Just as the depletion of stratospheric ozone threatens life on Earth today, its accumulation was one of the crucial processes that allowed life to develop in prehistoric times: 3O2(g) → 2O3(g) At a given instant, the reaction rate in terms of [O2] is 2.17 × 10−5 mol/L·s. What is it in terms of [O3]? Enter your answer in scientific notation.

Answer: The correct answer is Option c.

Explanation:

We are given:

Mass percentage of [tex]CH_4[/tex] = 20 %

So, mole fraction of [tex]CH_4[/tex] = 0.2

Mass percentage of [tex]C_2H_4[/tex] = 30 %

So, mole fraction of [tex]C_2H_4[/tex] = 0.3

Mass percentage of [tex]C_2H_2[/tex] = 35 %

So, mole fraction of [tex]C_2H_2[/tex] = 0.35

Mass percentage of [tex]C_2H_2O[/tex] = 15 %

So, mole fraction of [tex]C_2H_2O[/tex] = 0.15

We know that:

Molar mass of [tex]CH_4[/tex] = 16 g/mol

Molar mass of [tex]C_2H_4[/tex] = 28 g/mol

Molar mass of [tex]C_2H_2[/tex] = 26 g/mol

Molar mass of [tex]C_2H_2O[/tex] = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

[tex]\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}[/tex]

where,

[tex]\chi_i[/tex] = mole fractions of i-th species

[tex]m_i[/tex] = molar masses of i-th species

[tex]n_i[/tex] = number of observations

Putting values in above equation:

[tex]\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}[/tex]

[tex]\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75[/tex]

Hence, the correct answer is Option c.

Answer:

Take a look to R, where the units are L . atm/K . mol.. your pressure is in Torr...so make the conversion to atm. (760 Torr is 1 atm) and then take the volume... as you have mL, remember that R is with L, so convert mL to L by making the division /1000. Pressure and volume are those you have to convert

To calculate the number of moles, convert the pressure from torr to atm and the volume from mL to L, in order to match the given gas constant's units of L⋅atm/(K⋅mol). Then, use the ideal gas equation.

To find the number of moles of gas, n, in the given sample using the ideal gas equation PV=nRT, the pressure and volume units must match the units of the gas constant R. The given R is 8.206x10^-2 L⋅atm/(K⋅mol) meaning that P should be in atmospheres (atm) and V should be in liters (L).

The necessary conversions you need are:

After these conversions are carried out, the ideal gas equation can be used to calculate the number of moles, n.

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Answer : The mass of 5.0 moles of ribose is 750 grams.

Explanation : Given,

Moles of ribose = 5.0 moles

Molar mass of ribose = 150 g/mole

Formula used :

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

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

[tex]\text{Mass of ribose}=5.0mole\times 150g/mole=750g[/tex]

Therefore, the mass of 5.0 moles of ribose is 750 grams.

Final answer:

To calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20, you need to determine the molar mass of the compound and then divide the mass of the sample by the molar mass. Finally, multiply the number of moles by 1000 to convert to millimoles.

Explanation:

To calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20, we first need to determine the molar mass of FeSO4•C2H4(NH3)2SO4.4H20.

The molar mass of FeSO4 is 55.85 g/mol. The molar mass of C2H4(NH3)2SO4.4H20 can be calculated by adding the molar masses of each element (12.01 g/mol for C, 1.008 g/mol for H, 14.01 g/mol for N, 32.06 g/mol for S, 16.00 g/mol for O, and 1.008 g/mol for H).

Next, we convert 500mg to grams. 500mg is equal to 0.5g.

Then, we divide the mass of the sample by the molar mass to calculate the number of moles. Finally, we multiply the number of moles by 1000 to convert to millimoles.

Therefore, the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20 can be calculated as follows:

Number of millimoles = (0.5g / molar mass) * 1000

Answer:

b) Cu2+

Explanation:

Ksp PbS = 3.4 E-28

Ksp CuS = 6.0 E-37

∴ Ksp = 3.4 E-28 = [ Pb2+ ] * [ S2- ]

∴ [ Pb2+ ] = 0.10 M

⇒ [ S2- ] = 3.4 E-28 / 0.10 = 3.4 E-27 M

∴ Ksp = 6.0 E-37 = [ Cu2+ ] * [ S2- ]

∴ [ Cu2+ ] = 0.10 M

⇒ [ S2- ] = 6.0 E-37 / 0.10 = 6.0 E-36 M

we have:

(1) [ S2- ] PbS >> [ S2- ] CuS

(2) Ksp PbS >> Ksp CuS

from (1) and (2) it can determined, that separation can be carried out and also the cation that precipitates first is the Cu2+

Answer:

The volume of feedstock needed to mantain an organic load rate of 2 kgVS/day is 0.055 m3/day of feedstock.

The HRT is 20.6 days.

Explanation:

First, we calculate how many kg is 1 m3 of feedstock. We know the density, so we can calculate the mass:

[tex]M=\rho*V=37.5\frac{lb}{ft^3}*1m^3*(\frac{3.281ft}{1m})  ^3=1324.5lb=600.7 kg[/tex]

If the VS are 6% in weight,

[tex]M_{vs}=0.06*M=0.06*600.7\,kg/m^3=36,0kgVS/m3[/tex]

The volume per day needed to feed 2 kg of VS/day is:

[tex]V=\frac{2kg}{36kg/m^3}= 0.055m3/day=5.5litres/day[/tex]

The HRT depends on the volume of the tank and the flow. Its equation is

[tex]HRT=\frac{V}{Q}=\frac{300gal}{0.055 m^3/day}*\frac{1m^3}{264.172gal}\\   \\HRT=20.6\,days[/tex]

Answer: Entropy

Explanation:

Entropy is the measure of randomness or disorder. It is a thermodynamic state function corresponding to the number of available microstates of a system.

Enthalpy is the difference between the energy of products and the energy of reactants.

Kinetics is the the branch of chemistry which deals with rates of chemical reactions.

Thermodynamics is the branch of chemistry which deals with energy and energy conversions.

Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=0.2829g[/tex]

Mass of [tex]H_2O=0.1159g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, [tex]\frac{12}{44}\times 0.2829=0.077g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, [tex]\frac{2}{18}\times 0.1159=0.0129g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = [tex]\frac{0.0064}{0.00066}=9.69\approx 10[/tex]

For Hydrogen  = [tex]\frac{0.0129}{0.00064}=19.54\approx 20[/tex]

For Oxygen  = [tex]\frac{0.00066}{0.00066}=1[/tex]

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

[tex]n=\frac{156.27g/mol}{156g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]

Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

Answer:

[tex]\boxed{\text{C$_{10}$H$_{20}$O}}[/tex]

Explanation:

In a combustion experiment, all the carbon ends up as CO₂, and all the hydrogen ends up as water.

Data:

Mass of menthol = 0.1005 g

     Mass of CO₂ = 0.2829 g

     Mass of H₂O = 0.1159   g

Calculations:

(a) Mass of each element

[tex]\text{Mass of C} = \text{0.2829 g CO$_{2}$} \times \dfrac{\text{12.01 g C}}{\text{44.01 g CO$_{2}$}} = \text{0.077 20 g C}\\\\\text{Mass of H} = \text{0.1159 g H$_{2}$O} \times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} =  \text{0.012 97 g H}\\\\\text{Mass of O} = \text{mass of menthol - mass of C - mass of H}\\= \text{0.1005 - 0.077 20 - 0.01297}= \text{0.010 33 g O}[/tex]

(b) Moles of each element

[tex]\text{Moles of C} = \text{0.077 20 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = 6.428 \times 10^{-3}\text{ mol C}\\\\\text{Moles of H} = \text{0.012 97 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{ 0.012 86 mol H}\\\\\text{Moles of O} = \text{0.010 33 g O} \times \dfrac{\text{1 mol O }}{\text{16.00 g O}} = 6.458 \times 10^{-4}\text{ mol O}[/tex]

(c) Molar ratios

Divide all moles by the smallest number of moles.

[tex]\text{C: } \dfrac{6.428 \times 10^{-3}}{6.458 \times 10^{-4}} = 9.954\\\\\text{H: } \dfrac{0.012 86}{6.458 \times 10^{-4}} = 19.92\\\\\text{O: } \dfrac{6.458 \times 10^{-4}}{6.458 \times 10^{-4}} = 1[/tex]

(d) Round the ratios to the nearest integer

C:H:O = 10:20:1

(e) Write the empirical formula

The empirical formula is C₁₀H₂₀O.

(f) Calculate the empirical formula mass

 10 × C = 10 × 12.01  = 120.1    u

20 × H = 20 × 1.008 =  20.16  u

  1 × O = 1 × 16.00    =   16.00 u

                EF  mass =  156.3    u

(g) Divide the molecular mass by the empirical formula mass.  

[tex]n = \dfrac{\text{MM}}{\text{EFM}} = \dfrac{156.27}{156.3} = 0.9998 \approx 1[/tex]

(h) Determine the molecular formula

[tex]\text{MF} = \text{(EF)}_{n} = \rm (C_{10}H_{20}O)_{1} = \textbf{C$_{10}$H$_{20}$O}\\\text{The molecular formula of menthol is } \boxed{\textbf{C$_{10}$H$_{20}$O}}[/tex]

Answer:

The correct answer is: alcohol (C-OH) functional group

Explanation:

Electronegativity is described as the ability or the tendency of an element to attract the electron density or shared bonding electrons towards itself.  

The electronegativity value of oxygen atom, nitrogen atom and, carbon atom is 3.44, 3.04, and 2.55.

From this data we can conclude that oxygen is more electronegative than nitrogen. Also, the electronegativity difference of oxygen and carbon (0.89) is greater than the electronegativity difference of nitrogen and carbon (0.49).

Therefore, the electronegativity of the oxygen containing alcohol functional group (C-OH) will be greater than the electronegativity of the nitrogen (C-NH) containing amine functional group.

Answer:

Enantiomers are the non-superimposable mirror images of each other.

Diastereomers are the stereisomers that are not a reflection or mirror images of each other.  

Explanation:

Stereoisomers are the chemical molecules having the same molecular formula and bond connectivity but different arrangement of atoms in space.

Stereoisomers are of two types: Enantiomers and Diastereomers

Enantiomers are the non-superimposable mirror images of each other. Enantiomers are also called optical isomers.

Diastereomers are the stereisomers that are not a reflection or mirror images of each other. Diastereomers include E-Z isomers, cis–trans isomers, meso compounds, non-enantiomeric optical isomers.

Answer:

ΔG = -1.53 kJ/mol

Explanation:

The given reaction is:

3-Phosphoglycerate → 2-Phosphoglycerate

The standard Gibbs free energy, ΔG°=+4.40 kJ

[2-Phosphoglycerate] = 0.290 mM

[3-Phosphoglycerate] = 2.90 mM

Temperature T = 37 C = 310 K

The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:

[tex]\Delta G =\Delta G^{0}+RTlnQ[/tex]

In this reaction:

[tex]\Delta G =\Delta G^{0}+RTln\frac{[2-Phosphoglycerate]}{[3-Phosphoglycerate]}[/tex]

[tex]\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln\frac{[0.290]}{[2.90]}=-1.53 kJ/mol[/tex]

Answer:

0.8 mL of protein solution, 9.2 mL of water

Explanation:

The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:

C₁V₁ = C₂V₂

We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.

V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL

Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:

(10 mL - 0.8 mL) = 9.2 mL

The statement that best describes a buffer is: A buffer resists change in pH by accepting hydrogen ions when acids are added to the solution and donating hydrogen ions when bases are added.

Why?

A buffer is a solution made by combining either:

The purpose of a buffer is to resist changes in pH when a strong acid or base is added to the solution.

If the buffer is composed of HA and A⁻ and a strong acid (e.g. HCl) is added, the buffer accepts hydrogen ions in the following way:

A⁻+HCl → HA+Cl⁻

If a strong base (e.g. NaOH) is added, the buffer donates hydrogen ions in the following way:

HA + NaOH → NaA + H₂O

The pH of the buffer at any given moment can be found by using the Henderson-Hasselbach equation, based on the equilibrium HA + H₂O ⇄ H₃O⁺ + A⁻

[tex]pH=pKa+log\frac{[A^{-}] }{[HA]}[/tex]

Have a nice day!

Answer:

A buffer resists change in pH by accepting hydrogen ions when acids are added to the solution and donating hydrogen ions when bases are added.

Explanation:

i took it

Answer:

a) Water removed = 8.6 kg

b) Dry air in the fresh air = 95.6 kg

c) Dry air in the recycled air = 27.3 kg

Explanation:

To solve this problem we have to make mass balances of the different streams.

1) Material balance for the dry solid

For every 100 kg of feed, we have 85 kg of dry solid and 15kg of water.

If the exit material has 7% of moisture content, the total dry solid represents 93% of the mass exiting the drier.

If the dry solid is 85 kg and represents 93% of the total exit material, the total amount of exit material is 85/0.93=91.4 kg. The difference (7%) is water, weighting (91.4-85)=6.4 kg.

The water removed for every 100 kg of feed is (15-6.4)=8.6 kg.

2) Material balance for the water

The water entering the system has to be the same that exit the system.

Let da be the amount of dry air. Then the water entering the drier is (15+0.01*da) and the water exiting the drier is (6.4+0.1*da). We can calculate the amount of dry air:

[tex]15+0.01*da=6.4+0.1*da\\(15-6.4)=(0.1-0.01)*da\\da=8.6/0.09=95.6[/tex]

For every 100 kg of feed, 95.6 kg of dry air is entering the drier.

3) Recycled air

Let rda be the amount of dry air in the recycled stream. We can balance the water content like:

water in the fresh air + water in the recycled air = water in the air entering the drier

[tex]0.01*da+0.1*rda=0.03*(da+rda)\\\\0.1*rda-0.03*rda=0.03*da-0.01*da\\\\0.07*rda=0.02*da\\\\rda=(0.02/0.07)*da=0.286*da=0.286*95.6=27.3 kg[/tex]

The amount of dry air in the recycled stream is 27.3 kg.

Explanation:

The sum of total number of protons present in an element is known as atomic number of the element.

And, it is known that for a neutral atom the number of protons equal to the number of electrons.

Since, no charge in present on given Cs atom it means that it is neutral in nature. Hence, number of protons and electrons present in Cs are 55.

Explanation:

It is known that in reversible isothermal compression, relation between work and pressure is as follows.

                     w = -2.303 RT log [tex]\frac{P_{2}}{P_{1}}[/tex]

                         = [tex]-2.303 \times 8.314 J/mol K \times log \frac{150 bar}{15 bar}[/tex]

                          = [tex]-5705.85 J /mol \times log (10)[/tex]

                          = -5705.85 J /mol

According to first law of thermodynamics, q = -w

Hence,                            q = -(-5705.85 J /mol)

                                            = 5705.85 J /mol

As 1000 J = 1 kJ. Hence, convert 5705.85 J/mol into kJ/mol as follows.

                             [tex]\frac{5705.85}{1000} kJ /mol[/tex]

                           = 5.7058 kJ/mol

Thus, we can conclude that heat  removal is 5.7058 kJ/mol and work required is -5705.85 J /mol.

Final answer:

In an isothermal and reversible compression of helium gas from 15 bar to 150 bar at 298 K, the work done can be calculated using the formula W = nRT ln([tex]P_1/P_2[/tex]), where n is the number of moles. The heat removed is equal to the work done, but with opposite sign.

Explanation:

Isothermal Compression of Helium Gas

When helium is isothermally and reversibly compressed in a piston/cylinder at a constant temperature of 298 K from an initial pressure of 15 bar to a final pressure of 150 bar, the work (W) done on the gas is found using the formula for isothermal processes for an ideal gas:

W = nRT ln([tex]P_1/P_2[/tex])

Where n is the number of moles of helium, R is the ideal gas constant (8.314 J/(mol K)), T is the temperature in Kelvin, and [tex]P_1[/tex] and [tex]P_2[/tex] are the initial and final pressures.

However, without knowing the number of moles of helium, we cannot compute the exact value of the work. As for the heat removal, for an isothermal process in an ideal gas, the amount of heat removed (Q) is equal to the work done on the gas:

Q = -W

Therefore, the heat removed would be numerically equal to the work required but opposite in sign since the work is done on the gas and the heat is released by the gas.

Answer:

49.26L

Explanation:

Hello,

Considering Boyle's law:

[tex]P_1V_1=P_2V_2[/tex]

The volume in the second state is given by (solving for it):

[tex]V_2=\frac{P_1V_1}{P_2} =\frac{25L*6.7kPa}{3.4kPa} \\V_2=49.26L[/tex]

Best regards.

Final answer:

By applying Boyle's Law, we can calculate that when the pressure of the nitrogen gas in a 25 L tank is decreased from 6.7 kPa to 3.4 kPa, the volume increases to 49.26 L.

Explanation:

The subject of the question involves applying Boyle's Law, which is a principle in Chemistry related to the behavior of gases under pressure. Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to the pressure. To calculate the new volume when the pressure is decreased, we can set up the equation as follows:

P1 × V1 = P2 × V2

Given that P1 = 6.7 kPa and V1 = 25 L (the initial state), and P2 = 3.4 kPa (the final pressure), we want to find V2, the final volume. Using Boyle's Law, we can calculate the new volume:

V2 = (P1 × V1) / P2

V2 = (6.7 kPa × 25 L) / 3.4 kPa

V2 = 49.26 L

Therefore, when the pressure of the nitrogen gas is decreased to 3.4 kPa, the volume increases to 49.26 L.

Answer:

A catalyst decreases the activation energy of a reaction.

Explanation:

A catalyst allows a reaction to go through a new pathway that has a lower activation energy. Therefore, it will recquire less energy to reach the transition state (activated complex) which increases the rate of the reaction. This is the reason why adding a catalyst makes a reaction faster.

The figure attached shows the graphical representation of the effect of a catalyst on the activation energy.

Answer:

c. A catalyst reduces the activation energy required for a reaction.

Explanation:

plato

Answer:

a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. It is Closed because the freezer only exchanges energy, Isothermal since the freezer maintain the temperature constant and Isothermal and Isobaric because the ice cube remains with volume and pressure constant.

b. The air inside the tire of a Nascar during the first minute of driving in a race. Closed because the tire only exchange energy at first and Isochoric since the volume of the tire remain constant.

c. Your body over the last week. Open because the body exchange matter and energy.

Explanation:

The open, closed, adiabatic and isolated systems are defined considering if exchange matter or energy, as the definitions below:

- An open system exchange matter and energy.

- A closed system exchange only energy.

- An adiabatic system only exchange matter.

- An isolated system not exchange matter and energy

The isothermal, isobaric, isochoric, or steady-state are defined as follows:

- Isothermal is a process at a constant temperature.

- Isobaric is a process at constant pressure.

- Isochoric is a process at a constant volume.

- A steady-state refers to a reaction in which the concentrations of the reactants, intermediaries, and products don't change over time.

Answer: Option (a) is the correct answer.

Explanation:

A spontaneous reaction is defined as the reaction which occurs in the given set of conditions without any disturbance from any other source.

A spontaneous reaction leads to an increase in the entropy of the system. This means that degree of randomness increases in a spontaneous reaction.

For example, [tex]H_{2}(g) \rightarrow 2H(g)[/tex]

Here, 1 mole of hydrogen is giving 2 moles of hydrogen. This means that degree of randomness is increasing on the product side due to increase in number of moles.

Hence, there will also be increase in entropy.

Whereas in the reaction, [tex]CO_{2}(s) \rightarrow CO_{2}(g)[/tex] here number of moles remain the same. Hence, the reaction is not spontaneous.

Thus, we can conclude that the reaction [tex]H_{2}(g) \rightarrow 2H(g)[/tex] is spontaneous at STP.

Answer:

a. 0.01 mol/L

b. 1382 mg/L

c. 1382 ppm

d. 0.138%

Explanation:

Hello,

I'm attaching two photos showing the numerical procedure for this exercise.

Note: since we don't have the density of the solution and the solvent isn't specified, we could assume it is water (universal solvent).

Best regards.

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ[tex]\frac{1mol}{102g}[/tex] = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ[tex]\frac{1mol}{36,5g}[/tex] = 0,274 moles

For a total reaction of 0,274 moles of HCl you need:

0,274×[tex]\frac{1molesAl_{2}O_3}{6 mole HCl}[/tex] = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×[tex]\frac{2 moles AlCl_{3}}{6 moles HCl}[/tex] × 133[tex]\frac{g}{mol}[/tex] = 12,1 g of AlCl₃

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles = 0,0523 moles

And its mass is:

0,0523 molesₓ[tex]\frac{102g}{1mol}[/tex] = 5,33 g of Al₂O₃

I hope it helps!

Answer : The mass of oxygen present in the flask is 0.03597 grams.

Explanation :

First we have to determine the moles of [tex]CO_2[/tex] gas.

[tex]\text{ Moles of }CO_2=\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}=\frac{1.100g}{44g/mole}=0.025moles[/tex]

Now we have to calculate the moles of the oxygen gas.

Using ideal gas equation:

[tex]PV=nRT[/tex]

As, the moles is an additive property. So,

[tex]PV=(n_{O_2}+n_{CO_2})RT[/tex]

where,

P = pressure of gas = 608 mmHg = 0.8 atm

(conversion used : 1 atm = 760 mmHg)

V = volume of gas = 1 L

T = temperature of gas = 373 K

[tex]n_{O_2}[/tex] = number of moles of oxygen gas = ?

[tex]n_{CO_2}[/tex] = number of moles of carbon dioxide gas = 0.025 mole

R = gas constant = [tex]0.0821L.atmK^{-1}mol^{-1}[/tex]

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

[tex](0.8atm)\times (1L)=(n_{O_2}+0.025)mole\times (0.0821L.atmK^{-1}mol^{-1})\times (373K)[/tex]

[tex]n_{O_2}=0.001124mole[/tex]

Now we have to calculate the mass of oxygen gas.

[tex]\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2[/tex]

[tex]\text{Mass of }O_2=0.001124mole\times 32g/mole=0.03597g[/tex]

Therefore, the mass of oxygen present in the flask is 0.03597 grams.

Final answer:

To find the mass of oxygen in a flask with CO2, convert the pressure to atm, calculate moles of CO2, and use the ideal gas law to find CO2's partial pressure. Subtract this from the total to find oxygen's partial pressure, calculate its moles, and use its molar mass to find the oxygen's mass.

Explanation:

To find the mass of the oxygen in the flask, we can apply the ideal gas law and use Dalton's Law of Partial Pressures. Given that 1.100g of CO2 was introduced into a 1L flask containing oxygen gas at 373K with a total pressure of 608mmHg, if CO2 and O2 were the only gases present, we have enough information to solve for the mass of the oxygen gas.

First, convert the total pressure to atm (608 mmHg = 0.8 atm) because standard gas law constants are typically given in atmospheres. The first step in solving for the mass of oxygen involves finding the moles of CO2 introduced, using its molar mass (44.01 g/mol), and applying the ideal gas law:

Calculate moles of CO2: Moles = mass / molar mass = 1.100g / 44.01 g/mol = 0.025 moles of CO2.

Apply Ideal Gas Law for CO2: Use P = nRT/V, where R = 0.0821 L atm/mol K, to find the partial pressure of CO2.

Since the total pressure is known, we can subtract the partial pressure of CO2 to find the partial pressure of O2.

Finally, use the ideal gas law again with the partial pressure of O2 to find the moles of O2 present, and then use the moles of O2 along with its molar mass (32.00 g/mol) to calculate the mass of oxygen in the flask.

This approach accounts for the behavior of the gas mixture under the given conditions, utilizing pressure, volume, temperature, and the molar mass of gases to solve for the unknown mass of oxygen.

Answer : The final temperature of gas will be, 149 K

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 1.55 L

[tex]V_2[/tex] = final volume of gas = 753 ml  = 0.753 L

[tex]T_1[/tex] = initial temperature of gas = [tex]32^oC=273+32=305K[/tex]

[tex]T_2[/tex] = final temperature of gas = ?

Now put all the given values in the above formula, we get the final temperature of the gas.

[tex]\frac{1.55L}{0.753L}=\frac{305K}{T_2}[/tex]

[tex]T_2=149K[/tex]

Therefore, the final temperature of gas will be, 149 K

Answer:

[tex]d=0.92\frac{kg}{m^{3}}[/tex]

Explanation:

Using the Ideal Gas Law we have [tex]PV=nRT[/tex] and the number of moles n could be expressed as [tex]n=\frac{m}{M}[/tex], where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

[tex]PV=\frac{m}{M}RT[/tex]

If we pass the V to divide:

[tex]P=\frac{m}{V}\frac{RT}{M}[/tex]

As the density is expressed as [tex]d=\frac{m}{V}[/tex], we have:

[tex]P=d\frac{RT}{M}[/tex]

Solving for the density:

[tex]d=\frac{PM}{RT}[/tex]

Then we need to convert the units to the S.I.:

[tex]T=100^{o}C+273.15[/tex]

[tex]T=373.15K[/tex]

[tex]P=1bar*\frac{0.98atm}{1bar}[/tex]

[tex]P=0.98atm[/tex]

[tex]M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}[/tex]

[tex]M=0.0289\frac{kg}{mol}[/tex]

Finally we replace the values:

[tex]d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}[/tex]

[tex]d=9.2*10^{-4}\frac{kg}{L}[/tex]

[tex]d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}[/tex]

[tex]d=0.92\frac{kg}{m^{3}}[/tex]

AnswerAnswer:

The purity of the sample is 67.14 %

Explanation:

The titration reaction is as follows:

NaOH + aspirin-H → Na⁺ + aspirin⁻ + H₂O

When no more aspirin-H is left, the addition of more NaOH raises the pH and the color of the indicator turns, in this case to a pink color. This is the reaction at the endpoint that indicates that no more aspirin-H is left:

NaOH + aspirin⁻ → Na⁺ + aspirin⁻ + OH⁻

Then, the moles of NaOH added until the turn of the indicator must be equal to the number of moles of aspirin present in the solution since NaOH reacts with aspirin in a 1:1 ratio.

Then:

moles of aspirin in the solution = moles of added NaOH

moles of aspirin in the solution = Concentration of NaOH * volume

moles of aspirin in the solution = 0.1105 mol/l * 0.01522 l = 1.682 x 10⁻³ mol

Knowing the molar mass of aspirin, we can calculate the mass of aspirin present in the solution:

1.682 x 10⁻³ mol aspirin *(180.16 g / mol) = 0.3030 g.

Since the sample contained 0.4513 g, the percent of aspirin in the sample will be: 0.3030 g * (100 % / 0.4513 g) = 67.14 %

We will get the same result if we convert the mass of the sample to mol and calculate the purity using moles instead of mass:

moles of aspirin in the sample:

0.4513 g * ( 1 mol / 180.16g) = 2.505 x 10⁻³ mol aspirin

The purity will be then:

1.682 x 10⁻³ mol * ( 100 % / 2.505 x 10⁻³ mol) = 67.14 %

Explanation:

The given data is as follows.

         Mass of hexane = 10 kg,        Mass of octane = 30 kg

Formula to calculate average density of mixture is as follows.

                      [tex]\sum x_{i} \rho_{i}[/tex]

where,    [tex]x_{i}[/tex] = mass fraction of i component in mixture

              [tex]\rho_{i}[/tex] = density of i component

Hence, calculating the mass fraction of both hexane and octane as follows.

           [tex]x_{hexane}[/tex] = [tex]\frac{10}{(10 + 30)}[/tex] = 0.25

           [tex]x_{octane}[/tex] = [tex]\frac{30}{(10 + 30)}[/tex] = 0.75

Therefore, calculate the average density as follows.

          [tex]\rho_{mixture}[/tex] = [tex](0.26 \times 660) + (0.75 \times 703)[/tex]

                         = 692.25 [tex]kg/m^{3}[/tex]

Thus, we can conclude that the average density of given hexane/octane mixture is 692.25 [tex]kg/m^{3}[/tex].

Answer : The answer will be 20.68

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The rule apply for the addition and subtraction is :

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

The given expression is:

[tex](17.543+2.19)\times 1.04821[/tex]

In the given expression, 17.543 has 5 significant figures and 2.19 has 3 significant figures. From this we conclude that least precise number present after the decimal point is 2.  So, the answer will be:

[tex](19.73)\times 1.04821[/tex]

In the given expression, 19.73 has 4 significant figures and 1.04821 has 6 significant figures. From this we conclude that 4 is the least significant figures in this problem. So, the answer should be in 4 significant figures.

[tex]\Rightarrow 20.68[/tex]

Thus, the answer will be 20.68

Answer:  [tex](27.81,\ 29.39)[/tex]

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : [tex]\alpha: 1-0.95=0.05[/tex]

Critical value: [tex]z_{\alpha/2}=1.96[/tex]

Sample mean : [tex]\overline{x}=28.6[/tex]

Standard deviation : [tex]\sigma=2.2[/tex]

The formula to find the confidence interval is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

i.e. [tex]28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}[/tex]

i.e. [tex]28.6\pm 0.787259889321[/tex]

[tex]\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)[/tex]

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = [tex](27.81,\ 29.39)[/tex]

Answer:

The amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are 5 kg of A

Explanation:

A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent, and knowing the solubility of component A at 80°C it is possible to know their amount, thus:

10Kg of water ×[tex]\frac{0,8 kg A}{1 kgWater}[/tex] = 8 kg of A

The maximum concentration that water can dissolve at 30°C is:

10Kg of water ×[tex]\frac{0,3 kg A}{1 kgWater}[/tex] = 3 kg of A

Thus, the amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are:

8 kg of A - 3 kg of A = 5 kg of A

I hope it helps!