Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius

Answer:

Average rate of reaction is 0.000565 M/min

Explanation:

Applying law of mass action for the given reaction:

Average rate = [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}[/tex]

Where, [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex] represents average rate of disappearance of [tex]N_{2}O_{5}[/tex], [tex]\frac{1}{4}\frac{[NO_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]NO_{2}[/tex] and [tex]\frac{[O_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]O_{2}[/tex]

Here,[tex]-\frac{[N_{2}O_{5}]}{\Delta t}[/tex] = [tex]-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min[/tex]

So average rate of reaction = [tex][tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex][/tex] = [tex]\frac{1}{2}\times (0.00113M/min)=0.000565M/min[/tex]

Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = 12.0 M

[tex]V_1[/tex] = volume of stock [tex]HCl[/tex]solution = 5.0 ml

[tex]M_2[/tex] = molarity of dilute [tex]HCl[/tex] solution = 0.2 M

[tex]V_2[/tex] = volume of  dilute [tex]HCl[/tex]  solution = ?

Putting in the values we get:

[tex](12.0M)\times 5.0=0.2\times V_2[/tex]

[tex]V_2=300ml[/tex]

Therefore, 300 ml of 0.2M HCl can be made from 5.0mL of a 12.0M [tex]Hcl[/tex] solution.

Answer:

1) 0.022 lbm

2) 276.253 RPM

3) 0.287 lbf

Explanation:

Given data:

mass = 10 kg

acceleration - [tex]13 g = 13\times 9.81 m/s^2 = 12[/tex]

7.53 m/s^2

r =6 inches = 0.1524 m

1) mass in lbm  [tex]= 0.01\times 2.2 = 0.022 lbm[/tex]

as 1 kg = 2.2 lbm

2) acceleration [tex] =  r \omega ^2[/tex]

[tex]127.53 = 0.1524 \times \omega^2[/tex]

[tex]\omega^2 = 836.811[/tex]

[tex] \omega = 28.927 rad/s[/tex]

[tex]1 rad/s  = 9.55 RPM[/tex]    

[tex][ 1 revolution = 2\pi,    1 rad/s = 1/2\pi RPS = \frac{60}{2\pi} RPM][/tex]

SO IN [tex]28.927 rad/s = \frac{60}{2\pi} \times 28.297 = 276.253 RPM[/tex]

3) Force in [tex]N = mass \times a = 0.01\times 127.53 = 1.2753 N[/tex]

                                                 [tex]=  1.2753\times 0.225 lbf = 0.287 lbf[/tex]

Answer: Scientific notation with 4 significant figures is [tex]4.565\times 10^{-5}[/tex]

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.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s preceding the first integers are never significant.

All zero’s after the decimal point are always significant.

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

As we are given that the value is 0.00004565

This number is written in scientific notation as : [tex]4.565\times 10^{-5}[/tex]

Therefore the scientific notation with 4 significant figures is [tex]4.565\times 10^{-5}[/tex]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =[tex]25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)[/tex]

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= [tex]25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}[/tex]

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x [tex]\frac{40.67 KJ}{mol H20}[/tex] x [tex]\frac{1 mol H20}{18 g}[/tex]

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

By calculating the moles of Na2CO3, we can infer the moles of HNO3 due to their 1:1 ratio in the reaction. Dividing the moles of HNO3 with the volume of the solution in liters gives the molarity, which is 0.39 M. The absolute uncertainty is calculated using a specific formula and found to be 0.001 M.

The molarity, M, of a solution is the amount of solute divided by the volume of solution (in Liters). We'll start by calculating the moles of Na2CO3, which is obtained by dividing the given mass of the Na2CO3 by its molar mass.

Calculate moles of Na2CO3 = mass (g) / molar mass (g/mol) = 0.9616 g / 105.988 g/mol = 0.00907 mol.

Since the reaction between HNO3 and Na2CO3 is a 1:1 ratio, the moles of HNO3 is equal to the moles of Na2CO3. Now we know amount of solute (HNO3) is 0.00907 mol. To express the volume of the HNO3 solution in liters, we divide by 23.45 mL by 1000 mL/L, giving 0.02345 L.

Molarity of HNO3 = moles of solute / volume of solution in L = 0.00907 mol / 0.02345 L = 0.387 mol/L or M.

The absolute uncertainty can be found using the following formula: (Delta M/M) = sqrt((Delta m / m)^2 + (Delta V / V) ^2) = sqrt((0.0009 / 0.9616)^2 + (0.05 / 23.45)^2) = 0.002. Therefore, absolute uncertainty = M * (Delta M/M) = 0.387 M * 0.002 = 0.000774 M.

The final answer with significant figures: The molarity of HNO3 = 0.39 ± 0.001 M.

https://brainly.com/question/34662977

#SPJ12

Answer:

The least amount of energy emitted in this case is 0.6 eV.

The corresponding quantum number n would be n=4.

The wavelenght asociated to the emitted photon would be 2.06 [tex]\mu[/tex]m, corresponding to the Infrared spectrum.

Explanation:

For calculating the energy of an electron emitted/absorbed in an electronic transition of the hydrogen atom, the next equation from the Bohr model can be used:

[tex]E=E_{0} Z^2 [\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]

, where E is the photon energy, [tex]E_0[/tex] is the energy of the first energy level (-13.6 eV), Z is the atomic number, [tex]n_1[/tex] is the quantum number n of the starting level and [tex]n_2[/tex] the quantum number n of the finishing level. In this case, [tex]n_2=3[/tex], and [tex]n_1=4[/tex], because this excited level is the next in energy to n=3.

Considering that [tex]1 eV= 1.60217662x10^{-19} J[/tex], and using the Planck equation [tex]E=h\nu=\frac{hc}{\lambda}[/tex], you can calculate the wavelenght or the frequency associated to that photon. Values in the order of [tex]\mu[/tex]m in wavelenght belong to the Infrared spectrum, wich can not being seen by humans.

Answer:

1. [tex]E=1.059x10^{-19}J[/tex]

2. [tex]n=4[/tex]

3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.

Explanation:

Hello,

1. At first, according to the equation:

[tex]E=E_oZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]

Whereas [tex]E[/tex] is the emitted energy, [tex]E_o[/tex] the first level energy, [tex]Z[/tex] the atomic number, [tex]n_1[/tex] the first level and [tex]n_2[/tex] the second level. In such a way, the closest the [tex]n[/tex]'s are, the least the amount of emitted energy, therefore, [tex]n_1=3[/tex] (based on the statement) and [tex]n_2=4[/tex], thus, the least amount of energy turns out being:

[tex]E=(2.179x10^{-18}J)(1^2)(\frac{1}{3^2}-\frac{1}{4^2})\\E=1.059x10^{-19}J[/tex]

2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily [tex]n=4[/tex]

3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:

[tex]E=\frac{hc}{\lambda} \\\lambda=\frac{hc}{E}=\frac{(6.62607004x10^{-34} m^2 kg / s)(299 792 458m/s)}{1.059x10^{-19}m^2 kg / s^2}  \\\lambda=1.88x10^{-6}m[/tex]

Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.

Best regards.

Answer:

P=atm

[tex]b=\frac{L}{mol}[/tex]

Explanation:

The problem give you the Van Der Waals equation:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

First we are going to solve for P:

[tex](P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}[/tex]

[tex]P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}[/tex]

Then you should know all the units of each term of the equation, that is:

[tex]P=atm[/tex]

[tex]n=mol[/tex]

[tex]R=\frac{L.atm}{mol.K}[/tex]

[tex]a=atm\frac{L^{2}}{mol^{2}}[/tex]

[tex]b=\frac{L}{mol}[/tex]

[tex]T=K[/tex]

[tex]V=L[/tex]

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

[tex]P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}[/tex]

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

[tex]P=\frac{L.atm}{L-L}-atm[/tex]

Then operate the fraction subtraction:

P=[tex]P=\frac{L.atm-L.atm}{L}[/tex]

[tex]P=\frac{L.atm}{L}[/tex]

And finally you can find the answer:

P=atm

Now solving for b:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

[tex](V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}[/tex]

Replacing units:

[tex]b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}[/tex]

Multiplying and dividing units,(please see the second photo below), we have:

[tex]b=\frac{L-\frac{L.atm}{atm}}{mol}[/tex]

[tex]b=\frac{L-L}{mol}[/tex]

[tex]b=\frac{L}{mol}[/tex]

Answer:

option A= electrons

Explanation:

The molecules is formed when two or more atoms combine together through chemical bond. The atoms can be from same elements or from different elements. When atoms shares the electron a covalent bond is formed.

For example:

In hydrogen molecule H2 the two atom of hydrogen share their electrons and form a bond.

H· + H· → H-H  OR  ( H:H)

The force of attraction is electrostatic, between these two atoms.

The molecule can be formed by the sharing of electrons between the atoms different elements.

For example:

In the molecule of ammonia (NH3) there are two different kind of atoms i.e, nitrogen and hydrogen. Ammonia consist of total four atoms. The one atom is nitrogen and others three atoms are of hydrogen.The nitrogen atom consist of five valance electrons, three electrons are used to form three covalent bond with hydrogen. While the one electron pairs is still present on nitrogen atom.The three hydrogen atoms are bonded with one nitrogen atom through single covalent bond for each atom.

We determined the moles of methanol in a 200.0 kg mixture by using the given weight percentage and the molar mass of methanol. We also found the total flow rate of the mixture using the given flow rate of propyl acetate and the weight percentage of propyl acetate in the mixture.

(a) To determine the g-moles of methanol in 200.0 kg of the mixture, you first need to find the mass of methanol in the mixture. As we know, the weight % of methanol in the mixture is 25.0. Hence, the weight of methanol in the mixture is 25.0 wt% of 200.0 kg, which equals 50,000 g. The molar mass of methanol (CH3OH) is approximately 32.04 g/mole. Thus, the moles of methanol in the mixture would be determined by dividing the mass of methanol by the molar mass of methanol, which comes out to be approximately 1562.4 moles (50,000 g / 32.04 g/mole).

(b) The flow rate of the total mixture can be obtained by using the wt% of propyl acetate in the mixture. We know that the mixture is 25.0 wt% methanol, so it is 75.0 wt% propyl acetate. Given the flow rate of propyl acetate is 100 lb-mole/h, the flow rate of the total mixture would be 100 lb-mole/h / 0.75, which equals approximately 133.3 lbm/h.

https://brainly.com/question/32648436

#SPJ12

Answer:

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Explanation:

Calorie intake of an adult in a day =[tex]2.2\times 10^3 calorie[/tex]

Calorie burnt by an adult in a day = [tex]2.0\times 10^3 calorie[/tex]

Excess  nutritional energy in day=

[tex](2.2\times 10^3 calorie)-(2.0\times 10^3 calorie)[/tex]

[tex]=2.0\times 10^2 calorie[/tex]

1 kcal = 4.184 kJ

So , 1000 cal = 4.184 kJ

[tex]1 cal = 4.184\times 10^{-3} kJ[/tex]

[tex]2.0\times 10^2 calorie=2.0\times 10^2\times 4.184\times 10^{-3} kJ[/tex]

[tex]=0.8368 kJ[/tex]

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

[tex]14.6\times 10^3[/tex] kilo joules of excess nutritional energy = 1 lb fat

The 1 kilo joule of excess nutritional energy  = [tex]\frac{1}{14.6\times 10^3}[/tex] lb fat

Excessive nutritional energy of an adult per day = 0.8368 kiloJoules

Amount of fat gained by an adult per day =

= [tex]0.8368 kiloJoules\times \frac{1}{14.6\times 10^3}=5.7315\times 10^{-5} lb[/tex] of fat

In 1 day an adult gains = [tex]5.7315\times 10^{-5} lb[/tex] of fat

Time taken to gain 1 lb fat:

[tex]\frac{1}{5.7315\times 10^{-5}} day=17,447.41 days[/tex]

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Final Answer:

An adult who consumes an excess of 0.2 x 10³ Cal per day, equivalent to 0.8368 x 10³ kJ, will gain 1 lb of fat in approximately 17.5 days, as 1 lb of fat is stored for each 14.6 x 10³ kJ of excess nutritional energy consumed.

Explanation:

An adult consumes approximately 2.2 x 10³ Calories (Cal) per day and burns 2.0 x 10³ Cal per day. To find the excess nutritional energy in kilojoules, we need to subtract the energy burned from the energy consumed and then convert the result from Calories to kilojoules:

Excess energy (Cal) = 2.2 x 10³ Cal - 2.0 x 10³ Cal = 0.2 x 10³ Cal

Excess energy (kJ) = 0.2 x 10³ Cal x 4.184 kJ/Cal = 0.8368 x 10³ kJ

To calculate how long it will take for the adult to gain 1 lb of fat, we divide the amount of kilojoules that corresponds to 1 lb of fat by the daily excess kilojoules:

Days to gain 1 lb = (14.6 x 10³ kJ per 1 lb) / (0.8368 x 10³ kJ/day)

Days to gain 1 lb = 17.4512 days

Therefore, it will take approximately 17.5 days for the adult to gain 1 lb of fat, given the daily excess of 0.2 x 10³ Cal.

Answer:

We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Explanation:

As we know that buffer solution is the combination of weak acid with strong base or strong acid with weak base.

We know that CH3COOH is weak acid .Acetate buffer is the combination of weak acid CH3COOH and conjucate base CH3COO- (from salt).

So we have to add acid to achieve the proper final pH value of mixture.We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Answer:

To convert tons to kg you need to multiply it by 1000. So you have 45,000,000,000 kg of sulfuric acid produced each year in the world, you also know that the world population was 7,530,000,000 in 2017. So if you divide the amount of sulfuric acid by the world population you obtain the production rate, that is 5.976 kg/person. For the b), you know that 91.44 cm are 1 yd, and that 100 g are 1 kg, so you convert 1.82g/cm to 0.16642 kg/yd, now you just multiply it by 45,000,000,000 kg/year, so the annual rate in yd is 7,488,900,000 yd/year.

Answer : The unit of (D) in metric system is [tex]m^2/s[/tex]

Explanation :

The given expression is:

[tex](dm/dt)=(-2\pi)\times (d)\times (D)\times (\frac{M}{RT})\times P[/tex]

where,

m = mass

t = time

d = diameter

D = diffusion coefficient

M = molar mass

R = universal gas constant

T = temperature

P = partial pressure

In metric system,

The unit of mass is, kg

The unit of time is, s

The unit of diameter is, m

The unit of molar mass is, kg/mol

The unit of universal gas constant is, [tex]Nm/^oC.mol[/tex]

The unit of temperature is, [tex]^oC[/tex]

The unit of partial pressure is, [tex]N/m^2[/tex]

The unit of diffusion coefficient will be:

[tex]D=\frac{(dm/dt)}{(-2\pi)\times (d)\times (\frac{M}{RT})\times P}[/tex]

or,

[tex]D=\frac{(dm)\times (R)\times (T)}{2\pi \times (d)\times (dt)\times (M)\times (P)}[/tex]

[tex]D=\frac{(dm)\times (R)\times (T)}{(d)\times (dt)\times (M)\times (P)}[/tex]

Now put all the unit in this expression, we get:

[tex]D=\frac{(kg)\times (Nm/^oC.mol)\times (^oC)}{(m)\times (s)\times (kg/mol)\times (N/m^2)}[/tex]

[tex]D=m^2/s[/tex]

Therefore, the unit of (D) in metric system is [tex]m^2/s[/tex]

Answer: Option (b) is the correct answer.

Explanation:

According to Fourier's equation,

                           Q = [tex]kA(\frac{\Delta T}{\Delta x})[/tex]

Also,  [tex]\frac{Q}{A}[/tex] = q

So,  q = [tex]k(\frac{\Delta T}{\Delta x})[/tex]

where,     Q = quantity of heat transferred

                A = area of heat transferred

                q = rate of heat transfer

               [tex]\Delta T[/tex] = change in temperature

               [tex]\Delta x[/tex] = thickness

                  k = thermal conductivity of the solid material

Since rate of heat transfer is directly proportional to k, which is also known as thermal conductivity.

Therefore, it means that higher is the value of k higher will be rate of heat transfer (q). And, lower is the value of k lower will be the rate of heat transfer (q).

Thus, we can conclude that heat transfer through solid composites depend on higher composite heat conductivity.

Answer:

Sodium laurate, also known as sodium dodecanoate, is a soap. It is the salt of lauric acid. It is an amphiphilic organic molecule which is composed of a hydrophilic head (polar ) and a hydrophobic tail (non-polar fatty acid).

In a aqueous solution, it leads to the formation of a micelle. The hydrophilic head of the molecule interacts with the surrounding polar solvent molecules. Thereby, making the micelle soluble in the solution. Whereas, the hydrophobic tails present in the core of micelle, interacts with the non-polar oil particles.

Answer : The reaction must shift to the product or right to be in equilibrium.  The equilibrium concentration of [tex]H_2[/tex] is 7.0 M

Explanation :

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

First we have to determine the concentration of [tex]CH_4,H_2S,CS_2\text{ and }H_2[/tex].

[tex]\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=5M[/tex]

[tex]\text{Concentration of }H_2S=\frac{\text{Moles of }H_2S}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=2.5M[/tex]

[tex]\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=2.5M[/tex]

[tex]\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=5M[/tex]

Now we have to determine the value of reaction quotient (Qc).

The given balanced chemical reaction is,

[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]

The expression for reaction quotient will be :

[tex]Q_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

[tex]Q_c=\frac{(2.5)\times (5)^4}{(2.5)times (5)^2}=25[/tex]

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.

When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.

When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.

The given equilibrium constant value is, [tex]K_c=225[/tex]

From the above we conclude that, the [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored that means reaction must shift to the product or right to be in equilibrium.

Now we have to calculate the concentration of [tex]H_2[/tex] at equilibrium.

The given balanced chemical reaction is,

                     [tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]

Initial conc.   2.5            5               2.5          5

At eqm.       (2.5-x)       (5-2x)       (2.5+x)    (5+4x)

The concentration of [tex]CH_4[/tex] at equilibrium = 2.0 M

As we know that, at equilibrium

(2.5-x) = 2.0 M

x = 0.5 M

The concentration of [tex]H_2[/tex] at equilibrium = (5+4x) = 5 + 4(0.5) = 7.0 M

Therefore, the equilibrium concentration of [tex]H_2[/tex] is 7.0 M

Final answer:

The reaction CH4 (g) + 2 H2S (g) ⇋ CS2 (g) + 4 H2 (g) will proceed to the left to reach equilibrium since the reaction quotient Qc is greater than the given Kc. The equilibrium concentration of H2 when the concentration of methane, CH4, at equilibrium is 2.0 M is 7.0 M.

Explanation:

To determine in which direction the reaction CH4 (g) + 2 H2S (g) ⇄ CS2 (g) + 4 H2 (g) will proceed to reach equilibrium, we need to calculate the reaction quotient Qc and compare it with the equilibrium constant Kc. The provided Kc is 225. The initial concentrations are the moles of each substance divided by the volume of the container. Given initial amounts: CH4 = 1.0 mol, H2S = 2.0 mol, CS2 = 1.0 mol, and H2 = 2.0 mol in a 0.4 L container, at 960°C.

To find the concentrations, we divide the moles by the volume (in liters): [CH4] = 1.0 mol/0.4 L = 2.5 M, [H2S] = 2.0 mol/0.4 L = 5.0 M, [CS2] = 1.0 mol/0.4 L = 2.5 M, and [H2] = 2.0 mol/0.4 L = 5.0 M. The reaction quotient Qc is calculated using the expression Qc = [CS2][H2]⁴ / ([CH4][H2S]²). Plugging in the initial concentrations, we get Qc = (2.5 × 5.0⁴) / (2.5 × 5.0²).
As calculated, Qc turns out to be higher than the given Kc, indicating the reaction will shift to the left to reach equilibrium.

When it comes to the equilibrium concentration of H2, the student provided that the concentration of methane (CH4) at equilibrium is 2.0 M. The ratio of H2 to CH4 in the balanced chemical equation is 4:1, so for every 1 M of CH4 that reacts, 4 M of H2 is produced. Thus, if CH4 decreases to 2.0 M from 2.5 M (a decrease of 0.5 M), then H2 would have to increase by 4 times that amount (2.0 M). The equilibrium concentration of H2 would be the initial concentration plus this change: H2(eq) = 5.0 M + 2.0 M = 7.0 M.

Answer: Option (A) is the correct answer.

Explanation:

The given data is as follows.

  pH = 7.40,         [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex]

[tex]pK_{a}[/tex] = 6.10

We have to find [tex]\frac{[HCO_^{-}{3}]}{[CO_{2}]}[/tex] = ?

According to Henderson-Hasselbalch equation,

                 pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]

Hence, putting the given values into the above equation as follows.

                 pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]

or,              pH = [tex]pK_{a} + log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]

                 7.40 = 6.10 + [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]

             [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = 1.30

          [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = antilog (1.30)    

                                         = 20

Since, it is given that [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex].

Therefore, [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] or [tex]\frac{[HCO^{-}_{3}]}{[CO_{2}]}[/tex] = [tex]\frac{20}{1}[/tex]

Thus, we can conclude that the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40 is 20:1.

Ionic bonding occurs when a metal loses electrons to a nonmetal, forming ions. Contrary to the given propositions, these compounds typically have high, not low, melting points due to the high electronegativity differences between the bonding atoms, leading to ionic, not covalent, bonds. Not all the provided statements about ionic bonding are false.

In evaluating the statements about ionic bonding, we can identify the following truths: Ionic bonding occurs when there's a transfer of electrons from a metal to a nonmetal, producing ions—a metal tends to lose electrons, becoming a cation (positive ion), while the nonmetal gains these electrons, morphing into an anion (negative ion)

Statement a, therefore, is incorrect because it flips the properties of metals and nonmetals. As for statement b, it's true: CaCl2 does form due to Ca²+ being more stable than a single calcium atom. Exploring statement c, ionic compounds—like CaCl2—notably possess high (not low) melting points due to the strong electrostatic forces (ionic bonds) holding the ions together. Lastly, statement d is false because the electronegativity difference between the bonding atoms of ionic compounds is large, not small. This large difference leads to the transfer of electrons instead of sharing them (like in covalent bonds where electronegativity differences are smaller). Statement e falsely claims all previous propositions are incorrect—the truth is mixed!

https://brainly.com/question/18297125

#SPJ11

Answer:

a) n-butane has a higher boiling point

b) 1-butanol has a higher boiling

Explanation:

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to greater molar mass and longer chain (more interactions between each molecule).

Given the molecules, diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has a higher boiling point mainly due to hydrogen bonding forces. The OH group is more electronegative than ether group.

The larger and more complex n-butane has a higher boiling point due to London dispersion forces, while 1-butanol has a higher boiling point due to hydrogen bonding.

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to its larger size and increased London dispersion forces, a type of intermolecular force.

London dispersion forces increase with increased molecular size and shape because larger and more complex molecules tend to have more electrons, leading to larger temporary fluctuations in charge distribution.

Given the molecules diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has the higher boiling point mainly due to the presence of hydrogen bonding, another type of intermolecular force. Hydrogen bonding is generally stronger than other types of intermolecular forces, leading to higher boiling points.

https://brainly.com/question/9328418

#SPJ6

Answer:

A liquid, at any temperature, is in equilibrium with its own steam. This means that on the surface of the liquid or solid substance, there are gaseous molecules of this substance. These molecules exert a pressure on the liquid phase, a pressure known as vapor pressure.

In chemistry, when we talk about dry basis, we talk about a state in which the presence of water in a gaseous state is denied for the calculation. So vapor pressure equals zero.

When we talk about the wet basis, the presence of water in the steam is considered for the calculation, which normally is expressed as a percentage or moisture.

In summary, for a gas mixture steam:

So, in wet basis we have an extra component (water).

Assuming we only have 2 components in our steam, and being X the molar fraction of eact component:

For dry basis the mole fraction of A it is obtained by subtracting the molar fraction of B from one. And for wet basis, we have to substract the molar fraction of B AND the molar fraction of water vapor. So, logically, the mole fraction Xa will be less for wet basis.

Answer:

1384 kJ/mol

Explanation:

The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:

Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J

Extra significant figures are kept to avoid round-off errors.

We then calculate the moles of the organic compound:

(0.6654 g)(mol/46.07) = 0.0144432 mol

We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)

(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol

Explanation:

It is known that like dissolves like because the solubility of any solute basically depends on its solvents polarity.

Water loving solutes are known as hydrophilic in nature.

For example, glucose or sugar is able to dissolve in polar solvents like water because they are polar themselves.

Whereas molecules that repel water molecules are known as hydrophobic in nature. So, non-polar molecules do not dissolve in water because they form aggregates and hence, non-polar molecules do not dissolve in polar solvents.

But non-polar solvents dissolve in non-polar solvents.

Hence, when vinegar and oil are mixed together then they will not dissolve because they are immiscible. As vinegar is like water so, it is able to dissolve hydrophilic molecules.

And, oil being non-polar in nature will not dissolve in polar solvents.

Whereas glucose and sodium sitrate are both hydrophilic in nature which means that they will dissolve in water but not in any organic solvent.

Thus, we can conclude that when we add glucose and sodium citrate to a mixture of vinegar then both the solutes will dissolve in it.

Answer:

lets set the ratio -A/HA as R:

pH = pKa + log(R,10) => pKa + log10(R)

pH = 5.5

pKa = 4.76

R => 10^(pH - 4.76)

10^(pH - 4.76) => 5.4954

Given R (-A/HA) a number bigger than 1, then the concentration of  -A is bigger than HA

Explanation:

Answer:

There is more A⁻ than HA in the solution  

Explanation:

The equation for the ionization of a weak acid is

HA + H₂O ⇌H₃O⁺ + OH⁻

When HA and A⁻ are present in comparable amounts, we can use the Henderson-Hasselbalch equation:

[tex]\begin{array}{rcl}\text{pH} &=& \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\5.5 & = & 4.76 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\0.74 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\end{array}[/tex]

0.74 is a positive number, and a number must be greater than one for its logarithm to be positive. That is,

[tex]\begin{array}{rcl}\dfrac{[\text{A}^{-}]}{\text{[HA]}} & > & 1\\\\\textbf{[A]}^{-} & > & \textbf{[HA]}\\\end{array}[/tex]

Final answer:

To calculate the approximate enthalpy change for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The approximate enthalpy change is -3818 kJ.

Explanation:

To calculate the approximate enthalpy change (ΔH) for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The reaction is:

QX3 + 3H20 => Q(OH)3 + 3HX

We can calculate the energy absorbed or released by breaking and forming these bonds.

First, we calculate the energy required to break the bonds in the reactants:

Then, we calculate the energy released when the bonds in the products are formed:

Finally, we sum up the energy changes:

Therefore, the approximate enthalpy change for the reaction is -3818 kJ.

The reaction order with respect to reactant A in a rate law equation, such as rate=k[A]^m[B]^n is symbolized by the exponent 'm'. This indicates the dependency of the reaction rate on the concentration of A. It commonly takes integer values with variations understood experimentally.

In the context of a rate law equation, like rate=k[A]^m[B]^n, the reaction order with respect to a reactant A is represented by the exponent m. This means items present in reactant A would contribute in a mathematical manner towards the reaction rate. It's important to note that this value is typically an integer.

For instance, if m=1, it implies that the reaction is first order with respect to A, meaning there is a linear relationship between the concentration of A and the reaction rate. If m=2, the reaction is second order in A, suggesting that the reaction rate is proportional to the square of A's concentration.

To establish the rate constant (k) and the reaction order (m and n), experimental techniques are employed to observe and tabulate the rate of reaction as the concentrations of reactants change. Bear in mind that k doesn't depend on concentration of reactants, but does fluctuate with temperature.

https://brainly.com/question/34445133

#SPJ6

The reaction order with respect to reactant A in the rate law equation rate = k[A]^m[B]^n is represented by the exponent m, which must be determined experimentally.

The reaction order with respect to a reactant in a rate law equation tells us how the rate of the reaction depends on the concentration of that reactant. It is expressed as an exponent in the rate law equation. For reactant A, given the rate law equation rate = k[A]^m[B]^n, the reaction order with respect to A is the exponent m. This value of m must be determined from experimental data; it is not necessarily related to the stoichiometric coefficient of A in the overall balanced equation.

Answer:

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.

Q = 10x4x(40 - 90)

Q = -2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling(r) will be:

r = 1.955x(90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes

Answer:

0.7681

Explanation:

Given that the mass percentage of methanol = 85.5 %

which means that 85.5 g of methanol is present in 100 g of the solution.

Thus, mass of water = 100 g - 85.5 g = 14.5 g

The formula for the calculation of moles is shown below:

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

Mass of methanol = 85.5 g

Molar mass of methanol = 32.04 g/mol

Moles of methanol = 85.5 g / 32.04 g/mol = 2.6685 moles

Mass of water = 14.5 g

Molar mass of methanol = 18 g/mol

Moles of methanol = 14.5 g / 18 g/mol = 0.8056 moles

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}[/tex]

Applying values as:

[tex]Mole\ fraction\ of\ methanol=\frac {2.6685}{2.6685+0.8056}[/tex]

Mole fraction of methanol in the solution = 0.7681

Explanation:

1.) 175 km to μm

[tex]1 km=10^9 \mu m[/tex]

[tex]175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m[/tex]

3.) 385 nm to dm

[tex]1 nm=10^{-8} dm[/tex]

[tex]385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm[/tex]

5.) 492 μm  to m

1 μm =  [tex]10^{-6} m[/tex]

[tex]492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m[/tex]

7.) [tex]52\times 10^3[/tex] dm to mm

1 dm = 100 mm

[tex]52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm[/tex]

9.) [tex]321\times 10^{35}[/tex] mm to km

[tex]1 mm = 10^{-6} km[/tex]

[tex]321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km[/tex]

11.) [tex]456\times 10^3[/tex] m to km

m = 0.001 km

[tex]456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km[/tex]

13.) [tex]422\times 10^3[/tex] m to nm

[tex]1 m = 10^{9} nm[/tex]

[tex]422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm[/tex]

15.) [tex]4.87\times 10^{30}[/tex] m to pm

[tex]1 m = 10^{12} pm[/tex]

[tex]4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm[/tex]

17.) [tex]5.26\times 10^3[/tex] m to um

1 m =  [tex]10^{6} \mu m[/tex]

[tex]5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m[/tex]

19.) [tex]1.25\times 10^{35}[/tex]m to Mm

1 m =  [tex]10^{-6} Mm[/tex]

[tex]1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm[/tex]

21.) [tex]4.22\times 10^3[/tex] Tm to nm

[tex]1 Tm = 10^{21} nm[/tex]

[tex]4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm[/tex]