The solubility of CO2 in water at 25°C and 1 atm is 0.034 mol/L. What is its solubility under atmospheric conditions? (The partial pressure of CO2 in air is 0.0003 atm.) Assume that CO2 obeys Henry’s law.

Answer:

the molecular formula of the compound is N2O4

Explanation:

- Find the empirical formula

mole of N present = mass of N divided by molar mass of N = 0.140/14 = 0.01 mole

mole of O present = mass of O divided by molar mass of O = 0.320/16 = 0.02 mole

Divide both by the smallest number of mole to determine the coefficient of each, the smallest number of mole is 0.01 thus:

quantity of N = 0.01/0.01 = 1

quantity of O = 0.02/0.01= 2

thus the empirical formula = NO2

- Now determine the molecular formula by finding the ratio of molecular formula and empirical formula

Molar mass of molecular formula = 92.02 amu = 92.02 g/mole

Molar mass of empirical formula NO2 = (14 + (16 x 2)) = 46 g/mole

the x factor = 92.02/46 = 2

Molecular formula = 2 x NO2 = N2O4

Final answer:

The molecular formula of the compound with 0.140 grams of N and 0.320 grams of O and a molecular mass of 92.02 amu is N₂O₄.

Explanation:

To determine the molecular formula of the compound, we first need to calculate the number of moles of nitrogen (N) and oxygen (O) in the sample. Using the atomic masses of N (14.0 amu) and O (16.0 amu), we divide the mass of each element in the sample by its respective atomic mass to find the moles:

Moles of O = 0.320 g / 16.0 g/mol = 0.02 mol

Next, we determine the simplest whole number ratio of the moles of each element, which gives us the empirical formula. In this case, the ratio of N to O is 1:2, so the empirical formula is NO₂.

Since we know the molecular mass of the compound is 92.02 amu, we can compare it to the mass of the empirical formula to find the molecular formula. The mass of NO₂ is 14.0 + (2 x 16.0) = 46.0 amu. Thus, the molecular mass of our compound (92.02 amu) is approximately twice that of NO₂, so the molecular formula is N₂O₄.

Answer:The isoelctric point of phenylalanine is 5.91. Kindly refer the attachment for structure.

Explanation:

Isoeectronic point is the pH at which the amino acid do not migrate in presence of an electric field.Hence at isoelectric point the molecule exists as a neutral molecule.

So for a molecule to not migrate in electric field it must be neutral. At isoelectric point the net charge on the amino acid molecule is zero.

The amino acid at isoelectric point exists as a zwitter ion which has a equal positive and equal negative charge making the net charge of the amino acid to be zero.

since the amino acid molecule has two kinds of functional group present which are an amine group and a carboxylic acid group.

So to calculate the isoelectric point we must know the pKa of these functional groups.

The formula for calculating the isoelectic point (pI) is :

[tex]pI=[pKa1+pKa2]\times 1/2[/tex]

pI=Isoelectric point

pKa1=pKa of carboxy group

pKa2=pKa of amine group

we have the pKa1 and pKa2 given in the question so putting these values in above formula we can calculate the isoelectric point.

pKa1=2.58

pKa2=9.24

pI=1/2×[pKa1+pKa2]

pI=1/2×[2.58+9.24]

pI=5.91

The isoelectric point is 5.91 hence at a pH of 5.91 phenylalanine would exist as a zwitterion and would not migrate in presence of a electric field.

Hey there!:

Ksp =(Mg⁺²)(2F⁻)²

Ksp = (1.18*10⁻³)(2*1.18*10⁻³)²

Ksp = 6.57*10⁻⁹

Hope that helps!

Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, the chlorine-containing compound that is predicted to have the greatest (most exothermic) lattice energy is; MgCl₂

     The lattice energy is defined as the energy required to dissociate one mole of an ionic compound to its constituent gaseous ions.

       Thus; F =  (q₁  ×  q ₂)/r²

        Now, Lattice energy is inversely proportional to the size of the ions. This implies that as the size of the ions increases, then the lattice energy will decrease. The size of the ions is also called the atomic radii.

Thus, as atomic radii increases, lattice energy will decrease.

      From the formula above and definition, we can tell that Lattice energy decreases down a group. Also, lattice energy will increase as the charges increase.

Now, the compounds we are dealing with are;

CaCl₂ - Calcium Chloride

NaCl - Sodium Chloride

MgCl₂ - Magnesium Chloride

KCl - Potassium Chloride

CCl₄ - Carbon tetrachloride

The charges of the metals that form the chlorides above are;

Ca = +2

Na = +1

Mg = +2  

K =  +1

The highest charges are Ca and Mg.

Now, they both belong to group two of the periodic table with Ca below Mg in the periodic table and as such Mg will have a greater Lattice energy.

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From the list of chlorine-containing compounds,  MgCl₂ would have the greatest lattice energy. Mg²⁺ has a greater charge than the other cations, resulting in stronger electrostatic attraction and most exothermic energy. CCl₄ is a covalently bonded molecule not an ionic compound, so does not possess lattice energy.

Based on Coulomb's Law of electrostatic attraction, the strongest (most exothermic) lattice energy would be present in the compound with the highest positively charged ion and smallest ionic radii, which results in a significant increase in lattice energy. This comes from the idea that lattice energy increases with higher charges and decreases with larger ionic radii.

In this case, it would be MgCl₂, because Mg⁺ has a greater charge than the other cations listed (Ca²⁺, Na⁺, K⁺). Greater charge results in stronger electrostatic attraction and hence the most exothermic lattice energy.

It's worth noting that CCl₄ is a covalently bonded molecule and not an ionic compound, thus it does not have lattice energy in the context of ionic compounds.

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Answer:

70.3824 grams of carbondioxide is produced.

Explanation:

[tex]C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)[/tex]

Moles of glucose =  [tex]\frac{48.0 g}{180 g/mol}=0.2666 mol[/tex]

According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.

Then 0.2666 mol of gluocse will reactwith :

[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of oxygen

Mass of 1.5996 moles of oxygen gas:

[tex]1.5996 mol\times 32 g/mol = 51.1872 g[/tex]

51.1872 grams of oxygen are required to convert 48.0 grams of glucose to [tex]CO_2[/tex] and [tex]H_2O[/tex].

According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.

Then 0.2666 moles of glucose will give:

[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of carbon dioxide

Mass of 1.5996 moles of carbon dioxide gas:

[tex]1.5996 mol\times 44 g/mol = 70.3824 g[/tex]

70.3824 grams of carbondioxide is produced.

Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.

Let's consider the overall equation for the combustion of glucose in the human body.

C₆H₁₂O₆(aq) + 6 O₂(g) ⟶ 6 CO₂(g) + 6 H₂O(l)

We can calculate the grams of oxygen required to react with 48.0 g of glucose considering the following relations.

[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molO_2}{1molGlucose} \times \frac{32.00 gO_2}{1 molO_2} = 51.2 gO_2[/tex]

We can calculate the grams of carbon dioxide produced from 48.0 g of glucose considering the following relations.

[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molCO_2}{1molGlucose} \times \frac{44.01 gCO_2}{1 molCO_2} = 70.4 gCO_2[/tex]

Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.

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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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Answer : The pH of the solution is, 7.4

Explanation : Given,

pOH = 7.1

[tex]K_w=2.93\times 10^{-15}[/tex]

First we have to calculate the value of [tex]pK_w[/tex].

The expression used for the calculation of [tex]pK_w[/tex] is,

[tex]pK_w=-\log [K_w][/tex]

Now put the value of [tex]K_w[/tex] in this expression, we get:

[tex]pK_w=-\log (2.93\times 1-^{-15})[/tex]

[tex]pK_w=15-\log (2.93)[/tex]

[tex]pK_w=14.5[/tex]

Now we have to calculate the pH of the solution.

As we know that,

[tex]pH+pOH=pK_w[/tex]

Now put all the given values in this formula, we get:

[tex]pH+7.1=14.5[/tex]

[tex]pH=7.4[/tex]

Therefore, the pH of the solution is, 7.4

A solution has a pOH of 7.1  and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.

pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values are more alkaline.

A solution has a pOH of 7.1 at 10 °C. We can calculate the pH at this temperature using the following expression.

pH + pOH = pKw

pH = pKw - pOH

pH = -log (2.93 × 10⁻¹⁵) - 7.1 = 7.4

where,

A solution has a pOH of 7.1  and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.

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Answer:

[tex]P=5.6*10^{6} Pa[/tex]

Explanation:

Consider that, as the system is adiabatic, [tex]U_{1}= U_{2}[/tex] where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: [tex]U=H-PV[/tex], and that the internal energy of the first state is the sum of the internal energy of each tank.

So, [tex]H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}[/tex]

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: [tex]H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}[/tex]

So, [tex]P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}[/tex]

Isolating [tex]P_{2}[/tex],

[tex]P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}[/tex]

[tex]V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}[/tex]

So,

[tex]P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa[/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:

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

The question involves calculating the CO2 emissions from the annual production of calcium oxide in the U.S., based on the stoichiometry of the reaction converting CaCO3 to CaO, which releases CO2.

The question is about calculating the yearly release of CO2 to the atmosphere resulting from the annual production of calcium oxide (CaO) in the United States, given that the reaction for producing CaO from calcium carbonate (CaCO3) releases CO2. Since the reaction is CaCO3(s) → CaO(s) + CO2(g), for each mole of CaCO3 decomposed, one mole of CO2 is released. Calcium carbonate (mol weight = 100.09 g/mol) decomposes to give calcium oxide (mol weight = 56.08 g/mol) and carbon dioxide (mol weight = 44.01 g/mol). Given the annual production of CaO is 8.6 × 10¹° kg, we first convert this mass to moles (using CaO's molar mass), and then calculate the corresponding moles (and mass) of CO2 released.

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:

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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Answer:

=71.76 grams

Explanation:

At Room temperature and pressure, 1 mole of an ideal gas occupies a volume of 24 liters.

Therefore, the number of moles occupied by 28.7 L is:

(28.7×1)/24=1.196 moles.

Mass=Number of moles× RMM

RMM of N₂O₂ is 60

mass=1.196 moles× 60 grams/mol

=71.76 grams

Answer:

Li

Explanation:

The group 1 elements are the alkali elements, and they are the ones that have one valence electron int heir outer shell, along wiht them Hydrogen is also an element that has one valence electron but is not considered group 1 because it is not a solid in temperature room, but a gas and has different properties, from the options the most different from the others is litium, because it can react with the same elements than the other elements in group one, plus nitrogen in a very violent way.

The group 1 element that exhibits slightly different chemistry from the others is: Lithium (Li).

The chemical elements found in group 1 of the periodic table are highly electro positive metals and have a single (1) valence electron in its outermost shell.

Some examples of these chemical elements (alkali metals) are;

Li is the symbol for the chemical element referred to as Lithium.

Lithium (Li) is an alkali metal and as such it is found in group 1 of the periodic table with a single (1) valence electron in its outermost shell.

Also, the electronic configuration of Lithium (Li) is written as;

As a result of the small size of Lithium (Li), it is exhibits slightly different chemistry from the others such as:

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Answer:

the concentration remains = 0.138 M

Explanation:

Let us calculate the charge passed:

The charge passed during the electrolysis = current X time(s)

charge passed = 2.40X30X60= 2160 C

we know that

96485 C = 1 F

Therefore 2160 C = 0.0224 F

according to Faraday's law of electrolysis if we pass one Faraday of charge through an electrolytic solution, 1 gram equivalent of metal will be deposited.

the gram equivalent of nickel deposited = 0.0224

The moles of Nickel deposited = 0.0224/2 = 0.0112 mol

the initial moles of Nickel ions present in solution is

[tex]moles=molarityXvolume(L)=0.250X0.1=0.0250mol[/tex]

The moles of nickel ion gets consumed = 0.0112

So moles of nickel ion left after electrolysis = 0.0138

[tex][Ni^{+2}]=\frac{moles}{volume(L)}\frac{0.0138}{0.1}=0.138M[/tex]

The concentration of Ni2+ ions remaining in the solution can be found by applying Faraday's first law of electrolysis and subtracting the moles of Ni2+ discharged from the initial moles present in the solution.

The electrolysis process involves a chemical reaction in which electricity is used to break down a substance into its constituent elements. In this scenario, we are determining the concentration of Ni2+ in the solution after the electrolysis process of NiSO4.

As per Faraday's first law of electrolysis, the mass of any element discharged during electrolysis is directly proportional to the quantity of electricity (current) passed. Thus, using Faraday's constant (96485  C/mol e-) and the given current and time, we can calculate how many moles of Ni2+ are plated out.

The number of moles of Ni2+ discharged would be equal to the number of moles initially present in the 100.0 mL of 0.250 M NiSO4 solution. By subtracting the moles of Ni2+ that were discharged from the total initial moles of Ni2+, we can obtain the number of moles of Ni2+ remaining in the solution. This value can be converted back into molarity by dividing the moles of Ni2+ remaining by the volume of the solution in liters.

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Answer:

35.5 amu is the average atomic mass of chlorine

Explanation:

Fractional abundance of chlorine‑35 = 75.5%

Fractional abundance of chlorine‑37 = 24.5%

Average atomic mass is equal to summation of products of all isotopes masses into their fractional abundance.

Average atomic mass =  Σ(Mass of an isotope × fractional abundance)

Average atomic of chlorine :

[tex]35 amu\times 0.755+37 amu\times 0.245=35.49 amu\approx 35.5 amu[/tex]

35.5 amu is the average atomic mass of chlorine

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:

False

Explanation:

Atomic bonding can be of two types:

Ionic bonds: The bond formed between two opposite charged ions. The strong electrostatic force acting between the ions make them compounds of high stability . Example: NaCl

Covalent bonds: The bond formed by sharing of the electrons between the atoms. The balance between the attractive force (nucleus, electron) and repulsive force (electron, electron or nucleus, nucleus) makes the compound stable. Example: [tex]CH_4[/tex]

The bonds are either formed by transferring (losing and gaining) or sharing of electrons. Thus, compound forming these bonds share or transfer electrons to attain noble gas configuration. Thus, they lack free electrons.

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 value of [tex]K_c[/tex] for the following reaction will be, 1024

Explanation :

[tex]K_c[/tex] is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants

The given balanced equilibrium reaction is,

[tex]P_4O_{10}(s)\rightleftharpoons P_4(s)+5O_2(g)[/tex]

As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.

The expression for equilibrium constant for this reaction will be,

[tex]K_c=[O_2]^5[/tex]

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

[tex]K_c=(4.000)^5[/tex]

[tex]K_c=1024[/tex]

Therefore, the value of [tex]K_c[/tex] for the following reaction will be, 1024

The value of the equilibrium constant for the given reaction is 1,024.

Equilibrium constant of any chemical reaction at the equilibrium state is calculated as:

Equilibrium constant Kc = Concentration of product / Concentration of reactant

Given chemical reaction is:

P₄O₁₀(s) → P₄(s) + 5O₂(g)

Equilibrium constant for the above reaction is written as:

Kc = [P₄][O₂]⁵ / [P₄O₁₀]

For this reaction only concentration of oxygen gas is taken into consideration, as all other quantities are present in solid form and their value at equilibrium is 1. So, value of Kc is calculated as:

Kc = [O₂]⁵

Given concentration of O₂ = 4M

Kc = (4)⁵ = 1,024

Hence, 1,024 is the value of Kc.

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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:

True

Explanation:

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

This statement is true.

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex] Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

[tex]PV=nRT[/tex]

[tex]n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol[/tex]

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

[tex]\frac{2}{1}\times 12.1597 mol=24.3194 mol[/tex] of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

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.

Hey there!:

240 mL bottle of the oral solution , so 43.2% alcohol

Therefore:

240 mL *  43.2 / 100

= 2.4 * 43.2

=> 103.68 mL of  alchohol

Hope this helps!

Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. The quantity of alcohol in a 240 ml bottle of the oral solution is 103.68 ml of alcohol.

Alcohol is a drink that is made up of rotten fruits and vegetables, It is made by fermentation. They cause unconsciousness to the brain and the body. It creates hallucinations.

Ritonavir is taken with meals twice a day. Ask your doctor or pharmacist to explain any instructions on your prescription label that you are unsure about following.

Given the 43.2% alcohol and 26.57% propylene glycol. The bottle is of 240 ml.

240 mL x  43.2 / 100

2.4 x 43.2 = 103.68 ml of  alcohol.

Thus, the quantity of alcohol in a 240 ml bottle is 103.68 ml of alcohol.

To learn more about the quantity of alcohol, refer to the below link:

https://brainly.com/question/485204

#SPJ2

Answer:

Part A: 36 MBq; Part B: 18 MBq

Explanation:

The half-life is the time it takes for half the substance to disappear.

The activity decreases by half every half-life

A =Ao(½)^n, where n is the number of half-lives.

Part A

3.0 da = 1 half-life

A =  Ao(½) = ½ × 72 MBq = 36 MBq

Part B

6.0 da = 2 half-lives

A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq

Final answer:

Part A: The activity of 201Tl after 3.0 days is 36 MBq. Part B: The activity of 201Tl after 6.0 days is 18 MBq.

Explanation:

Part A: To calculate the activity of 201Tl after 3.0 days, we need to determine the number of half-lives that have passed. Since the half-life of 201Tl is 3.0 days, dividing 3.0 days by the half-life gives us 1 half-life. Each half-life reduces the activity of the isotope by half, so after 1 half-life, the activity would be half of the initial activity. Therefore, the activity after 3.0 days would be 72 MBq / 2 = 36 MBq.

Part B: To calculate the activity of 201Tl after 6.0 days, we can use the same method as in Part A. Since the half-life is 3.0 days, 6.0 days is equivalent to 2 half-lives. Each half-life reduces the activity by half, so after 2 half-lives, the activity would be 72 MBq / 2 / 2 = 18 MBq.

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.

Learn more: https://brainly.com/question/24900805

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: 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.