The Henry’s law constant for CO2 is 3.4 × 10−2 M/atm at 25 °C. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?

Final answer:

The average atomic mass of titanium on the hypothetical planet is calculated by using the weighted average of the abundances and masses of its isotopes, resulting in 46.8989 amu.

Explanation:

The average atomic mass of titanium on that hypothetical planet can be calculated by multiplying the abundance of each isotope by its mass (in atomic mass units, amu), then summing these products. The calculation will look as follows:

To find the average atomic mass, we add these values together to get the sum which should give us the average atomic mass of titanium on the planet.

The correct calculation would be:

Hence, the average atomic mass of titanium for this planet is 46.8989 amu. It's important to note that the average atomic mass of an element is the weighted average of all the isotopes of that element.

Answer:

the entropy change of the copper block = - 117.29 J/K

the entropy change of the water = 138.01 J/K

the entropy change of the universe = 20.72 J/K

Explanation:

For Copper block:

the mass of copper block [tex](m_c)[/tex] = 1.00 kg

Temperature of block of copper [tex](T_c)[/tex] = 100°C

= (100+273)K

= 373K

Standard Heat capacity for copper [tex](C_c)[/tex] = 386 J/kg.K

For water:

We know our volume of liquid water to be = 4.00 L

At 0.0°C Density of liquid water  = 999.9 kg/m³

As such; we can determine the mass since : [tex]density = \frac{mass}{volume}[/tex]

∴ the mass of 4.00 L of liquid water at 0.0°C will be its density × volume.

= 999.9 kg/m³ × [tex]\frac{4}{1000}m^3[/tex]

= 3.9996 kg

so, mass of liquid water [tex](m_w)[/tex] = 3.9996 kg

Temperature of liquid water [tex](T_w)[/tex] at 0.0°C = 273 K

Standard Heat Capacity of  liquid water [tex](C_w)[/tex] = 4185.5 J/kg.K

Let's determine the equilibruium temperature between the copper and the liquid water. In order to do that; we have:

[tex]m_cC_c \delta T_c =m_wC_w \delta T_w[/tex]

[tex]1.00*386*(373-T_\theta)=3.996*4185.5*(T _\theta-273)[/tex]

[tex]386(373-T_\theta)=16725.26(T_\theta-273)[/tex]

[tex](373-T_\theta)=\frac{16725.26}{386} (T_\theta-273)[/tex]

[tex](373-T_\theta)=43.33 (T_\theta-273)[/tex]

[tex](373-T_\theta)=43.33 T_\theta-11829.09[/tex]

[tex]373+11829.09=43.33 T_\theta+T_\theta[/tex]

[tex]12202.09 =43.33T_\theta[/tex]

[tex]T_\theta= 275.26 K[/tex]

∴ the equilibrium temperature = 275.26 K

NOW, to determine the Entropy change of the copper block; we have:

[tex](\delta S)_{copper}=m_cC_cIn(\frac{T_\theta}{T_c} )[/tex]

[tex](\delta S)_{copper}=1.0*386In(\frac{275.26}{373} )[/tex]

[tex](\delta S)_{copper}=-117.29 J/K[/tex]

The entropy change of the  water can also be calculated as:

[tex](\delta S)_{water}=m_wC_wIn(\frac{T_\theta}{T_w} )[/tex]

[tex](\delta S)_{water}=3.9996*4185.5In(\frac{275.26}{373} )[/tex]

[tex](\delta S)_{water}=138.01J/K[/tex]

The entropy change of the universe is the combination of both the entropy change of copper and water.

[tex](\delta S)_{universe}=(\delta S)_{copper}+(\delta S)_{water}[/tex]

[tex](\delta S)_{universe}=(-117.29+138.01)J/K[/tex]

[tex](\delta S)_{universe}=20.72J/K[/tex]

Answer:

The density is 10.25 [tex]\frac{atom}{nm^{2} }[/tex]

Explanation:

From the question we are given that the lattice spacing  nm

        V = [tex](0.546)^{3} }[/tex] × [tex]nm^{3}[/tex]

            = 0.1628 [tex]nm^{3}[/tex]

           zinc blende lattice has 4 atom per unit cell.

           This means that 4  atoms is contained in 0.15056 [tex]nm^{3}[/tex]

            Density of the cubic unit in terms of atom  is given as =

                                                                                                      =26.565 [tex]\frac{atoms}{nm^{3} }[/tex]

            For plane  (110) the spacing between the cubic unit in the crystal denoted by  [tex]d_{hkl}[/tex] is given as =  [tex]\frac{a}{\sqrt{1^{2} + 1^{2} +0^{2} } }[/tex] Where a is the lattic spacing = 0.546

                                               =   [tex]\frac{0.546}{\sqrt{2} }[/tex]  = 0.376 nm

               The density of the lattice in (110) plane  =26.565 [tex]\frac{atoms}{nm^{3} }[/tex] × 0.386 nm

                                                                                 = 10.25 [tex]nm^{2}[/tex]

Answer:

2.29 × 10⁴ times

Explanation:

A single penny is 1.52 mm thick. The distance covered by 1 mole of pennies (6.02 × 10²³ pennies) is:

6.02 × 10²³ p × (1.52 mm/1 p) = 9.15 × 10²³ mm = 9.15 × 10²³ × 10⁻³ m = 9.15 × 10²⁰ m

The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. Considering 1 ly = 9.46 × 10¹⁵ m, this distance in meters is:

4.22 ly × (9.46 × 10¹⁵ m/1 ly) = 3.99 × 10¹⁶ m

The times that the stack would go between the earth and Alpha Centauri are:

9.15 × 10²⁰ m / 3.99 × 10¹⁶ m = 2.29 × 10⁴

Answer:

Net work done overall = sum of work done for all the processes = 16,995.84 J

Explanation:

From the start, P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = ?

We can obtain V from PV = nRT; n = 2, R = 8.314 J/mol.K

V = 2 × 8.314 × 600/(1000000) = 0.009977 m³

P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = 0.009977 m³

For an adiabatic process for an ideal gas,

P(V^γ) = constant

γ = ratio of specific heats = Cp/CV = 1.4,

P₂ = 1 bar = 10⁵ Pa

P₁ (V₁^1.4) = P₂ (V₂^1.4) = k

10⁶ (0.009977^1.4) = 10⁵(V₂^1.4) = 1579.75 = k

V₂ = 0.0517 m³

Work done for an adiabatic process

W = k((V₂^(1-γ)) - (V₁^(1-γ))/(1-γ)

W = 1579.75 ((0.0517^0.4) - (0.009977^0.4))/0.4

W = 582.25 J

We still need T₂

PV = nRT

T₂ = P₂V₂/nR = 100000×0.0517/(2×8.314) = 310.92K

Step 2, constant volume heating,

Work done at constant volume is 0 J.

T₂ = 310.92K, T₃ = 600K

V₂ = 0.0517 m³, V₃ = V₂ = 0.0517 m³ (Constant volume)

P₂ = 1bar, P₃ = ?

PV = nRT

P₃ = nRT₃/V₃ = 2 × 8.314 × 600/0.0517 = 192974.85 Pa = 1.93bar

Step 3, isothermally returned to the initial state.

P₃ = 1.93bar, P₄ = P₁ = 10bar

T₃ = 600K, T₄ = T₁ = 600K (Isothermal process)

V₃ = 0.0517 m³, V₄ = V₁ = 0.009977 m³

Work done = nRT In (V₃/V₁) = 2 × 8.314 × 600 In (0.0517/0.009977) = 16413.59 J

Net work done = W₁₂ + W₂₃ + W₃₁ = 582.25 + 0 + 16413.59 = 16995.84 J

Hope this helps!!

Answer:

No, (a) Ti²⁺ is only paramagnetic

Explanation:

Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.

(a) Ti²⁺

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{2}[/tex]

The electrons in 3d orbital = 2 (Unpaired)

Thus, the ion is paramagnetic as the electrons are unpaired.

(b) Zn²⁺

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}[/tex]

The electrons in 3d orbital = 10 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

(c) Ca²⁺

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^6[/tex]

The electrons in 3p orbital = 6 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

(d) Sn²⁺

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^2[/tex]

The electrons in 5s orbital = 2 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

The compounds of these ground-state ions (a) Ti²⁺, (b) Zn²⁺, (c) Ca²⁺, and (d) Sn²⁺ are not paramagnetic.

The compounds of these ground-state ions are not paramagnetic.

Paramagnetic substances have unpaired electrons in their atoms or ions, which cause them to be attracted to a magnetic field. To determine whether an ion is paramagnetic, we need to consider its electron configuration.

(a) Ti²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d². The 3d² sublevel has two unpaired electrons, so Ti²⁺ is paramagnetic.

(b) Zn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰. All the sublevels in its electron configuration are filled, so Zn²⁺ is not paramagnetic.

(c) Ca²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². All the sublevels are filled, so Ca²⁺ is not paramagnetic.

(d) Sn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰. All the sublevels are filled, so Sn²⁺ is not paramagnetic.

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Answer:

                     Ethynylcyclopropane is the stable isomer for given alkyne.

Explanation:

                     In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,

                                                    CₙH₂ₙ₋₂

Putting value on n = 5,

                                                    C₅H₂.₅₋₂

                                                    C₅H₈

Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.

                      C₅H₈   ------------(-2 H) ---------->  C₅H₆

Hence, the molecular formula for given compound is C₅H₆

Below, 4 different isomers with molecular formula C₅H₆ are attached.

                                      The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.

                                      Compounds 3-ethylcycloprop-1-yne, cyclopentyne and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

[tex]t_{1/2}=\frac{0.693}{k}}[/tex] .....(1)

Rate law expression for first order kinetics is given by the equation:

[tex]t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}[/tex] ......(2)

where,  

k = rate constant

[tex]t_{1/2}[/tex] =Half life of the reaction = [tex]2.42\times 10^3 s[/tex]

t = time taken for decay process = ?

[tex][A_o][/tex] = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of [tex][A_o][/tex]

[A]=[tex]\frac{66.8}{100}\times 0.163 M=0.108884 M[/tex]

[tex]k=\frac{0.693}{2.42\times 10^3 s}[/tex]

[tex]t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}[/tex]

t = 1,409.19 s

1 minute = 60 sec

[tex]t=\frac{1,409.19 }{60} min=23.49 min[/tex]

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Answer:

14.45 mL

Explanation:

The rule apply for the addition and subtraction for significant digits is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

Initial Burette reading = 0.75 mL ( 2 significant digits)

Final Burette reading = 15.20 mL ( 4 significant digits)

Liquid dispensed = 15.20 mL - 0.75 mL = 14.45 mL ( Answer to two decimal places )

To find the amount of liquid dispensed from the buret, subtract the initial volume (0.75 mL) from the final volume (15.20 mL), resulting in 14.45 mL of liquid dispensed. Ensure that the answer is reported with the same level of precision as the measurements from the buret.

To calculate the amount of liquid dispensed from the buret, you subtract the initial measurement from the final measurement. In this case, the initial measurement is 0.75 mL and the final measurement after the liquid is added to the flask is 15.20 mL. Therefore, the amount of liquid dispensed is 15.20 mL - 0.75 mL = 14.45 mL.

Burets are commonly used in titration analyses and allow volume measurements to the nearest 0.01 mL, which is why our final answer should be reported with the same level of precision as the buret readings. According to the proper significant figures rules, the result should be reported with the same number of decimal places as the measurement with the fewest decimal places. Since the readings from the buret are given to two decimal places (0.75 mL and 15.20 mL), our calculated volume of dispensed liquid also should be reported to two decimal places as 14.45 mL.

Answer: 12.92g of CoSO4

Explanation:

Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol

Molarity of CoSO4 = 0.303mol/L

Mass conc. In g/L = Molarity x molar Mass

= 0.303x155 = 46.965g/L

275 grams of water = 0.275L of water

46.965g of CoSO4 dissolves in 1L

Therefore Xg of CoSO4 will dissolve in 0.275L i.e

Xg of CoSO4 = 46.965x0.275 = 12.92g

Therefore 12.92g of CoSO4 is needed

Answer:

1.   [tex]R=k[A]^1[B]^2[/tex]

2.  [tex]R=k[B]^1[/tex]

3.  [tex]R=k[A]^0[B]^0=k[/tex]

4.  [tex]R=k[A]^1[B]^{-1}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

(1) is second order in B and overall third order.

2A + B → C

Order of the reaction = sum of stoichiometric coefficient

= x + 2 = 3

x = 1

Rate of the reaction =R

[tex]R=k[A]^1[B]^2[/tex]

(2) is zero order in A and first order in B.

2A + B → C

Rate of the reaction =R

[tex]R=k[A]^0[B]^1=k[B]^1[/tex]

Order of the reaction = sum of stoichiometric coefficient

= 0 + 1 = 1

(3) is zero order in both A and B .

2A + B → C

Order of the reaction = sum of stoichiometric coefficient

= 0 + 0 = 0

Rate of the reaction =R

[tex]R=k[A]^0[B]^0=k[/tex]

(4) is first order in A and overall zero order.

2A + B → C

Order of the reaction = sum of stoichiometric coefficient

= 1 + x = 0

x = -1

Rate of the reaction = R

[tex]R=k[A]^1[B]^{-1}[/tex]

Answer:

Magnesium (Mg)

Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]

Explanation:

Ionization Energy:

It is the amount of energy required to remove valance shell electron of an atom in gaseous phase. The more the electron is closer to the nucleus the more energy is required to remove the electron.

Trend:

Increases from left to right in period.

Decreases generally from top to bottom in group.

In our case the data is:

IE₁ = 738

IE₂ = 1450

IE₃ = 7732

IE₄ = 10,539

IE₅ = 13,628

This data matches the Ionization energies of Magnesium (Mg). Magnesium has 12 electrons with 2 electrons in valance shell. It has 12 Ionization energies.

Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]

The electron configuration of the Period 3 element with the given successive IEs is;

The ionization energy data given is in synchrony with that of Magnesium, Mg.

In essence, the element in question is Magnesium, Mg.

The electronic configuration of Magnesium is therefore;

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The weight of the water in the original solution is 2.78 grams. The percentage of water in the original 4.00 grams solution of sodium chloride, once the water was evaporated, is therefore approximately 69.5%.

The weight of the water in the original solution can be found by subtracting the weight of the remaining sodium chloride from the total weight of the solution. So, water = total weight - weight of sodium chloride = 4.00 grams - 1.22 grams = 2.78 grams.

The percentage of water in the solution can be found by the following formula:

Percentage = (Weight of water / Total weight) * 100 = (2.78 grams / 4.00 grams) * 100 = 69.5%

So, the percentage of water in the 4.00 gram solution of sodium chloride after the water was evaporated, is approximately 69.5%.

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Answer:

a. Smallest  atomic  radius  in  6A – Oxygen  (O)

b. Largest  atomic  radius  in  Period
6 – Cesium
(Cs)

c. Smallest  metal  in  period
3 – Aluminum
(Al)

d. Highest  IE1  in  Group
4A –Carbon
(C)

e. Lowest  IE1  in  period
5 – Rubidium
(Rb)

f. Most  metallic  in  Group
5A – Bismuth
(Bi)
or
element
115

g. Group
3A  element  that  forms  the  most  basic  oxide – Thallium  

(Tl)  or  element
113

h. Period
4  element  with  filled  outer
level – Krypton
(Kr)

i. Condensed  gound  state  configuration  is  [Ne]3s23p2 –

Germanium
(Ge)

j. Condensed  ground
state  configuration  is  [Kr]5s24d6 –

Ruthenium
(Ru)

k. Forms
2+
ion  with  electron  configuration  of  [Ar]3d3 – Vanadium  

(V)

l. Period
5  element  that  forms
3+
ion  with  pseudo‐noble  gas  

configuration – Indium
(In)

m. Period
4
transition  element  that  forms
3+  diamagnetic  ion –

Scandium
(Sc)

n. Period
4  transition  element  that  forms
2+
ion  with  
half‐filled  d  

sublevel – Manganese
(Mn)

o. Heaviest  Lanthanide – Lutetium
(Lu)

p. Period
3  element  whose
2‐
ion  is  isoelectronic  with  Ar – Sulfur  

(S)

q. Alkali
earth  metal  whose  cation  is  isoelectronic  with  Kr –

Strontium
(Sr)

r. Group
5 A
metalloid  with  the  most  acidic  oxide – Arsenic
(As)
or  

Antimony
(Sb)

Elements in a group are chemically similar to each other.

The periodic table is an arrangement of elements in groups and periods. The elements in the same group share a lot of chemical similarity with each other. The elements that are in the same period only have the same number of valence shells.

The elements described by each statement is;

(a) Smallest atomic radius in Group 6A(16) - oxygen

(b) Largest atomic radius in Period 6 - cesium

(c) Smallest metal in Period 3 - Aluminium

(d) Highest IE₁ in Group 4A(14) - carbon

(e) Lowest IE₁ in Period 5 -  Bismuth

(g) Group 3A(13) element that forms the most basic oxide -  Thallium

(h) Period 4 element with filled outer level - krypton

(i) Condensed ground-state electron configuration of [Ne] 3s²3p² - silicon

(j) Condensed ground-state electron configuration of [Kr] 5s²4d⁶- xenon

(k) Forms 2+ ion with electron configuration [Ar] 3d³ - vanadium

(l) Period 5 element that forms 3+ ion with pseudo–noble gas configuration - Indium

(m) Period 4 transition element that forms 3+ diamagnetic ion - Scandium

(n) Period 4 transition element that forms 2+ ion with a halffilled d sublevel - Manganese

(o) Heaviest lanthanide - Lutetium

(p) Period 3 element whose 2- ion is isoelectronic with Ar - sulfur

(q) Alkaline earth metal whose cation is isoelectronic with Kr - Strontium

(r) Group 5A(15) metalloid with the most acidic oxide - nitrogen

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Answer:

a) Element = Chlorine

b) Element = Arsenic

Explanation:

The knowledge of Orbitals and Quantum number and the electronic configuration is applied as shown in the analysis in the attached file.

The equilibrium constant, Kp, for the reaction COCl2(g) → CO(g) + Cl2(g) at 600 K is calculated to be 0.634.

To calculate the equilibrium constant, Kp, for the reaction COCl2(g) → CO(g) + Cl2(g) at 600 K, we must first use the given equilibrium partial pressure of COCl2(g) to determine the change in pressure for CO and Cl2 since the reaction started with only COCl2 present. The initial pressure of COCl2 was 0.822 atm, and the equilibrium partial pressure was 0.351 atm, which implies the pressure change (∆P) for CO and Cl2 is 0.822 atm - 0.351 atm = 0.471 atm.

Since the reaction shows that 1 mole of COCl2 produces 1 mole of CO and Cl2 each, the equilibrium partial pressures of CO and Cl2 are also 0.471 atm each. Now, we can write the expression for Kp, which is Kp = (PCO)(PCl2) / (PCOCl2). Plugging in the values, we get Kp = (0.471 atm)(0.471 atm) / (0.351 atm), and thus Kp is calculated to be 0.634 at 600 K.

Answer:

Molality for the solution is 1.57 m

Explanation:

Molality is mol of solute in 1kg of solvent.

20.3 % by mass means that 20.3 g of solute (FeCl₃) are contained in 100 g of solution..

Let's determine the mass of solvent.

Mass of solution = Mass of solvent + Mass of solute

100 g = Mass of solvent + 20.3 g

100 g - 20.3 g = Mass of solvent → 79.7 g

Let's convert the mass in g to kg

79.7 g . 1kg / 1000 g = 0.0797 kg

Let's determine the moles of solute (mass / molar mass)

20.3 g / 162.2 g/mol = 0.125 mol

Molality = 0.125 mol / 0.0797 kg → 1.57 m

Final answer:

The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.

Explanation:

This question demands basic understanding of molatility.

To calculate the molality of the solution, we need to determine the moles of solute (iron(III) chloride) and the mass of the solvent (water).

First, convert the mass percent to grams of solute:

Mass of solute = (20.3%)(1.280 g/mL)(1000 mL) = 260.48 g

Next, calculate the moles of solute:

Moles of solute = (260.48 g)/(162.2 g/mol) = 1.603 mol

Finally, calculate the molality:

Molality = (1.603 mol)/(1 kg) = 1.60 m

Therefore, The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.

The molarity of the 10.5% by mass glucose solution is 0.600 M.

To find the molarity of the glucose solution, we need to determine the number of moles of glucose present in the solution first. We can use the percent by mass to calculate this.

Given that the solution is 10.5% by mass, we know that 10.5 grams of glucose is present in a 100 gram solution. We can convert this to moles by dividing the mass of glucose by its molar mass, which is 180.16 g/mol.

So, the number of moles of glucose is 10.5 g / 180.16 g/mol = 0.0583 mol. To find the molarity, we divide the number of moles by the volume of the solution in liters. The volume of the solution can be determined by multiplying the density of the solution by its mass: 100 g / (1.03 g/mL) = 97.09 mL = 0.0971 L.

Therefore, the molarity of the glucose solution is 0.0583 mol / 0.0971 L = 0.600 M.

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The molarity of a 10.5% by mass glucose solution with a density of 1.03 g/mL can be calculated by assuming a 100 g sample of the solution. The mass of glucose per liter is found to be 108.1395 g, and with the molar mass of glucose, the molarity is determined to be 0.600 M.

The question pertains to determining the molarity of a 10.5% by mass glucose solution with a given density of 1.03 g/mL. To calculate the molarity, we need to use the given mass percentage and density to find out how many moles of glucose are present in a liter of solution.

First, assume you have 100 g of this solution. Because it's a 10.5% by mass solution, this means there are 10.5 g of glucose (C6H12O6) and 89.5 g of water in the mixture.

Using the density, we find the volume of 100 g of solution:
100g / 1.03g/mL = 97.09 mL
Because we want to know the molarity per liter, it's important to work with a liter of the solution:
(1000 mL/L) / (97.09 mL) = 10.299 L^-1 multiplication factor
Now, we will use the multiplication factor to scale up the mass of glucose to what would be in one liter:
10.5 g * 10.299 = 108.1395 g glucose per liter

The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol, so the number of moles in one liter would be:
108.1395 g / 180.16 g/mol = 0.600 mol/L

Therefore, the molarity of the glucose solution is 0.600 M.

Answer:

a). Nepthalene is a D2h molecule because it has 3 C2 axes which are perpendicular and it has a mirror plane which is horizontal. Moreover C2 is taken as principal axis.

b). 1,8-dichloronaphthalene is a C2v molecule beccause it has 1 C2 axis, two rings are joined by the C-C bond and it also has two mirro planes.

c). 1,5-dichloronaphthalene is a C2h molecule because it has only 1 C2 axis which is pependicular to the plane, it has an inversion center and also a mirror plane which is horizontal in position.

d). 1,2-dichloronaphthalene is a Cs molecule because it has only a mirror plane.

Explanation:

The given reaction equation is as follows.

       [tex]2HI(g) \rightarrow H_{2}(g) + I_{2}(g)[/tex]

       [tex]\frac{-d[HI]}{dt} = k[HI]^{2}[/tex]

 [tex]-\int_{[HI]_{o}}^{[HI]_{t}} \frac{d[HI]}{[HI]^{2}} = k \int_{o}^{t} dt[/tex]

    [tex]-[\frac{-1}{[HI]}]^{[HI]_{t}}_{[HI]_{o}} = kt[/tex]

        [tex]\frac{1}{[HI]_{t}} - \frac{1}{[HI]_{o}} = kt[/tex] .......... (1)

where,   [tex][HI_{o}][/tex] = Initial concentration

             [tex][HI]_{t}[/tex] = concentration at time t

                  k = rate constant

                  t = time

Now, we will calculate the initial concentration of HI as follows.

            Initial rate = [tex]1.6 \times 10^{-7} mol/sec[/tex]

                k = [tex]6.4 \times 10^{-9}[/tex]

           R = [tex]k[HI]^{2}_{o}[/tex]

    [tex][HI]^{2}_{o} = \frac{R}{k}[/tex]

                = [tex]\frac{1.6 \times 10^{-7}}{6.4 \times 10^{-9}}[/tex]

      [tex][HI]_{o}[/tex] = 5 M

Now, we will calculate the concentration of [tex][HI]_{t}[/tex] at t = [tex]1.01 \times 10^{10}[/tex] sec as follows.

Using equation (1) as follows.

           k = [tex]6.4 \times 10^{-9}[/tex]

         [tex]\frac{1}{[HI]_{t}} - \frac{1}{5}[/tex] = [tex](6.4 \times 10^{-9}) \times 1.01 \times 10^{10}[/tex]

           [tex]\frac{1}{[HI]_{t}}[/tex] = 64.44

              [tex][HI]_{t}[/tex] = 0.0155 M

Thus, we can conclude that concentration of HI at t = [tex]1.01 \times 10^{10}[/tex] sec is 0.0155 M.

A reception subservient on the second and first-order reactants is called a second-order reaction.

The correct answer is:

The concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.

The equation according to the question is:

2 HI (g) ⇒ H₂ (g) + I₂ (g)

[tex]\dfrac{\text{-d}\left[\begin{array}{ccc}\text{HI}\end{array}\right] }{\text{dt}} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2}[/tex]

[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{t}} } - \dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{o}} } &= \text{kt}[/tex] .......equation (1)

Where, the initial concentration can be represented as: [tex]\left[\begin{array}{ccc}\text{HI}_{o} \end{array}\right][/tex]

The concentration at time t = [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex]

Rate constant will be = k

The time will be = t

The initial concentration of HI can be calculated as:

The initial rate = 1.6 × 10⁻⁷ mol/sec

k = 6.4 × 10⁻⁹

[tex]\text{R} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]

[tex]\dfrac{\text{R}}{\text{k}} & =\left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]

= [tex]\dfrac{1.6 \times 10^{-7} }{6.4 \times 10^{-9} }[/tex]

[tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right]\end{array}\right] _{0} &= 5 \;\text{M}[/tex]

To calculate the concentration  [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex] at time (t) = 1.01 × 10 ¹⁰ sec.

Now, using the above equation: (1)

k = 6.4 × 10⁻⁹

[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} - \dfrac{1}{5}[/tex]        

= (6.4 × 10⁻⁹) × 1.01 × 10¹⁰

[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 64.44[/tex]

[tex]{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 0.0155 \;\text{M}[/tex]

Therefore, concentration of HI at t =  1.01 × 10¹⁰ sec is 0.0155 M.

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Answer: The statement is true

Explanation:

The half-life of a radioactive isotope is the time taken for half of the total number of atoms in a given sample of the isotope to decay.

For instance

The half-life of radium is 1622 years. This means that if we have 1000 radium atoms at the beginning, then at the end of 1622 years, 500 atoms would have disintegrated, leaving 500 undecayed radium atoms

Thus, the statement is true

Answer:

0.60 mole of formic acid contains the greatest mass of oxygen

(19.2 grams)

Explanation:

Step 1: Data given

Moles of ethanol = 0.75 moles

Molar mass ethanol = 46.07 g/mol

Moles of formic acid = 0.60 moles

Molar mass of formic acid = 46.03 g/mol

Moles of H2O = 1.0

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles of oxygen in each

Ethanol: For 1 mol ethanol we have 1 mol oxygen

For 0.75 moles ethanol we have 0.75 moles O

Mass O = 0.75 moles * 16.0 g/mol = 12.0 grams O

Formic acid: For 1 mol formic acid, we have 2 moles O

For 0.60 moles formic acid, we have 2*0.60 = 1.20 moles O

Mass O = 1.20 moles * 16.0 g/mol = 19.2 grams O

H2O: for 1.0 mol H2O we have 1 mol O

Mass O = 18.02 g/mol * 1.0 mol = 18.02 grams

0.60 mole of formic acid contains the greatest mass of oxygen

(19.2 grams)

Formic acid (HCO₂H) contains the greatest mass of oxygen among the three compounds with 19.2 grams of oxygen.

Step 1: Molar Mass of Oxygen

The molar mass of oxygen (O) is approximately 16 grams per mole (g/mol).

Step 2: Identify the Number of Oxygen Atoms in Each Compound

Step 3: Calculate the Mass of Oxygen in Each Compound

Ethanol (C₂H₅OH):  

Formic Acid (HCO₂H):  

Water (H₂O):  

Step 4: Compare the Masses of Oxygen

Explanation:

Since, the atomic number of nitrogen is 7 and its electronic distribution is 2, 5. So, in order to attain stability it needs to gain 3 electrons.

Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.

Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed [tex]N_{2}[/tex] is non-polar covalent in nature.

Answer:

non polar bond

Explanation:

Answer:

We need 0.894 grams of Na2CrO4

Explanation:

Step 1: Data given

Volume of a 0.150 M AgNO3 = 73.6 mL = 0.0736 L

Step 2: Calculate moles of Ag+

Moles Ag+ = moles AgNO3

Moles Ag+ = volume * molarity

moles Ag+ = 0.0736 L x 0.150 M = 0.01104 moles

Step 3: Calculate moles Na2CrO4

2AgNO3 + Na2CrO4 → Ag2CrO4 (s) + 2NaNO3

For 2 moles AgNO3 we need 1 mol Na2CRO4

For 0.01104 moles AgNO3 we need 0.01104/2 = 0.00552 moles Na2CrO4

Step 4: Calculate mass of Na2CrO4

Mass Na2CrO4 = moles * molar mass

Mass Na2CrO4 = 0.00552 moles * 161.97 g/mol

Mass Na2CrO4 = 0.894 grams

We need 0.894 grams of Na2CrO4

To precipitate all the silver ions, 0.893 g of Na2CrO4 is required.

To calculate the mass of Na2CrO4 required to precipitate all of the silver ions from the solution, we need to use the stoichiometry of the reaction. The balanced equation for the reaction is:

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to form 1 mole of Ag2CrO4. The molarity of AgNO3 is given as 0.150 M and the volume is 73.6 mL. First, we need to convert the volume to liters:

73.6 mL × (1 L / 1000 mL) = 0.0736 L

Next, we can use the molarity and volume to calculate the number of moles of AgNO3:

moles of AgNO3 = molarity × volume = 0.150 M × 0.0736 L = 0.01104 moles

Since the reaction ratio is 2 moles of AgNO3 to 1 mole of Na2CrO4, we can calculate the number of moles of Na2CrO4:

moles of Na2CrO4 = (0.01104 moles of AgNO3) / 2 = 0.00552 moles

Finally, we can use the molar mass of Na2CrO4 to calculate the mass:

mass of Na2CrO4 = moles of Na2CrO4 × molar mass of Na2CrO4

From the periodic table, the molar mass of Na2CrO4 is:

molar mass of Na2CrO4 = 22.99 g/mol + (2 × 35.45 g/mol) + 4 × 16.00 g/mol = 161.97 g/mol

Therefore, the mass of Na2CrO4 required to precipitate all of the silver ions is:

mass of Na2CrO4 = 0.00552 moles × 161.97 g/mol = 0.893 g

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Answer:

The age of the ore is 4.796*10^9 years.

Explanation:

To solve this question, we use the formula;

A(t) =A(o)(1/2)^t/t1/2

where;

A(t) =3.22mg

A(o) = 6.731mg

t1/2 = 4.51*70^9 years

t = age of the ore

So,

A(t) =A(o)(1/2)^t/t1/2

3.22 = 6.73 (1/2)^t/4.51*10^9

Divide both sides by 6.73

3.22/6.73= (1/2)^t/4.51*10^9

0.47825= (0.5)^t/4.51*10^9

Log 0.4785 = t/4.51*10^9 • log 0.5

Log 0.4785/log 0.5 • 4.51*10^9 = t

t = 1.0634 * 4.51*10^9

t = 4.796*10^9

So therefore, the age of the ore is approximately 4.796*10^9 years.

Each of the following quantum numbers is related to an orbital characteristic in the following ways: n, the primary quantum number

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Final answer:

The principal quantum number determines the energy range and distance of an electron from the nucleus. The angular momentum quantum number determines the shape or type of the orbital. The magnetic quantum number determines the orientation of the orbital in space.

Explanation:

The feature of an orbital related to the principal quantum number (n) is the general range for the value of energy and the probable distances that the electron can be from the nucleus.

The feature of an orbital related to the angular momentum quantum number (l) is the shape or type of the orbital. The values of 1 range from 0 to (n-1), and orbitals with the same principal quantum number and the same l value belong to the same subshell.

The feature of an orbital related to the magnetic quantum number (m₁) is the orientation of the orbital in space. The values of m₁ range from -l to +l, and orbitals with the same l value have different orientations.

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

1.[tex]Li(s)+N_2(g)\rightarrow Li_3N(s)[/tex]

The balanced equation is:

[tex]6Li(s)+N_2(g)\rightarrow 2Li_3N(s)[/tex]

2. [tex]TiCl_4(l)+H_2O(l)\rightarrow  TiO_2(s)+HCl(aq)[/tex]

The balanced equation is:

[tex]TiCl_4(l)+2H_2O(l)\rightarrow TiO_2(s)+4HCl(aq)[/tex]

3. [tex]NH_4NO_3(s)\rightarrow N_2(g)+O_2(g)+H_2O(g)[/tex]

The balanced equation is:

[tex]2NH_4NO_3(s)\rightarrow 2N_2(g)+O_2(g)+4H_2O(g)[/tex]

4.[tex] Ca3P2(s)+H2O(l)\rightarrow Ca(OH)2(aq)+PH3(g)[/tex]

The balanced equation is:

[tex] Ca_3P_2(s)+6H_2O(l)\rightarrow 3Ca(OH)_2(aq)+2PH_3(g)[/tex]

5. [tex]Al(OH)_3(s)+H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+H_2O(l)[/tex]

The balanced equation is:

[tex]2Al(OH)_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+6H_2O(l)[/tex]

6.[tex] AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+NaNO_3(aq)[/tex]

The balanced equation is:

[tex] 2AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+2NaNO_3(aq)[/tex]

7. [tex]C_2H_5NH_2(g)+O_2(g)\rightarrow  CO_2(g)+H_2O(g)+N_2(g)[/tex]

The balanced equation is:

[tex]4C_2H_5NH_2(g)+15O_2(g)\rightarrow  8CO_2(g)+14H_2O(g)+2N_2(g)[/tex]

A balanced chemical equation is a representation of a chemical reaction using chemical formulas and symbols, where the number of atoms of each element on the left side (reactants) is equal to the number of atoms of the same element on the right side (products). In other words, it obeys the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction

1. Li(s)+N2(g)→Li3N(s)

The balanced equation is :

[tex](Li(s) + N_2(g) \rightarrow Li_3N(s)\)[/tex]

2. TiCl4(l)+H2O(l)→TiO2(s)+HCl(aq)

The balanced equation is :

[tex]\(TiCl_4(l) + 4H_2O(l) \rightarrow TiO_2(s) + 4HCl(aq)\)[/tex]

3. NH4NO3(s)→N2(g)+O2(g)+H2O(g)

The balanced equation is :

[tex]\(NH_4NO_3(s) \rightarrow N_2(g) + 2O_2(g) + 2H_2O(g)\)[/tex]

4. Ca3P2(s)+H2O(l)→Ca(OH)2(aq)+PH3(g)

The balanced equation is :

[tex]\(Ca_3P_2(s) + 6H_2O(l) \rightarrow 3Ca(OH)_2(aq) + 2PH_3(g)\)[/tex]

5. Al(OH)3(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2O(l)

The balanced equation is :

[tex]\(2Al(OH)_3(s) + 3H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 6H_2O(l)\)[/tex]

6.AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+NaNO3(aq)

The balanced equation is :

[tex]\(2AgNO_3(aq) + Na_2SO_4(aq) \rightarrow Ag_2SO_4(s) + 2NaNO_3(aq)\)[/tex]

7. C2H5NH2(g)+O2(g)→CO2(g)+H2O(g)+N2(g)

The balanced equation is :

[tex]\(2C_2H_5NH_2(g) + 9O_2(g) \rightarrow 4CO_2(g) + 10H_2O(g) + 2N_2(g)\)[/tex]

These equations show the balanced chemical reactions along with the respective phases of the substances involved.

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Answer: The wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]

Explanation:

To calculate the threshold wavelength for a given work function, we use the equation:

[tex]\phi =h\nu_o[/tex]

where,

[tex]\phi[/tex] = work function of the potassium metal = [tex]2.29eV=3.66\times 10^{-19}J[/tex]      (Conversion factor:  [tex]1eV=1.6\times 10^{-19}[/tex] )

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

[tex]\nu_o=\frac{c}{\lambda _o}[/tex]

c = speed of light = [tex]3\times 10^9m/s[/tex]

[tex]\lambda_o[/tex] = wavelength of light

Putting values in above equation:

[tex]3.66\times 10^{-19}J=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda_o}\\\\\lambda_o=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.66\times 10^{-19}J}=5.43\times 10^{-7}m[/tex]

Hence, the wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]

Explanation:

a)

Moles of hydrogen gas = [tex]n_1=\frac{1.00 g}{2 g/mol}=0.5 mol[/tex]

Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.5 mol+0.03125 mol=0.53125 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.53125 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=6.54 atm[/tex]

Partial pressure of the hydrogen gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=6.54 atm\times \frac{0.5 mol}{0.53125 mol}=6.16 atm[/tex]

Partial pressure of the oxygen gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=6.54 atm\times \frac{0.03125 mol}{0.53125 mol}=0.38 atm[/tex]

hydrogen

= [tex]\frac{n_1}{n}\times 100=\frac{0.5 mol}{0.53125 }\times 100[/tex]

= 94.12%

oxygen :

= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.53125  mol}\times 100[/tex]

= 5.88%

b)

Moles of nitrogen gas = [tex]n_1=\frac{1.00 g}{28 g/mol}=0.03571 mol[/tex]

Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.03571 mol+0.03125 mol=0.06696 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.06696 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=0.82 atm[/tex]

Partial pressure of the nitrogen gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=0.82 atm\times \frac{0.03571 mol}{0.06696 mol}=0.44 atm[/tex]

Partial pressure of the oxygen gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=0.82 atm\times \frac{0.03125 mol}{0.06696 mol}=0.38 atm[/tex]

Composition of each in mole percent :

nitrogen

= [tex]\frac{n_1}{n}\times 100=\frac{0.03571 mol}{0.06696 }\times 100[/tex]

= 53.33%

oxygen :

= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.06696 mol}\times 100[/tex]

= 46.67%

c)

Moles of methane gas = [tex]n_1=\frac{1.00 g}{16 g/mol}=0.0625 mol[/tex]

Moles of ammonia gas = [tex]n_2=\frac{1.00 g}{17g/mol}=0.0588 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.0625 mol+0.0588 mol=0.1213 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.1213 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=1.49 atm[/tex]

Partial pressure of the methane gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=1.49 atm\times \frac{0.0625 mol}{0.1213mol}=0.77 atm[/tex]

Partial pressure of the ammonia gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=1.49 atm\times \frac{0.0588 mol}{0.1213mol}=0.72 atm[/tex]

Composition of each in mole percent :

Methane :

= [tex]\frac{n_1}{n}\times 100=\frac{0.0625 mol}{0.1213mol}\times 100[/tex]

= 51.52%

Ammonia

= [tex]\frac{n_2}{n}\times 100=\frac{0.0588 mol}{0.1213mol}\times 100[/tex]

= 48.47%

To calculate the partial pressure of each gas and the composition of the mixture, we can use the ideal gas law. By dividing the mass of each gas by its molar mass, we can calculate the number of moles. The total pressure can be obtained by summing up the partial pressures, and the mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

To calculate the partial pressure of each gas, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. To calculate the total pressure, we can sum up the partial pressures of each gas. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

a) 1.00 g H2 and 1.00 g O2:

To calculate the number of moles, we divide the mass of each gas by its molar mass. The molar mass of H2 is 2 g/mol and the molar mass of O2 is 32 g/mol. So, the number of moles of H2 is 1 g / 2 g/mol = 0.5 mol, and the number of moles of O2 is 1 g / 32 g/mol = 0.03125 mol. The total number of moles is 0.5 mol + 0.03125 mol = 0.53125 mol.

The partial pressure of H2 can be calculated by multiplying the number of moles of H2 by the gas constant R and the temperature in Kelvin, and then dividing by the volume. The same process can be applied to calculate the partial pressure of O2. Finally, the total pressure can be calculated by summing up the partial pressures. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

b) 1.00 g N2 and 1.00 g O2:

Using the same process as before, we can calculate the number of moles of N2 and O2. The molar mass of N2 is 28 g/mol, so 1 g of N2 is equal to 1 g / 28 g/mol = 0.03571 mol of N2. The molar mass of O2 is 32 g/mol, so 1 g of O2 is equal to 1 g / 32 g/mol = 0.03125 mol of O2. The total number of moles is 0.03571 mol + 0.03125 mol = 0.06696 mol.

Using the ideal gas law, we can calculate the partial pressures of N2 and O2, the total pressure, and the mole percent composition.

c) 1.00 g CH4 and 1.00 g NH3:

Following the same calculations as above, the molar mass of CH4 is 16 g/mol, so 1 g of CH4 is equal to 1 g / 16 g/mol = 0.0625 mol of CH4. The molar mass of NH3 is 17 g/mol, so 1 g of NH3 is equal to 1 g / 17 g/mol = 0.05882 mol of NH3. The total number of moles is 0.0625 mol + 0.05882 mol = 0.12132 mol.

By using the ideal gas law, we can calculate the partial pressures of CH4 and NH3, the total pressure, and the mole percent composition.

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Answer:

Rate = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹

Explanation:

The equation of the study of biochemical reaction mechanisms is:

³²P  →  ³²S + e⁻  

The rate of that reaction is:

Rate = k [³²P]

where k: is the rate constant and [³²P]  is the concentration of ³²P.  

Hence, having that: k = 4.85x10⁻² day⁻¹ and, [³²P] = 0.0031 M, the instantaneous rate of production of electrons is:

Rate = k [³²P]  = (4.85x10⁻² day⁻¹)(0.0031 mol*L⁻¹) = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹

Therefore, the instantaneous rate of production of electrons is 1.50x10⁻⁴ mol*day⁻¹*L⁻¹.

I hope it helps you!