Calculating the Molecular Weight and Subunit Organization of a Protein From Its Metal Content The element molybdenum (atomic weight 95.95) constitutes 0.08% of the weight of nitrate reductase. If the molecular weight of nitrate reductase is 240,000, what is its likely quaternary structure?

Answer:

The boiling point of a solution of 500.0 g of ethylene glycol dissolved in 500.0 g of water is 108.258°C.

Explanation:

Elevation in boiling point : [tex]\Delta T_b[/tex]

[tex]\Delta T_b=T_b-T[/tex]

[tex]\Delta T_b=K_b\times m[/tex]

[tex]T_b[/tex] = Boiling point of the solution

T = Boiling point of pure solvent

[tex]K_b[/tex]= Molal elevation constant of solvent

m = molality of the solution

Molality of the  ethylene glycol solution:

[tex]molality=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}[/tex]

Moles of ethylene glycol = [tex]\frac{500.0 g}{62 g/mol}=8.0645 mol[/tex]

Mass of solvent that uis water = 500.0 g = 0.5000 kg

[tex]m=\frac{8.0645 mol}{0.5000 kg}=16.1290 m[/tex]

Molal elevation constant of water =[tex]K_b=0.512^oC/m[/tex]

[tex]\Delta T_b=0.512^oC/m\times 16.1290 m=8.258^oC[/tex]

Boiling point of the solution =[tex]T_b[/tex]

Boiling point of pure water = T = 100°C

[tex]T_b=T+\Delta T_b=100^oC+8.258^oC=108.258^oC[/tex]

The boiling point of a solution of 500.0 g of ethylene glycol dissolved in 500.0 g of water is 108.258°C.

Final answer:

To calculate the boiling point of the solution, use the equation ΔT = Kb * molality, where ΔT is the boiling point elevation, Kb is the boiling point elevation constant, and molality is the molal concentration of the solution. Calculate the molality by dividing the number of moles of ethylene glycol by the mass of water. Substitute the molality into the equation to calculate the boiling point elevation, and add this elevation to the boiling point of pure water (100°C) to find the boiling point of the solution.

Explanation:

To calculate the boiling point of the solution, we need to use the equation:

ΔT = Kb * molality

Where ΔT is the boiling point elevation, Kb is the boiling point elevation constant, and molality is the molal concentration of the solution.

First, we need to calculate the molality of the solution by dividing the number of moles of ethylene glycol by the mass of water. The number of moles of ethylene glycol can be found by dividing the mass of ethylene glycol by its molar mass, and the mass of water is given as 500.0 g.

Once we have the molality, we can substitute it into the equation to calculate the boiling point elevation. Finally, we add this elevation to the boiling point of pure water (100°C) to find the boiling point of the solution.

Answer: Option (D) is the correct answer.

Explanation:

According to Kinetic theory of gases, molecules of a gas move in rapid and random motion. That is, particles are constantly in linear motion.

When these molecules colloid with each other then no energy is gained or lost by them. Also, these molecules occupy negligible amount of volume as compared to the volume of the container in which they are placed.

Moreover, as there is no energy loss taking place so, these molecules of gas undergo perfect elastic motion.

Therefore, we can conclude that all of the above given options are true according to the kinetic theory of gases.

Answer:Option c is incorrect

Explanation:

Diels-Alder reaction is a 4+2 cycloaddition reaction .

The reaction occurs between a diene and a dieneophile .

Generally In the Diels-alder reaction HOMO of the diene and LUMO of the dienophile react with proper orbital symmetry. But vice-versa can also be used.

HOMO-Highest occupied molecular orbital

LUMO-Lowest unoccupied molecular orbital

The primary driving force for the diels alder reaction is is the conversion of 2pi bonds into 2 sigma bonds. The sigma bonds are energetically more favorable than pi bonds.

The diels alder reaction happens through orbital interaction and hence the substituents on either the diene or dienophile do not change there stereochemistry in the product so  it is a stereospecific reaction.Since predominantly only isomer would be produced. so a is correct.

The Diels alder reaction is a concerted( more  bonds form at a time) reaction which means it is just a one step reaction. so statement b is correct.

The option c is incorrect as diels alder reaction occurs through orbital interaction in a pericyclic manner. Since Diels alder reactions are pericyclic in nature and occur through orbital symmetry they do not involve polar intermediates like carbocation or radicals.

The dienes must be conjugated as on account of conjugation stability of a diene also increases. Also since we know that due to conjugation the energy of LUMO decreases and that of HOMO increases and so HOMO is more reactive and generally we involve HOMO of the diene and LUMO of the dienophile. So conjugated dienes are important.

So for a bond formation to take place in a  Diels-Alder reaction HOMO and LUMO with proper symmetry must overlap.

Answer:

C) in either position

Explanation:

There are two kinds of membrane protein:

a) Integral proteins:  They have a fixed or permanent association with the membrane. They are majorly embedded in the middle layer of membrane.   A kind of integral proteins are transmembrane proteins, they can cross the membrane and are path for transport of ions or molecule in and out cell.

b) Peripheral proteins: The are confined to the surface of membrane and are boned with ionic interactions. they are more in number as compared to integral proteins.

Answer : The given equation are not balanced equation.

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on the product side.

The given chemical reaction is,

[tex]HNO_3+NaHSO_3\rightarrow NaNO_3+H_2O[/tex]

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of oxygen are not balanced and the molecule [tex]SO_2[/tex] are missing on the product side.

So, in order to balanced the chemical reaction, the molecule [tex]SO_2[/tex] are added on the product side.

Thus, the balanced chemical reaction will be,

[tex]HNO_3+NaHSO_3\rightarrow NaNO_3+H_2O_SO_2[/tex]

To determine the molality of the commercial grade HCl solution, we calculate the mass of HCl and water in the solution, and the moles of HCl. We plug these values into the molality formula, which gives us a molality of approximately 17.5 m (mol/kg).

To determine the molality of the HCl solution, we first need to understand that molality (m) is defined as the number of moles of solute per kg of solvent. Therefore we need to know the amount of HCl in moles and the weight of water in kg.

Commercial grade HCl solutions are 39.0% HCl by mass. This means that in 1000g of solution there is 390g of HCl. Since the molar mass of HCl is roughly 36.5 g/mol, this means we have about 10.68 moles of HCl in 1000g solution.

The HCl solution has a density of 1.20 g/mL. This means that 1000g (1kg) solution occupy 1000/1.20 = 833.3 mL. Since the solution is made up of HCl and water, we can say that the mass of water = total mass of solution - mass of HCl = 1000g - 390g = 610g (or 0.61 kg).

Then we find the molality of HCl solution by using the formula: Molality (m) = moles of HCl/kg of water = 10.68 moles/0.61 kg = 17.5 m (mol/kg).

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Answer:

For a: The balanced chemical equation is given below.

For b: The mass of carbon dioxide produced will be [tex]1.07\times 10^3g[/tex]

Explanation:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor can be destroyed but it can only be transformed from one form to another form.

This law also states that the total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the given reaction, the balance chemical equation follows:

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]

All the substances are present in gaseous state.

By Stoichiometry of the reaction:

1 mole of propane gas produces 3 moles of carbon dioxide gas.

So, 8.11 moles of propane gas will produce = [tex]\frac{3}{1}\times 8.11=24.33mol[/tex] of carbon dioxide gas.

Now, calculating the mass of carbon dioxide using equation:

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

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 24.33 mol

Putting values in above equation, we get:

[tex]24.33mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=1070.52g[/tex]

Hence, the amount of [tex]CO_2[/tex] produced in the given reaction and expressed in scientific notation is [tex]1.07\times 10^3g[/tex]

Answer: The number of [tex]Fe^{2+}[/tex] ions in one molecule of hemoglobin are 4.

Explanation:

According to mole concept:

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

We are given:

Mass of 1 mole of hemoglobin = [tex]6.8\times 10^4g[/tex]

[tex]6.022\times 10^{23}[/tex] number of molecules have a mass of [tex]6.8\times 10^4g[/tex]

So, 1 molecule of hemoglobin will have a mass of [tex]\frac{6.8\times 10^4g}{6.022\times 10^{23}}\times 1=1.129\times 10^{-19}g[/tex]

It is also given that 0.33 mass % of hemoglobin has [tex]Fe^{2+}[/tex] ions

So, mass of [tex]Fe^{2+}[/tex] ions will be = [tex]\frac{0.33}{100}\times 1.129\times 10^{-19}g=3.7257\times 10^{-22}g[/tex]

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

Given mass of iron ion = [tex]3.7257\times 10^{-22}g[/tex]

Molar mass of iron ion = 55.85 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }Fe^{2+}\text{ ion}=\frac{3.7257\times 10^{-22}g}{55.85g/mol}=6.67\times 10^{-24}mol[/tex]

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

So, [tex]6.67\times 10^{-24}[/tex] moles of hemoglobin will contain = [tex]6.022\times 10^{23}\times 6.67\times 10^{-24}=4[/tex]

Hence, the number of [tex]Fe^{2+}[/tex] ions in one molecule of hemoglobin are 4.

There are [tex]4[/tex] Fe2+ ions in one molecule of hemoglobin. Each hemoglobin molecule contains 4 iron atoms, and all these iron atoms are in the ferrous ion (Fe2+) form

To determine how many Fe2+ ions are present in one molecule of hemoglobin, we can proceed with the following steps:

1. Calculate the mass of Fe2+ in one molecule of hemoglobin:

  Given:

  - Iron (Fe) makes up 0.33 mass % of hemoglobin.

  - Molar mass of hemoglobin (Hb) = 6.8 × 10^4 g/mol.

  Mass of Fe in one mole of hemoglobin:

[tex]\[ \text{Mass of Fe in 1 mol Hb} = 0.33\% \times \text{Molar mass of Hb} \] \[ \text{Mass of Fe in 1 mol Hb} = 0.33 \times 10^{-2} \times 6.8 \times 10^4 \text{ g} \] \[ \text{Mass of Fe in 1 mol Hb} = 2244 \text{ g} \][/tex]

2. Calculate the number of moles of Fe in one mole of hemoglobin:

Now, determine the number of moles of Fe:

  [tex]\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} \][/tex]

  The molar mass of Fe (Fe2+) is approximately 55.845 g/mol (atomic mass of Fe).

[tex]\[ \text{Moles of Fe} = \frac{2244 \text{ g}}{55.845 \text{ g/mol}} \] \[ \text{Moles of Fe} \approx 40.17 \text{ mol} \][/tex]

3. Calculate the number of Fe2+ ions in one molecule of hemoglobin:

  Hemoglobin (Hb) contains 4 iron atoms per molecule, and each iron atom in Hb is present as Fe2+.

  Therefore, the number of Fe2+ ions in one molecule of hemoglobin:

[tex]\[ \text{Number of Fe}^{2+} \text{ ions per molecule of Hb} = 4 \][/tex]

Answer:

Here's what I get.

Explanation:

The Antoine equation is

[tex]\log p = A - \dfrac{B }{C+T}[/tex]

A = 13.7819

B = 2726.81

C = 217.572

I did the calculations and the plots in Excel.

Figure 1 shows the calculations, Figure 2 is the linear plot, and Figure 3 is the log plot.

Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is 15.3 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

[tex]2C(s)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)[/tex]      [tex]\Delta H^o_{rxn}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)[/tex]    [tex]\Delta H_1=131.3kJ[/tex]    ( ×  2)

(2) [tex]CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)[/tex]     [tex]\Delta H_2=-41.2kJ[/tex]

(3) [tex]CH_4(g)+H_2O(g)\rightarrow 3H_2(g)+CO(g)[/tex]     [tex]\Delta H_3=206.1kJ[/tex]

The expression for enthalpy of the reaction follows:

[tex]\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times \Delta H_2]+[1\times (-\Delta H_3)][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(2\times (131.3))+(1\times (-41.2))+(1\times (-206.1))]=15.3kJ[/tex]

Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is 15.3 kJ.

Using the Nernst equation with given concentrations and standard cell potential, the [tex]\(H^+\)[/tex] concentration in the cathode compartment is approximately 7916 M.

To solve this problem, we can use the Nernst equation, which relates the cell potential[tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential[tex](\(E^\circ\)),[/tex] the reaction quotient ((Q)), the temperature ((T)), and the gas constant ((R)).

The Nernst equation is given by:

[tex]\[E_{\text{cell}} = E^\circ - \frac{0.0592}{n} \log(Q)\][/tex]

Where:

[tex]- \(E_{\text{cell}}\)[/tex]= cell potential under non-standard conditions

[tex]- \(E^\circ\)[/tex]= standard cell potential

- (n) = number of moles of electrons transferred in the balanced redox reaction

- (Q) = reaction quotient

- (R) = gas constant[tex](\(8.314 \, \text{J/mol} \cdot \text{K}\))[/tex]

- (T) = temperature in Kelvin

Given:

[tex]- \(E^\circ = 0.76 \, \text{V}\)[/tex]

[tex]- \(E_{\text{cell}} = 0.53 \, \text{V}\)[/tex]

[tex]- \(P_{\text{H}_2} = 1.0 \, \text{atm}\)[/tex]

[tex]- \([\text{Zn}^{2+}] = 1.0 \, \text{M}\)[/tex]

- We know that the number of moles of electrons transferred (\(n\)) is 2 because of the balanced equation.

First, let's find the reaction quotient (\(Q\)) using the given concentrations:

[tex]\[Q = \frac{[\text{Zn}^{2+}][\text{H}_2]}{[\text{H}^+]^2}\][/tex]

Given [tex]\([\text{Zn}^{2+}] = 1.0 \, \text{M}\), \([\text{H}_2] = 1.0 \, \text{atm}\)[/tex], and[tex]\([\text{H}^+]\)[/tex]as the unknown concentration in the cathode compartment, we can substitute these values into the equation.

[tex]\[Q = \frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\][/tex]

Now, we can use the given cell potentials and the Nernst equation to solve for [tex]\([\text{H}^+]\).[/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]

Let's solve this equation step by step:

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0296}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \log\left(\frac{1.0}{[\text{H}^+]^2}\right)\][/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \left(\log(1.0) - \log([\text{H}^+]^2)\right)\][/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]

Now, let's simplify and solve for [tex]\([\text{H}^+]\):[/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]

[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0592 \log([\text{H}^+])\][/tex]

Now, let's isolate \(\log([\text{H}^+])\):

[tex]\[0.53 \, \text{V} - 0.76 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]

[tex]\[-0.23 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]

Now, divide by (-0.0592):

[tex]\[\frac{-0.23 \, \text{V}}{-0.0592} = \log([\text{H}^+])\][/tex]

[tex]\[3.89189 \approx \log([\text{H}^+])\][/tex]

Now, we can find[tex]\([\text{H}^+]\)[/tex] by taking the antilog of (3.89189):

[tex]\([\text{H}^+] = 10^{3.89189}\)[/tex]

[tex]\([\text{H}^+] \approx 7915.82 \, \text{M}\)[/tex]

So, the concentration of [tex]\(H^+\)[/tex] in the cathode compartment is approximately[tex]\(7915.82 \, \text{M}\).[/tex]

The complete net ionic equation for the reaction between HCl(aq) and K₂CO₃(aq) is:

We'll begin by writing the dissociation equation for HCl and K₂CO₃. This is illustrated below:

HCl(aq) —> H⁺(aq) + Cl¯(aq)

K₂CO₃(aq) —> 2K⁺(aq) + CO₃²¯(aq)

In solution, the reaction will proceed as follow:

HCl(aq) + K₂CO₃(aq) —>

2H⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g) + 2Cl¯(aq) + 2K⁺(aq)

Cancel out the spectator ions (i.e Cl¯ and K⁺) to obtain the net ionic equation.

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Answer:

g= n 8.47 and you'll choose the answer...

Explanation:

[tex] \sqrt[x]{2} |3| { \sqrt[ log_{\%g}(3) ]{2} }^{3} {.}^{.83} \geqslant g \times \frac{.83}{0.83} \sqrt[ \geqslant ]{.83} 0.83 \times \frac{32e}{3} \geqslant log_{ \cos(?) }(?) \cos(?) log_{?}(?) e[/tex]

[tex] \sqrt[ \geqslant \sqrt[ log_{ \geqslant log_{ \cot( | log_{ \geqslant love \sqrt[ \geqslant | \sqrt[ \geqslant \geqslant \sqrt[ \geqslant \sqrt[ \geqslant ]{2.10} ]{3.8} ]{love} | ]{2 = 3} }(2 = 6) | ) }(love) }(.) ]{.} ]{.} love\%[/tex]

Answer : The expression for the fugacity coefficient [tex]\ln \phi[/tex], for the mixture is, -0.3669.

Explanation : Given,

Fugacity coefficient of component 1 = 0.784

Fugacity coefficient of component 2 = 0.638

Mole fraction of component 1 = 0.4

First we have to calculate the mole fraction of component 2.

As we know that,

[tex]\text{Mole fraction of component 1}+\text{Mole fraction of component 2}=1[/tex]

[tex]\text{Mole fraction of component 2}=1-0.4=0.6[/tex]

Now we have to calculate the expression for the fugacity coefficient [tex]\ln \phi[/tex].

Expression used :

[tex]\ln \phi=X_1\ln \phi_1+X_2\ln \phi_2[/tex]

where,

[tex]\phi[/tex] = fugacity coefficient

[tex]\phi_1[/tex] = fugacity coefficient of component 1

[tex]\phi_2[/tex] = fugacity coefficient of component 2

[tex]X_1[/tex] = mole fraction of of component 1

[tex]X_2[/tex] = mole fraction of of component 2

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

[tex]\ln \phi=0.4\times \ln(0.784)+0.6\times \ln(0.638)[/tex]

[tex]\ln \phi=-0.3669[/tex]

Therefore, the expression for the fugacity coefficient [tex]\ln \phi[/tex], for the mixture is, -0.3669.

The original sample of potassium phosphate octahydrate had a mass of 19.6 grams. When it was heated, it released 7.93 grams of water.

Further Explanation:

For every mole of the compound potassium phosphate octahydrate, there are 8 moles of water of hydration which can be removed from the crystal by heating without altering the chemical composition of the substance.

To determine how much original sample was used, the amount of water released upon heating may be used as well as the mole ratio of the water of hydration with the compound itself following the steps below:

STEP 1: Convert 7.93 g water to moles.

[tex]moles \ of\ H_{2}O \ = 7.93 \ g \ H_{2}O \ (\frac{1 \ mol \ H_{2}O}{18.00 \ g \ H_{2}O})\\\boxed {moles \ of \ H_{2}O \ = 0.4406 \ mol}[/tex]

STEP 2: Calculate the moles of original sample using the mole ratio: 1 mol K3PO4 8H2O : 8 mol H2O.

[tex]moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.4406 \ mol \ H_{2}O \ (\frac{1 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ }{8 \ mol \ H_{2}O})\\\\\boxed {moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.0551 \ mol}[/tex]

STEP 3: Convert the moles of original sample to mass.

[tex]mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 0.0551 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ (\frac{356.3885 \ g}{1 \ mol\ K_{3}PO_{4}\ 8H_{2}O})\\ mass \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 19.637 \ g[/tex]

Following the significant figures of the given, the final answer should be:

[tex]\boxed {mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 19.6 \ g}[/tex]

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Keywords: water of hydration, hydrate

Answer: The mass of original hydrate is 19.63 grams.

Explanation:

We are given:

Mass of water released = 7.93 grams

We are given a chemical compound known as potassium phosphate octahydrate having chemical formula of [tex]K_3PO_4.8H_2O[/tex]

Mass of [tex]K_3PO_4.8H_2O[/tex] = 356.4 grams

Mass of 8 water of crystallization = (8 × 18) = 144 grams

By applying unitary method, we get:

144 grams of water is released when 356.4 grams of salt is heated.

So, 7.93 grams of water will be released when = [tex]\frac{356.4g}{144g}\times 7.93g=19.63g[/tex] of salt is heated.

Hence, the mass of original hydrate is 19.63 grams.

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

[tex]K_{sp}[/tex] = [tex]5.02\times 10^{-6}[/tex]

First we have to calculate the solubility of [tex]OH^-[/tex] ion.

The balanced equilibrium reaction will be:

[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]

Let the solubility will be, 's'.

The concentration of [tex]Ca^{2+}[/tex] ion = s

The concentration of [tex]OH^-[/tex] ion = 2s

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]

Let the solubility will be, 's'

[tex]K_{sp}=(s)\times (2s)^2[/tex]

[tex]K_{sp}=(4s)^3[/tex]

Now put the value of [tex]K_[sp}[/tex] in this expression, we get the solubility.

[tex]5.02\times 10^{-6}=(4s)^3[/tex]

[tex]s=1.079\times 10^{-2}M[/tex]

The concentration of [tex]Ca^{2+}[/tex] ion = s = [tex]1.079\times 10^{-2}M[/tex]

The concentration of [tex]OH^-[/tex] ion = 2s = [tex]2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M[/tex]

First we have to calculate the pOH.

[tex]pOH=-\log [OH^-][/tex]

[tex]pOH=-\log (2.158\times 10^{-2})[/tex]

[tex]pOH=1.67[/tex]

Now we have to calculate the pH.

[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33[/tex]

Therefore, the pH of a saturated solution is, 12.33

Answer:

misteri Cell ini quest ia half-life of beauty of misteri best, of Cell can't answer =

Explanation:

[tex] \sqrt[ \geqslant { { | \geqslant | \geqslant \sqrt[ \gamma \% log_{ \tan( \sqrt[ < \pi \sqrt[ | \geqslant \sqrt[ < \leqslant |x| ]{y} | \times \frac{?}{?} ]{?} ]{?} ) }(?) ]{?} | | }^{2} }^{?} ]{ \sqrt[ < \gamma log_{ \frac{ | \geqslant y \sqrt[ |x \sqrt{ |?| } | ]{?} | }{?} }(?) ]{?} } [/tex]

A function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds is m (t) = 9 x 2⁻t/¹⁰.

Half life is defined as the amount of time it takes for a radioactive substance (or half its atoms) to break down or change. The time it takes for roughly half of the radioactive atoms in a sample to transform into a more stable form is known as the half-life. The half-life of each radioactive element varies. Half-life is the length of time it takes for a radioactive element to decay to half of its initial value. This suggests that a source's activity has a half-life when it takes time for it to decrease to half its initial value.

The half-life of krypton-91  = 10 s

At time t = 0 a heavy canister contains 7 g of this radioactive gas.
h = 10

m_0 = 7

m(t) = m₀ x 2⁻t/h

m(t) = 7 x 2⁻t/10

Thus, a function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds is m (t) = 9 x 2⁻t/¹⁰.

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Answer:

[tex]\boxed{\text{4.220 mol}}[/tex]

Explanation:

            3NO₂ + H₂O → 2HNO₃ + NO

n/mol:  12.66  

You get 1 mol of NO from 3 mol of NO₂

[tex]\text{Moles of NO} = \text{12.66 mol NO}_{2}\times \dfrac{\text{1 mol NO}}{\text{3 mol NO}_{2}} = \textbf{4.220 mol NO}\\\\\text{The reaction forms } \boxed{\textbf{4.220 mol}} \text{ of NO}[/tex]

Answer: The moles of NO produces are 4.22 moles

Explanation:

We are given:

Moles of nitrogen dioxide = 12.66 moles

The given chemical equation follows:

[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 1 mole of NO

So, 12.66 moles of nitrogen dioxide will produce = [tex]\frac{1}{3}\times 12.66=4.22mol[/tex] of NO

Hence, the moles of NO produces are 4.22 moles

Answer:

8 water molecules

Explanation:

The hydrogen bond may be H-O~H-N or  H-N~H-O; in the first one, the hydrogen bond is between an oxygen atom and a hydrogen which is covalently bonded to a nitrogen atom. The second one is the hydrogen bond of a nitrogen atom with a hydrogen covalently bonded to a oxygen one. The first case would be the hydrogen bonds that water may form with the hydrogen of the urea; the second ones would be the hydrogen bonds that urea may form with water molecules. So, for each nitrogen in urea there would be a hydrogen bond, and for each hydrogen too. Finally, the oxygen in the urea molecule may form hydrogen bonds with water as well, but it has two lone pairs to donate, so the oxygen atom may form hydrogen bond with 2 water molecules:

N=(2 because of the oxygen atom of the urea)+(4 because of the hydrogen bonded to nitrogen)+2(because of the nitrogens).

N=8.

One urea molecule can theoretically form a maximum of four hydrogen bonds with water molecules, two from its NH2 groups and two from the lone pairs of electrons on its oxygen atom.

This capacity to bond with water makes urea an effective compound in the formulation of agricultural fertilizers.

Urea is an organic compound that has the formula (NH2)2CO. It can form hydrogen bonds with water due to the presence of hydrogen atoms in NH2 groups and a lone pair of electrons on the oxygen atom. Through these groups, each urea molecule can form four hydrogen bonds with water molecules: two through the NH2 groups (each nitrogen can form a bond with water) and two through the oxygen atom (each lone pair can form a bond).

NH2 groups in urea can form a bond with water because nitrogen being a more electronegative element compared to hydrogen, can draw electrons towards itself and cause partial positive charge on hydrogen auxiliaries which can then attract the oxygen part of water molecules. Similarly, oxygen in urea can attract hydrogen parts from water molecules due to its lone pair of electrons on it.

Thus, understanding the interaction between urea and water molecules through hydrogen bonding is not only essential in chemistry but also has practical applications, such as in the formulation of fertilizers for agricultural use. It's through this principle that urea can deliver the necessary nutrients for plant growth when mixed with water and applied to soils.

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Answer:

a.

ΔF = w  = 2.94 kJ

ΔSuniv = 0

ΔG  =  0

b.

ΔS = 0.1000×8.315×1.099 = 0.913 J/mol K

ΔQ =0

ΔH = 0

Explanation:

a. You have  to find the ΔG, ΔF, who are two forms of free energy

G: Gibbs free energy

F: Helmholtz free energy

-G: Gibbs free energy:

For solve these, you have the following equation:

ΔG = ΔH   – T ΔS           with T constant (Eq 1)

Where:

ΔG = change in Gibbs free energy

ΔH= change in enthalpy

T = temperature

ΔS = change in entropy

This process is irreversible and isothermic, it last means that temperature doesn’t change.

For that reason:

ΔS  = q/T      with p constant.   (Eq 2)

Where:

q = heat

And, with p constant, it just making P-V work, so:

ΔH = qp =q   (Eq 3)

where:

qp = heat at constant pressure.

-F: Helmholtz free energy

To find ΔF, you have to use the following equation:

ΔF = ΔU   – T ΔS   With T constant,  (Eq 4)

Where:

ΔF = change in Helmholtz free energy

ΔU= change in internal energy

T = temperature

ΔS = change in entropy

And also, you have to use the equation for internal energy:

ΔU = q + w (Eq 5)

b. For this problem we have to establish two states, A and B, based on the data given from the problem:

State A:

V1 = 2 dm³

T1 = 300K

State B:

V2 = 6 dm³

T2 = ?

Due to the adiabatic properties of the process, this expansion  makes that change on heat “q” equals to 0:

Δq= 0

SO we have to ask ourselves what is the value of the change in entropy. But we don’t know if the process is reversible or not. Also, we don’t know if the process is static or not, and the volume could be hard to define.

Answer:

Acetic acid neutralizes added base, and sodium acetate neutralizes added acid.

Explanation:

Acetic acid is the acid that will dissociate to release H⁺ ion which will react and neutralize the  added base.  

CH₃COOH →  H⁺ + CH₃COO⁻

H⁺ + OH⁻ → H₂O

Sodium acetate will dissociate to release the acetate ion (CH₃COO⁻) which will react and neutralize the added acid.  

CH₃COONa →  Na⁺ + CH₃COO⁻

H⁺ + CH₃COO⁻ → CH₃COOH

Based on the information given, the correct option will be A. Acetic acid neutralizes added base, and sodium acetate neutralizes added acid.

In conclusion, the correct option is A.

Learn more about acids on:

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Answer: The enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.

Explanation:

To calculate the number of moles, we use the equation:

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

For ammonia:

Given mass of ammonia = [tex]1.26\times 10^4g=1260g[/tex]

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of ammonia}=\frac{1260g}{17g/mol}=74.11mol[/tex]

We are given:

Moles of ammonia = 74.11 moles

For the given chemical reaction:

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g);\Delta H^o_{rxn}=-92.6kJ[/tex]

By Stoichiometry of the reaction:

If 2 moles of ammonia produces -92.6 kJ of energy.

Then, 74.11 moles of ammonia will produce = [tex]\frac{-92.6kJ}{2mol}\times 74.11mol=-3431.3kJ[/tex] of energy.

Thus, the enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.

Answer:

-34317.56 Kj

Explanation:

=1.26 x 10^4/17

= 741.2 moles

If 2 moles of ammonia gives - 92.6 Kj/mol

What about 741.2 moles

741.2/2 x - 92.6

= - 34317.56 KJ

Answer:  A. Internal energy : May be viewed as the sum of the kinetic and potential energies of the molecules

B. Latent heat: The internal energy associated with the phase of a system.

C. Chemical (bond) energy : The internal energy associated with the atomic bonds in a molecule

D. Nuclear energy : The internal energy associated with the bonds within the nucleus of the atom itself

Explanation:

Internal energy is defined as the total energy of a closed system. Internal energy is the sum of potential energy of the system and the kinetic energy of the system. It is represented by symbol U.

Latent heat is the thermal energy released or absorbed by a thermodynamic system when the temperature of the system does not change. It is thus also called as hidden heat.

Chemical energy is the energy stored in the bonds of molecules.

Nuclear energy is the energy which is stored in the nucleus of an atom called as binding energy within protons and neutrons.

Answer:

Use Ka3

henderson-hasselbach equation: pH = pKa + log [base]/[acid]

pKa3 = - log Ka3 = - log 4.2 x 10^-13 = 12.38

therefore: pH = 12.38 + log (0.28/0.33) = 12.30

the pH is 12.30

Explanation:

phosphoric acid is a polyprotic acid meaning it donates more than one proton

weak Acid  ↔ conjugate Base

H3PO4      ↔   H2PO4^-  corresponding to Ka1

H2PO4^-     ↔   HPO4^2- corresponding to Ka2

HPO4^2-    ↔   PO4^3-   corresponding to Ka3

A buffer consist of a weak acid and its conjugate base, the given buffer has the combination of HPO4^2- and PO4^3- thud we used Ka3

knowing that we used henderson-hasselbach equation to get the pH which is 12.30

Answer:

The particles of the liquid slide around faster as the kinetic energy of the particles increases.

Explanation:

After all the bonds in the solid state are broken in part CD, the more free particles in the liquid state gain more kinetic energy with increase in energy supplied.

The increase in kinetic energy is indicated by the temperature increase thus the positive gradient of the part CD.

Kinetic energy means more vibrations thus the particles slide more and more against each other.

Answer:

For a: The equilibrium concentration of CO and [tex]H_2[/tex] are 0.24 M and 0.32 M.

For b: The value of [tex]K_c\text{ and }K_p[/tex] are 1.5625 and [tex]8.41\times 10^{-4}[/tex]

Explanation:

We are given:

Volume of container = 5 L

Initial moles of CO = 1.8 moles

Initial concentration of CO = [tex]\frac{1.8mol}{5L}[/tex]

Initial moles of [tex]H_2[/tex] = 2.2 moles

Initial concentration of [tex]H_2[/tex] = [tex]\frac{2.2mol}{5L}[/tex]

Equilibrium moles of [tex]CH_3OH[/tex] = 0.6

Equilibrium concentration of [tex]CH_3OH[/tex] = [tex]\frac{0.6mol}{5L}=0.12M[/tex]

The chemical equation for the formation of methanol follows:

          [tex]CO(g)+H_2(g)\rightleftharpoons CH_3OH(l)[/tex]

t = 0     [tex]\frac{1.8}{5}[/tex]       [tex]\frac{2.2}{5}[/tex]            0

[tex]t=t_{eq}[/tex]     [tex]\frac{1.2}{5}[/tex]     [tex]\frac{1.6}{5}[/tex]            [tex]\frac{0.6}{5}[/tex]

So, the equilibrium concentration of CO = [tex]\frac{1.2}{5}=0.24M[/tex]

The equilibrium concentration of [tex]H_2[/tex] = [tex]\frac{1.6}{5}=0.32M[/tex]

The expression of [tex]K_c[/tex] for the given chemical reaction follows:

[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]}[/tex]

We are given:

[tex][CH_3OH]=0.12mol/L[/tex]

[tex][CO]=0.24mol/L[/tex]

[tex][H_2]=0.32mol/L[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.12}{0.24\times 0.32}=1.5625[/tex]

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

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

Where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?

[tex]K_c[/tex] = equilibrium constant in terms of concentration = 1.5625

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature = 525 K

[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=0-2=-2[/tex]

Putting values in above equation, we get:

[tex]K_p=1.5625\times (0.0821\times 525)^{-2}\\\\K_p=8.41\times 10^{-4}[/tex]

Hence, the value of [tex]K_c\text{ and }K_p[/tex] are 1.5625 and [tex]8.41\times 10^{-4}[/tex]

Answer:

  0.64 kW

Explanation:

The potential energy of a mass (M) at some height (h) is computed from ...

  PE = Mgh

At 1 kg/liter, the available power is the rate at which that energy is available ...

  (490 kg/min)×(1 min/(60 s))×(9.8 m/s²)(8 m) ≈ 640.3 kg·m²/s³

  = 640.3 W

In kilowatts, that is 0.64 kW.

Final answer:

The water wheel could have produced approximately 0.641 kW of power by converting the gravitational potential energy of water. This is calculated using the water's mass flow rate and the height of the wheel, considering the density of water is 1000 kg/m³.

Explanation:

The power that the water wheel could have produced can be calculated using the principles of mechanical energy and gravitational potential energy (GPE). The formula for GPE is given by mgh, where m is the mass of the water, g is the acceleration due to gravity (9.8 m/s²), and h is the height.

The flow rate of water is 490 liters per minute, which we convert to cubic meters per second (m³/s) for our calculations, as 490 liters per minute is equal to 0.49 m³/min or 0.00817 m³/s, given that 1 cubic meter equals 1000 liters. Using the density of water, which is 1000 kg/m³, the mass flow rate of water is calculated as the product of the flow rate by the density (0.00817 m³/s * 1000 kg/m³ = 8.17 kg/s).

Thus, the power (P) in watts (W) is P = mgh which translates into P = 8.17 kg/s * 9.8 m/s² * 8 m. Upon calculation, this gives a power of approximately 640.88 watts, which we convert to kilowatts (kW) by dividing by 1000, resulting in approximately 0.641 kW.

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of [tex]MgCl_2[/tex] (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of [tex]MgCl_2[/tex] (solute) = 3.37 g

First we have to calculate the mass of solute.

[tex]\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2[/tex]

[tex]\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g[/tex]

Now we have to calculate the mass of solution.

[tex]\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g[/tex]

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg[/tex]

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

[tex]\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%[/tex]

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

[tex]\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%[/tex]

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

hey there!:

Volume in liters ( V ) = 111 mL / 1000 => 0.111 L

Pressure in atm ( P )  = 734 / 760 => 0.965789 atm

temperature in Kelvin ( K )  = 34+273.15 => 307.15 K

Molar gas constant ( R ) = 0.0826 atm*L/mol*K

ideal gas  equation :

p*V = n*R*T

moles of gas:

n =  p*v / R*T

n = 0.965789 * 0.111 / 0.0826 * 307.15

n = 0.107202 / 25.37059

n = 0.004225 moles

Therefore:

Molar mass =  mass / moles of gas

molar mass = 0.168 / 0.004225

molar mass = 39.76 g/mol

Hope this helps!

The molar mass of the gas was calculated by applying the ideal gas law, rearranging it for molar mass, and using the given quantities in the problem. Conversions to appropriate units were made, and the molar mass was found to be 31.1 g/mol.

The molar mass of a gas can be calculated by applying the ideal gas law, PV = nRT. By rearranging this to solve for the molar mass, using known quantities from the problem, and converting units appropriately, we find:

Applying these steps, we find that the molar mass of the gas is approximately 31.1 g/mol.

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hey there!:

As El Tation is located at higher altitude than Old Faithful, the atmospheric pressure would be less at Ei Tatio.  

Due to low pressure, the boiling point of water will be reduced at El Tatio. Hence, at lower temperature than Old Faithful, geysers will erupt at El Tatio due to depression of boiling point at reduced atmospheric pressure.

Hope this helps!

Geysers at El Tatio would erupt at a lower temperature than Old Faithful due to the significantly higher elevation and consequent lower atmospheric pressure.

The geysers at the El Tatio geyser field, which is located at an elevation of 4,320 meters above sea level, would erupt at a lower temperature compared to Old Faithful, which is at 2,240 meters. This is due to the decrease in atmospheric pressure with increasing altitude. As atmospheric pressure decreases, the boiling point of water also decreases. For example, at sea level with an atmospheric pressure of standard 760 mm Hg, water boils at 100°C, but at higher altitudes such as in Denver, Colorado (1,600 meters), water boils at approximately 95°C. Therefore, with El Tatio being significantly higher in altitude compared to Old Faithful, its geysers would erupt at a lower temperature, which also impacts activities such as cooking.