The value of the Solubility Product Constant for lead phosphate is ________Write the reaction that corresponds to this Ksp value.

_(Aq,S,L) +_(Aq,S,L) <-------->_(Aq,S,L) +_____(Aq,S,L)

Ksp values are found by clicking on the "Tables" link.

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Answers

Answer 1

Answer:

Answer: The reaction for the $K_(sp)$ value of lead phosphate is given below and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is expressed as $K_(sp)$

The chemical formula of lead phosphate is $Pb_3(PO_4)_2$

The equation for the hydration of the lead phosphate is given as:

$Pb_3(PO_4)_2(s)+H_2O(l)\rightarrow 3Pb^(2+)(aq.)+2PO_4^(3-)(aq.)$

The solubility product of lead phosphate is $3.0\rightarrow 10^(-44)$ . This means that it is highly insoluble in water as the solubility product is very very low.

Hence, the reaction for the $K_(sp)$ value of lead phosphate is given above and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

Answer:

The activation energy is $=8.1\,kcal\,mol^(-1)$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_(A)=kC_(A)$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^(3)$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_(o)=5dm^(3)s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= (v_(0))/(k)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]$

Rearrange the formula is as follows.

$k= (v_(0))/(V)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_(o)}$ is 1.

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^(-1)$

The rate constant in case of the CSTR can be calculated by using the formula.

$(V)/(v_(0))= (X(1+\epsilon X))/(k(1-X)).............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_(o)}$ is 0.5

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^(-1)$

The activation energy of the reaction can be calculated by using formula

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

Assuming that the distances between the two ions are the same in all cases, which of the following ion pairs has the greatest electrostatic potential energy (i.e., largest in magnitude)? Please explain your answer.a.) Na+ - Cl- b.) Na+ - O-2. c.) Al+3 - O-2. d.) Mg+2-O-2 e.) Na- -Mg+2

Answers

Answer:

Correct option: C

Explanation:

As given in the question that distance between two ions are same in all cases hence r is same for all.

potential energy:

$P.E =(k* q_(1) * q_(2))/(r)$

therefore potential energy depend on the two charge muliplication

so higher the charge multiplication higer will be the potential energy.

magnitude of charge multiplication follow as:

a. 1

b. 2

c. 6

d. 4

e. 2

in option C it is higher

so correct option is C

Which one of these could be in the unknown anion "X" in this acid: H3Xcarbonate
fluorate
nitrogen
nitrite

Could you explain how to find this? The process?

Answers

Answer:

the answer is nitrogen

(b) Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.
True/False

Answers

Answer:

True

Explanation:

Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.TRUE.

This is the complete combustion reaction of hexane, which proceeds according to the following equation.

C₆H₁₄(l) + 9.5 O₂(g) → 6 CO₂(g) + 7 H₂O(g)

If the combustion were incomplete, instead of carbon dioxide, the product would be carbon monoxide or carbon.

The steps in a reaction mechanism are as follows. Which species is acting as a catalyst? Step 1: Ag+(aq) + Ce4+(aq) <-----> Ag2+(aq) + Ce3+(aq) Step 2: Tl+(aq) + Ag2+(aq) -----> Tl2+(aq) + Ag+(aq) Step 3: Tl2+(aq) + Ce4+(aq) -----> Tl3+(aq) + Ce3+(aq)

Answers

The specie which is acting as a catalyst is; Ag+(aq).

Discussion:

The catalyst is a specie that exists in the same form at the beginning and end of the reaction.

The reaction's mechanism is as follows;

Step 1: Ag+(aq) + Ce⁴+(aq) <-----> Ag²+(aq) + Ce³+(aq)

Step 2: Tl+(aq) + Ag²+(aq) -----> Tl²+(aq) + Ag+(aq)

Step 3: Tl²+(aq) + Ce⁴+(aq) -----> Tl³+(aq) + Ce³+(aq)

Evidently, although Ag+(aq) was converted to Ag²+(aq) in Step 1 of the reaction; the Ag²+(aq) is reverted back to Ag+(aq) in Step 2 of the reaction.

brainly.com/question/22498090

Answer:

Ag⁺ acts as the catalyst.

Explanation:

Hello,

In this case, each step is reorganized:

- Step 1:

$Ag^+(aq) + Ce^(4+)(aq) \rightleftharpoons Ag^(2+)(aq) + Ce^(3+)(aq)$

- Step 2:

$Tl^+(aq) + Ag^(2+)(aq) \rightarrow Tl^(2+)(aq) + Ag^+(aq)$

- Step 3:

$Tl^(2+)(aq) + Ce^(4+)(aq) \longrightarrow Tl^(3+)(aq) + Ce^(3+)(aq)$

In such a way, Ag⁺ is converted to Ag²⁺ in the first step, but then it is regenerated to simple Ag⁺, therefore, Ag⁺ acts as the catalyst.

Best regards.

For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram

Answers

Answer:

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

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