A solution contains some or all of the following ions: Sn4+, Ag+, and Pb2+. The solution is treated as described below: Test 1) Addition of 6 M HCl causes a precipitate to form. Test 2) Addition of H2S and 0.2 M HCl to the liquid remaining from Test 1 produces no reaction. What conclusions can be drawn from the results of these two tests?

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]

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: Option (b) is the correct answer.

Explanation:

The given chemical reaction shows that hydrogen cyanide acid has been added to water which results in the formation of hydronium ion and cyanide ion.

Also, when we add a base like sodium hydroxide (NaOH) to HCN then it will help in accepting a proton ([tex]H^{+}[/tex]) from hydrogen cyanide. As a result, formation of [tex]CN^{-}[/tex] anion will be rapid and easy.

This will make the system not to do any extra work. So, amount of work done by system will decrease.

Thus, we can conclude that out of the given options, add solid NaOH to the reaction (assume no volume change) will decrease the amount of work the system could perform.

Boiling off water from the container will increase the concentration of all species and shift the equilibrium toward the reactants, thus decreasing the potential work the system could perform, aligning with Le Chatelier's Principle. Option A is correct.

Which, action would decrease the amount of work a system, involving the equilibrium reaction of HCN (aq) with water to produce CN⁻ (aq) and H₃O⁺ (aq), could perform. Various actions can shift the equilibrium of this reaction, affecting the system's potential work. According to Le Chatelier's Principle, the system will respond to minimize the effect of any change.

The action that will decrease the amount of work the system could perform is (a) boiling off water from the container, as it favors the formation of reactants over products.

Hence, A. is the correct option.

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:

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:

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 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:

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.

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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.

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:

True

Explanation:

For a stationary fluid, the pressure will vary in the x, y and z directions.

This statement is true.

Answer:

a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.

According to Raoult's law, the statement which is FALSE is:

Raoult's law states that if the pressure of the content of a liquid remains constant to the temperature, then it is proportional to the mole fraction of the mixture.

As a result of this, the vapor pressure of a solvent over a solution is less than the pure solvent and the greater the pressure of a gas over a solution, the greater the solubility.

Therefore, the correct answer is option A because it is false.

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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]

Answer:

a) 40 g

b) 58 g

c) 184 g

Explanation:

The calcium carbonate equivalent of any compound is calculated by calculating it molar mass.

The molar mass of the compound is the calcium carbonate equivalent as it corresponds to 1 mole of the compound and equals to one mole of calcium carbonate then.

a) MgO

Atomic mass of Mg = 24

Atomic mass of O = 16

Molar mass = CCE = 24+16 = 40 g

b) Mg(OH)₂

Atomic mass of Mg = 24

Atomic mass of O = 16

Atomic mass of H = 1

Molar mass of Mg(OH)₂= CCE = 24 + (2X16)+2(X1) = 58g

c) CaMg(CO₃)2

Atomic mass of Mg = 24

Atomic mass of O = 16

Atomic mass of C = 12

Atomic mass of Ca = 40

Molar mass = CCE = 40 + 24 + (2X12) + (6X16) = 184 g.

Answer:

B. Glycolysis involves the breakdown of glucose into pyruvate.

Explanation:

Glycolysis is the biological process whereby one glucose molecule is broken down to two pyruvate molecules. There are a lot of enzymatic processes that are involved in this. It is one of the most important reactions in the world because it allows living cells to harness chemical energy from organic molecules

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 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: 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:

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:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

To prepare 1L of a phosphate buffer at pH 7 with a total concentration of 0.1 M, one would need to mix equal molar amounts of NaH2PO4 and Na2HPO4, resulting in 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4.

The relevant pKa value for the pH target of 7.00 is 6.86, which is close to the second pKa of phosphoric acid. The Henderson-Hasselbalch equation is as follows:

pH = pKa + log([A-]/[HA]), where:

At pH 7.00, the ratio [A-]/[HA] is 1:1 because pH = pKa. Thus, we need equal molar amounts of NaH2PO4 and Na2HPO4. Since the total molarity is 0.1 M, this means we need 0.05 M of each component.
 

Therefore, you would need to weigh 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4 to prepare your buffer.

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

The vapor pressure of methanol at 25°C can be estimated using the Clausius-Clapeyron equation. By plugging in the values for methanol's boiling point, enthalpy of vaporization, and the desired temperature, we can solve for the vapor pressure. The estimated vapor pressure of methanol at 25°C is approximately 0.147 atm.

In order to estimate the vapor pressure of methanol at 25°C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure, temperature, and enthalpy of vaporization. The equation is:

ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1)

Where P1 is the known vapor pressure at a known temperature (T1), P2 is the vapor pressure at the desired temperature (T2), ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Plugging in the values for methanol:

P1 = vapor pressure at boiling point = 1 atm, T1 = boiling point = 65°C = 338 K, T2 = desired temperature = 25°C = 298 K, ΔHvap = 37,400 J/mol, R = 8.314 J/(mol·K)

the equation becomes:

ln(P2/1) = -37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K)

Solving for P2:

P2 = 1 × e ^ (-37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K))

P2 ≈ 0.147 atm

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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 volume of [tex]H_2[/tex] will be, 0.2690 L

Solution :

(a) Steps involved for this problem are :

First we have to calculate the moles of [tex]H_2O[/tex].

Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.

(b) First we have to calculate the moles of [tex]H_2O[/tex].

[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{15g}{18g/mole}=0.833moles[/tex]

The balanced chemical reaction is,

[tex]4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)[/tex]

From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.

Thus, the moles of hydrogen gas = 0.833 mole

Now we have to calculate the volume of hydrogen gas.

Using ideal gas equation,

[tex]PV=nRT[/tex]

where,

n = number of moles of gas  = 0.833 mole

P = pressure of the gas = [tex]745torr=\frac{745}{760}=0.98atm[/tex]

conversion used : (1 atm = 760 torr)

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

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = ?

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

[tex](0.98atm)\times V=(0.833mole)\times (0.0821Latm/moleK)\times (293K)[/tex]

[tex]V=0.2690L[/tex]

Therefore, the volume of [tex]H_2[/tex] will be, 0.2690 L

Answer:

V = 20.4 L

Explanation:

Step 1: Write the balanced equation

4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)

Step 2: Find the moles of H₂

We can establish the following relations.

The moles of H₂ obtained from 15.0 g of H₂O is:

[tex]15.0gH_{2}O.\frac{1molH_{2}O}{18.0gH_{2}O} .\frac{4molH_{2}}{4molH_{2}O} =0.833molH_{2}[/tex]

Step 3: Find the volume of H₂

We will use the ideal  gas equation.

P = 745 torr × (1 atm/760 torr) = 0.980 atm

V = ?

n = 0.833 mol

R = 0.08206 atm.L/mol.K

T = 20°C + 273 = 293 K

P × V = n × R × T

0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K

V = 20.4 L

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:

50 ltr 150 ltr

Explanation:

this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30

ratio of solution 1 and solution 2 =10:30

                                                     =1:3

total mixture is 200 liters

part of solution 1=[tex]\frac{1}{4}[/tex] ×200

                           =50 liters

part of solution 2=[tex]\frac{3}{4}[/tex] ×200

                           =150 liters

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:

[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

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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