36. Dimensional Analysis: A useful way to convert units using multiplication & division *Always start with what you are given! Practice: A. Convert 2.0 x 105 m to inches, given that 1 meter = 39.37 inches. B. Convert 1.004 kg to grams, given that 1 pound (lb) = 453.59 grams (g) and 1 kilogram (kg) = 2.205 pounds (lb). C. Convert 1.2 m/s to inches/minute. D. Convert 120 nm to mm.

Answer: Option (b) is the correct answer.

Explanation:

Sublimation is defined as the process in which a solid substance changes directly into gaseous phase without undergoing into liquid phase.

For example, when naphthalene balls are kept in a trunk or cupboard then after some days or months it changes into gas. Hence, they become consumed.

Also during this process energy is being absorbed by the solid substance because for breaking bonds energy is always absorbed by a substance. Only then it can change into liquid or vapor state.

So, a chemical process in which energy is being absorbed by a reaction is known as endothermic reaction. And, if heat energy is being released by the reaction then it is known as an exothermic reaction.

Therefore, we can conclude that when a solid sublimes, a solid converts to a gas. This phase change is endothermic because energy is absorbed when the intermolecular forces break.      

Explanation:

Rules for counting significant figures:

1) 101 cm - 99.1 cm = 1.9 cm ≈ 2.0 cm

1.9 cm ≈ 2.0 cm

There are two significant figures that is 2 and 0.

2) 0.5 atm + 9.8 atm = 10.3 atm ≈ 10.0 atm

There are three significant figures that is 3 , 0 and 0.

The atomic mass of Cr53 can be calculated by establishing an equation that takes into account the atomic masses and abundances of all isotopes of Chromium. Solve the equation for 'x' (representing the atomic mass of Cr53) to find your answer.

The atomic mass of a given isotope of an element is determined by the weighted average of the masses of its isotopes, each multiplied by the abundance of that isotope. Since the atomic masses and abundances of Cr50, Cr52, and Cr54 are already provided, we only need to account for Cr53's mass. Using Chromium's atomic mass (51.9961 u), we can establish an equation to solve for Cr53's atomic mass.

Here's our equation:

51.9961 u = (49.9460 u * Cr50's abundance) + (51.9405 u * 0.8379) + (x * Cr53's abundance) + (53.9389 u * 0.0237)

First calculate the contribution of Cr50, Cr52, and Cr54 to the atomic weight, then subtract this from the total atomic weight (51.9961 u). Divide this value by the Cr53's abundance to get the atomic weight of Cr53.

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The atomic mass of the Cr53 isotope is approximately 23.83 u, calculated based on its natural abundance and atomic mass in the given chromium isotope mixture.

To determine the atomic mass of the Cr53 isotope, we can use the information provided about the natural abundances and atomic masses of the chromium isotopes.

Let's denote the natural abundance of Cr50 as x. Since the ratio of Cr50 to Cr53 is 1:0.4579, the natural abundance of Cr53 would be 0.4579x. The total natural abundance of Cr50 and Cr53 is 1, so we have the equation:

x + 0.4579x = 1

Solving for x, we find that x is approximately 0.6852.

Now, we can calculate the total contribution of Cr53 to the atomic mass:

Atomic mass of Cr53 = Natural abundance of Cr53 * Atomic mass of Cr53

Atomic mass of Cr53 = 0.4579 * 51.9961

Calculating this gives us the atomic mass of the Cr53 isotope.

Atomic mass of Cr53 = 23.8256 u

In summary, the atomic mass of the Cr53 isotope is approximately 23.83 u.

Answer:

ΔG = 16.218 KJ/mol

Explanation:

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

Answer:

[tex]4.264*10^{6}[/tex]

Explanation:

Hello !

The correct answer is option C)  [tex]4.264*10^{6}[/tex]

To write in scientific notation, numbers between 0 and 9 are used before the point and the rest of the numbers afterwards.

Then we count how many places we ran the point. If you run to the right, the exponent of 10 is positive and if you run to the left, it is negative.

This way it is  [tex]4.264*10^{6}[/tex]

The density of the cylinder would be 3.652 gram/ cm³

.

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,

mass of the cylinder = 3.543 grams

the volume of the cylinder = πr²h

                                            = 3.14 ×.265²×4.4

                                            =0.97 cm³

By using the above formula for density

ρ = mass of the cylinder/volume of the cylinder

  = 3.543 grams/0.97 cm³

  =3.652 grams/ cm³

Thus,the density of the cylinder would be 3.652 grams/ cm³.

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To find the volume and density of a metal cylinder with a mass of 3.543 g, diameter of 0.53 cm, and height of 4.40 cm, we first calculate the volume as 3.08 cm³ using the formula V = πr²h. Then, we determine the density to be 1.15 g/cm³ by dividing the mass by the volume.

You just measured a metal cylinder and obtained the following information: mass - 3.543 g, diameter - 0.53 cm, height = 4.40 cm. To determine the volume and density of the cylinder, follow these steps:

Determining the Volume (V)

First, understand that the volume (V) of a cylinder can be calculated using the formula V = πr²h, where π (pi) is approximately 3.14, r is the radius of the cylinder (half of the diameter), and h is the height of the cylinder. Given a diameter of 0.53 cm, the radius is 0.53 cm / 2 = 0.265 cm. Substituting these values into the formula gives:

V = 3.14 * (0.265 cm)² * 4.40 cm = 3.08 cm³

Determining the Density (D)

Density can be found using the formula D = m / V, where m is the mass and V is the volume. Using the mass of 3.543 g and the volume of 3.08 cm³:

D = 3.543 g / 3.08 cm³ = 1.15 g/cm³

Answer:

C. fluid density & viscosity

Explanation:

In 1850, Darcy-Weisbach experimentally deduced an equation to calculate shear losses ("friction"), in a tube with permanent flow and constant diameter:

hf = (f x L x V^2) / (D x 2g)

where:

hf: shear losses

f:  shear loss factor (pipe roughness)

g: gravity acceleration

D: tube diameter

L: tube length  

V: fluid average speed in the tube

To calculate the loss factor “f” in the Poiseuille laminar region, he proposed in 1846 the following equation:

f = 64 / Re

Where:

Re: Reynolds number

The influence of the parameters on f is quantitatively different according to the characteristics of the current.

In any straight pipeline that transports a liquid at a certain temperature, there is a critical speed below which the regimen is laminar. This critical value that marks the transition between the two regimes, laminar and turbulent, corresponds to a Re = 2300, although in practice, between 2000 and 4000 the situation is quite inaccurate. Thus:

Re <2000: laminar regimen

2000 <Re <4000: critical or transition zone

Re> 4000: turbulent regime

Final answer:

The friction factor for fluid flow in a pipe does not depend upon the pipe length, pipe roughness, fluid density & viscosity, or mass flow rate of fluid.

Explanation:

The friction factor for fluid flow in a pipe does not depend upon the pipe length (A), pipe roughness (B), fluid density & viscosity (C), or mass flow rate of fluid (D).

This is because the friction factor, also known as the Darcy-Weisbach factor, is determined by the characteristics of the flow itself, such as the Reynolds number, which is a dimensionless quantity that relates the inertia of the fluid to the viscous forces acting on it.

The friction factor can be calculated using the Colebrook-White equation or obtained from Moody's diagram based on the relative roughness of the pipe and the Reynolds number.

It will take the microwave approximately 4.2 seconds to emit a total of 4650 J of energy, calculated by dividing the total energy by the power it emits per second, using two significant figures.

To calculate how long it will take for a microwave emitting 1,100 J of energy every second to emit a total of 4650 J, we can use the formula:

Time (s) = Total Energy (J) / Power (J/s).

Substituting the given values:

Time (s) = 4650 J / 1,100 J/s = 4.22727272727... seconds

Since the microwave emits energy at a rate of 1,100 J every second, we want to express the answer with the same number of significant figures as the least precise measurement, which is two significant figures. Therefore, the time should be rounded to 4.2 seconds.

Answer:

The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L

Explanation:

Molarity (M) means: moles of solute which are contained in 1 L of solution.  

In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.

0,2 moles ........... 1 L

5x10*-2 moles ........... x

x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L

(we can also say 250 mL)

Explanation:

(a)   Here, the assumption is that complete combustion of carbon is taking place. This means that there is no CO formation

Reaction equation when carbon is burning is as follows.

              [tex]C + O_{2} \rightarrow CO_{2}[/tex]

Molar mass of C = 12 kg/kmol

Molar mass of [tex]CO_{2}[/tex] = 44 kg/kmol

Hence, according to the stoichiometry,

1 Kmol of C reacted = 1 kmol of [tex]CO_{2}[/tex] produced

Therefore, 12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

It is given that coal contains 50 mass % Carbon

So, 1 kg of coal contains (0.50 × 1)kg Carbon

           Carbon in 1 kg Coal = 1 × 0.5 = 0.5 kg

As per the stoichiometry,

12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

0.5 kg of C reacted = x kg [tex]CO_{2}[/tex] produced

Therefore, value of x can be calculated as follows.

                    x = [tex]\frac{\text{44 kg of CO_{2} produced} \times \text{0.5 kg of C reacted}}{\text{12 kg of C reacted}}[/tex]

                       = 1.83 kg

This means that amount of [tex]CO_{2}[/tex] released is 1.83 kg.

(b)   It is assumed that coal contains 50 mass % carbon and 1 kg of coal burnt.

Since, it is given that energy density of coal is 24 Mj and efficiency of the power plant is 30%.

After burning 1 kg of coal amount of energy released = 24 Mj

Amount of energy converted to electricity = [tex]24 Mj \times 0.3[/tex] = 7.2 Mj

It is calculated that amount of [tex]CO_{2}[/tex] released per 1 kg of coal = 1.83 kg

Therefore, calculate the amount of [tex]CO_{2}[/tex] released in kg/Mj as follows.

        amount of [tex]CO_{2}[/tex] released in kg/Mj = [tex]\frac{\text{Amount of CO_{2} released in kg}}{\text{amount of energy converted to electricity}}[/tex]        

                        = [tex]\frac{1.83 kg}{7.2 mJ}[/tex]

                        = 0.2541 kg/Mj

Hence, the production of [tex]CO_{2}[/tex] in kg/MJ is 0.2541 kg/Mj.

To calculate the amount of CO2 produced from 1 kg of coal, first determine the mass of carbon, apply the conversion for the mass of CO2 produced per mass of carbon, and use the plant's efficiency and the coal's energy density to determine CO2 production per MJ.

The question involves consulting basic chemical stoichiometry and energy conversion to solve problems related to carbon dioxide emissions and energy efficiency in the context of coal combustion. Knowing the percentage of carbon in coal, we can determine the mass of carbon dioxide produced from burning coal. Additionally, we'll use the energy density of coal to calculate CO2 production per unit energy output of a coal-fired power plant.

To find the mass of CO2 produced from burning 1 kg of coal with 50% carbon content, we first calculate the mass of carbon that would burn:

The molecular weight of carbon is approximately 12 g/mol, and for CO2 it is approximately 44 g/mol. This implies that every 12 kg of carbon produces 44 kg of CO2. For 0.5 kg (or 500 g) of carbon, the mass of CO2 produced would be:

For the second part, to calculate the CO2 production in kg/MJ for a coal-fired power plant with 30% efficiency:

Answer:

A) 577.6 years

B) 1580.9 years

Explanation:

You can know the order of a reaction given the units of the rate constant:

Then we know that this is a first order reaction because years is a unit of time as well as seconds. The half life of a first order reaction is given by:

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

Here you can solve for the half life:

[tex]t_{1/2}=\frac{ln(2)}{0.0012 years^-1}=577.6 years[/tex]

Now for the rate law for a first order reaction is:

[tex][A]=[A_{o} ]*e^{-kt}[/tex]

Then, if you want to know how long does it take to reach a certain value you solve for time:

[tex]t=-ln([A]/[A]_{o}) *1/k[/tex]

The 15% of its original value is [tex]0.15*[A_{o}][/tex]

You solve for time:

[tex]t=-ln(0.15) *1/0.0012 years^-1=1580.9 years[/tex]

Hope it helps!

In the Decomposition reaction, a reactant gets split into two or more products. The half-life of the reaction is 577.6 years and the time taken to reach 15% of the original is 1580.9 years.

Half-life is the time taken by the radioactive isotope to decay one-half of its total amount.

From the units of the rate of the reaction, the order can be estimated as, for a first-order reaction unit is 1/s.

The half-life of the reaction can be given as,

[tex]\rm t\dfrac{1}{2} = \dfrac{\rm ln (2)}{k}[/tex]

The rate constant (k) for the decomposition reaction is 0.0012 per year.

Substituting values in the above equation:

[tex]\begin{aligned} \rm t\dfrac{1}{2} &= \dfrac{\rm ln (2)}{0.0012}\\\\&= 577.6\;\rm years\end{aligned}[/tex]

The rate law of the first-order reaction is given as,

[tex]\begin{aligned}\rm [A] = [A_{o}] \times e^{-kt}\\\\\rm t = -ln ([A][A_{o}]) \times \dfrac{1}{k}\end{aligned}[/tex]

Solving for time (t):

[tex]\begin{aligned}\rm t &= \rm -ln (0.15) \times \dfrac{1}{0.0012}\\\\& = 1580.9\;\rm years\end{aligned}[/tex]

Therefore, the half-life is 577.6 years and the time taken is 1580.9 years.

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Answer:

Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.

Explanation:

We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.

Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.

In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.

If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.

Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.

Answer:

14.20

Explanation:

The pH can be calculated as:

KOH is a strong base and it will dissociate completely in the solution as:

[tex]KOH\rightarrow K^++OH^-[/tex]

According to the equation and also mentioned that KOH will dissociate completely. So,

Molarity of the hydroxide ions furnished = Molarity of the KOH = 1.6 M

So,

pOH = -log[OH⁻] = -log(1.6) = -0.20

Also,  

pH + pOH = 14  

So, pH = 14 - ( - 0.20 ) = 14.20

Answer: The force that must be applied is 15 N.

Explanation:

Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.

Mathematically,

[tex]F=m\times a[/tex]

where,

F = force exerted = ?

m = mass of the object = 3 kg

a = acceleration of the object = [tex]5m/s^2[/tex]

Putting values in above equation, we get:

[tex]F=3kg\times 5m/s^2=15N[/tex]

Hence, the force that must be applied is 15 N.

Answer: The number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]

Explanation:

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

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

Given mass of bromine = 1.37 g

Molar mass of bromine = 79.904 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of bromine}=\frac{1.37g}{79.904g/mol}=0.0171mol[/tex]

According to mole concept:

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

So, 0.0171 moles of bromine will contain = [tex]0.0171\times 6.022\times 10^{23}=1.03\times 10^{22}[/tex] number of bromine atoms.

Hence, the number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]

Final answer:

To find the number of bromine atoms in 1.37 g of bromine, calculate the atomic mass using isotope information, convert grams to moles, and then use Avogadro's number, yielding approximately 1.03 × 10²² bromine atoms.

Explanation:

To calculate the number of bromine atoms in 1.37 g of bromine, we first need to determine the molar mass of bromine. The atomic mass of bromine is calculated based on the isotopic composition, considering Bromine has two isotopes, 79Br and 81Br with masses of 78.9183 amu and 80.9163 amu respectively, and relative abundances of 50.69% and 49.31%. Using these values, we calculate the average atomic mass of bromine:

Atomic mass = (0.5069 × 78.9183 amu) + (0.4931 × 80.9163 amu) = 79.904 amu.

We then convert grams of bromine to moles by dividing by the molar mass:

Number of moles = 1.37 g ÷ 79.904 g/mol = 0.01715 moles.

Using Avogadro's number, 6.022 × 10²³ atoms/mol, we multiply the number of moles of bromine by this constant to find the number of atoms:

Number of atoms = 0.01715 moles × 6.022 × 10²³ atoms/mol = 1.03 × 10²² atoms.

Thus, there are 1.03 × 10²² atoms of bromine in 1.37 g of bromine. Remember, when giving your answer in scientific notation, it is important to express it in the form of a number between 1 and 10 multiplied by a power of 10.

Answer: Yes, [tex]HCl[/tex] is a strong acid.

acid = [tex]HCl[/tex] , conjugate base = [tex]Cl^-[/tex] , base = [tex]H_2O[/tex], conjugate acid = [tex]H_3O^+[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes [tex]HCl[/tex] is a strong acid as it completely dissociates in water to give [tex]H^+[/tex] ions.

[tex]HCl\rightarrow H^++Cl^-[/tex]

For the given chemical equation:

[tex]HCl+H_2O\rightarrow H_3O^-+Cl^-[/tex]

Here, [tex]HCl[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]Cl^-[/tex] which is a conjugate base.

And, [tex]H_2O[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_3O^+[/tex] which is a conjugate acid.

Thus acid =  [tex]HCl[/tex]

conjugate base = [tex]Cl^-[/tex]

base = [tex]H_2O[/tex]

conjugate acid = [tex]H_3O^+[/tex].

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3OH[/tex] = 26.5 g

Molar mass of [tex]CH_3OH[/tex] = 32.04 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{26.5\ g}{32.04\ g/mol}[/tex]

[tex]Moles\ of\ CH_3OH= 0.8271\ moles[/tex]

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.8271\ moles}{0.735\ kg}[/tex]

Molality = 1.13 m

Answer: The final volume of the oxygen gas is 4.04 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

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

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K[/tex]

Putting values in above equation, we get:

[tex]\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L[/tex]

Hence, the final volume of the oxygen gas is 4.04 L

Answer:

I don't know if option e. is 22,5 g. because that is the correct answer for this. 15% NaCl (mm) gives you the information that in 100 g of solution, you have 15 g of solute. So, if in 100 g of solution you have 15 g of NaCl, in 150 g of solution, how much. Try the rule of three.

Explanation:

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

Answer:

See explanation

Explanation:

Fusion is the process of fusing two isotopes of Hydrogen namely Tritium  and Deuterium to produce Helium. To achieve this, tremendous heat required (about a million degrees Celsius). The same for the pressure. To achieve this, the hydrogen is made into a plasma through rarefaction so it becomes susceptible to magnetic fields. Magnetic confinement fusion is an approach to generating thermonuclear fusion power that uses magnetic fields to confine the hot fusion fuel in the form of a plasma. Electro-Magnets surround the chamber/reactor and are pulsed adiabatically (as in a bicycle pump) and the gas becomes extremely hot that may melt the surroundings.

As the ions in the plasma are charged (the plasma is so hot all the negatively-charged electrons are stripped off the atoms, leaving them with a positive charge) they respond to magnetic fields. Extra fields help shape the plasma and hold it stable.

Answer:

0.09 g of precipitate can be formed.

Explanation:

The chemical equation can be written as follows:

AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺

From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.

The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:

n° of moles of silver nitrate = concentration * volume

n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃

n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl

Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.

1.5 x 10⁻⁴ mol KCl is in excess.

The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.

Answer : The correct option is, (b) 20 kPa

Explanation :

The Raoult's law for liquid phase is:

[tex]p_A=x_A\times p^o_A[/tex]     .............(1)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p^o_A[/tex] = vapor pressure of pure substance A

[tex]x_A[/tex] = mole fraction of A

The Raoult's law for vapor phase is:

[tex]p_A=y_A\times p_T[/tex]      .............(2)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p_T[/tex] = total pressure of the mixture

[tex]y_A[/tex] = mole fraction of A

Now comparing equation 1 and 2, we get:

[tex]x_A\times p^o_A=y_A\times p_T[/tex]

[tex]p_T=\frac{x_A\times p^o_A}{y_A}[/tex]    ............(3)

First we have to calculate the total pressure of the mixture.

Given:

[tex]x_A=0.4[/tex] and [tex]x_B=1-x_A=1-0.4=0.6[/tex]

[tex]y_A=0.7[/tex] and [tex]y_B=1-y_A=1-0.7=0.3[/tex]

[tex]p^o_A=70kPa[/tex]

Now put all the given values in equation 3, we get:

[tex]p_T=\frac{0.4\times 70kPa}{0.7}=40kPa[/tex]

Now we have to calculate the vapor pressure of B.

Formula used :

[tex]x_B\times p^o_B=y_B\times p_T[/tex]

[tex]p^o_B=\frac{y_B\times p_T}{x_B}[/tex]

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

[tex]p^o_B=\frac{0.3\times 40kPa}{0.6}=20kPa[/tex]

Therefore, the vapor pressure of B is 20 kPa.

Answer:

Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17

Explanation:

Energy released during the complete combustion of 1 mole glucose = 2870 kJ

Energy required/muscle contraction cycle = 67 kJ/contraction

Energy conversion efficiency = 39%

Actual amount of energy converted to contraction per mole of glucose is:

[tex]=\frac{39}{100} *2870 kJ=1119.3 kJ[/tex]

Total contraction cycles fueled by the above energy is:

[tex]=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions[/tex]

Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

24.31 grs------------ 1 mol Mg

0.510 grs------------ x=0.02 mol Mg.

If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.

Then to calculate the pH we use the formula pH = - log [H+]

⇒ pH = -log [0.04]⇒ pH=1.4.

Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.

1 mol H2----------- 24.1 L

0.02 mol H2----- x= 0.482L.

Answer: The empirical and molecular formula for the given organic compound is [tex]C_2S[/tex] and  [tex]C_{16}S_8[/tex] respectively.

Explanation:

We are given:

Percentage of C = 42.83 %

Percentage of S = 57.17 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.83 g

Mass of S = 57.17 g

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{42.83g}{12g/mole}=3.57moles[/tex]

Moles of Sulfur = [tex]\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{57.17g}{32g/mole}=1.79moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.79 moles.

For Carbon = [tex]\frac{3.57}{1.79}=1.99\approx 2[/tex]

For Sulfur = [tex]\frac{1.79}{1.79}=1[/tex]

The ratio of C : S = 2 : 1

Hence, the empirical formula for the given compound is [tex]C_2S_1=C_2S[/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 = 448.70 g/mol

Mass of empirical formula = 56 g/mol

Putting values in above equation, we get:

[tex]n=\frac{448.70g/mol}{56g/mol}=8[/tex]

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

[tex]C_{(8\times 2)}S_{(8\times 1)}=C_{16}S_8[/tex]

Thus, the empirical and molecular formula for the given compound is [tex]C_2S[/tex] and  [tex]C_{16}S_8[/tex]

Final answer:

The empirical formula of 'sulflower' is CS2, found by using the given percentages to calculate the simplest whole number ratio of moles of carbon to sulfur. Dividing the compound's molar mass by the empirical formula's molar mass indicates that the molecular formula of 'sulflower' is C6S12.

Explanation:

To determine the empirical formula of the molecule known as 'sulflower', we can start by using the given percentages to find the moles of each element in the compound. Assuming we have 100 grams of the substance, we would have 57.17 grams of sulfur (S) and 42.83 grams of carbon (C). Using their molar masses (atomic sulfur is 32 g/mol and carbon is 12 g/mol), we can calculate the moles of each:

To find the empirical formula, we then find the simplest whole number ratio by dividing by the smallest number of moles, which gives us:

So, the empirical formula is CS2. To determine the molecular formula, we use the molar mass of the empirical formula and divide the given molar mass of the compound by this value. The empirical formula mass of CS2 is (12.01 g/mol × 2) + (32.07 g/mol × 1) = 76.09 g/mol. The molecular formula is then found by dividing the compound's molar mass by the empirical formula's molar mass:

448.70 g/mol / 76.09 g/mol = 5.9 ≈ 6

Therefore, the molecular formula is C6S12, as the empirical formula must be multiplied by 6 to get the molecular formula.

Answer: a) 0.08157 moles of [tex]Al^{3+}[/tex]

b) 0.24471 moles of  [tex]ClO_4^{-}[/tex]

c) 0.97884 moles of oxygen atoms

Explanation:-

The dissociation of the given compound is shown by the balanced equation:

[tex]Al(ClO_4)_3\rightarrow Al^{3+}+3ClO_4^{-}[/tex]

According to stoichiometry:

a) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 1 mole of [tex]Al^{3+}[/tex]

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{1}{1}\times 0.08157=0.08157moles[/tex] of [tex]Al^{3+}[/tex]

b) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 3 moles of [tex]ClO_4^{-}[/tex]

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{3}{1}\times 0.08157=0.24471moles[/tex] of  [tex]ClO_4^{-}[/tex]

c) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 12 moles of oxygen atoms

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{12}{1}\times 0.08157=0.97884moles[/tex] of oxygen atoms

Answer:

a. HNO₃

Explanation:

Nitric acid -

The chemical formula for nitric acid is - HNO₃ , with molecular mass of 63 g /mol , having one hydrogen atom, one nitrogen atom and 3 oxygen atoms .

Nirtic acid is a colorless solution , but the left over solution turns yellow as it get decomposed to nitrogen oxide and water .

The commercially used nitric acid is 68 % in water .

and more than 86% nitric acid is fuming nitric acid .

The reagent nitric acid is used a source of nitration .

Answer:

Ka = [tex]4.04 \times 10^{-11}[/tex]

Explanation:

Initial concentration of weak acid = [tex]4.5 \times 10^{-4}\ M[/tex]

pH = 6.87

[tex]pH = -log[H^+][/tex]

[tex][H^+]=10^{-pH}[/tex]

[tex][H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M[/tex]

HA dissociated as:

[tex]HA \leftrightharpoons H^+ + A^{-}[/tex]

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = [tex]1.35 \times 10^{-7}\ M[/tex]

[tex]Ka = \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}[/tex]

0.000000135 <<< 0.00045

[tex]Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}[/tex]