]

At equilibrium, the minimum concentration of Ag⁺ is

8.90x10⁻¹⁷ = [Ag⁺]³*0.760

[Ag⁺]³ = 1.171x10⁻¹⁶

[Ag⁺] = 4.89x10⁻⁶ M

3) The salt that is expected to precipitate first is Ag₃PO₄.

4) In order to determine if the separation is complete;

(1.092x10⁻¹²)/(4.89x10⁻⁶) * 100% = 2.23x10⁻⁵ %

This is less than 0.10% hence the separation is complete.

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Using the Combined Gas Law, the temperature is converted to Kelvin and pressure to atm, then the initial and final conditions are plugged into the formula to find that the final volume is approximately 0.808 liters.

The problem presented involves a gas law calculation where we need to find the final volume of the gas when measured at a different temperature and pressure while keeping the amount of gas constant. This is a straightforward application of the Combined Gas Law, which is stated as (P1 × V1) / T1 = (P2 × V2) / T2. Making sure to convert temperatures to Kelvin and pressures to atmospheres if they're not already can help in solving the problem accurately.

First, let's convert the temperatures from Celsius to Kelvin by adding 273.15. Therefore, 37 °C becomes 310.15 K and 18 °C becomes 291.15 K. Secondly, the pressure needs to be converted from mmHg to atm by dividing by 760. Thus, 695 mmHg becomes approximately 0.915 atm. Now we apply the Combined Gas Law:

P1 × (V1 / T1) = P2 × (V2 / T2)

(1.00 atm × 0.790 L) / 310.15 K = (0.915 atm × V2) / 291.15 K

After cross-multiplying and solving for V2, we find that the final volume of the gas when measured at 18°C and 695 mmHg, with a constant amount of gas, is approximately 0.808 liters.