List and discuss two major sources of errors in numerical methods

Answer:

a)CN=12

b)APF=74 %

c)a=0.35 nm

d)ρ=9090.9 [tex]Kg/m^3[/tex]

Explanation:

Given that

Nickel have FCC structure

We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .

Z=8 x 1/8 + 1/2 x 6 =4

Z=4

Coordination number (CN)

 The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.

CN=12

Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.

APF

[tex]APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}[/tex]

We know that for FCC

[tex]4R=\sqrt{2}\ a[/tex]

Now by putting the values

[tex]APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}[/tex]

APF=0.74

APF=74 %

[tex]4R=\sqrt{2}\ a[/tex]

[tex]4\times 0.124=\sqrt{2}\ a[/tex]

a=0.35 nm

Density

[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]

We know that M for Ni

M=58.69 g/mol

a=0.35 nm

[tex]N_A=6.023\times 10^{23}\ atom/mol[/tex]

[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]

[tex]\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3[/tex]

ρ=9090.9 [tex]Kg/m^3[/tex]

Answer:

Recrystallization refers to the method used for purifying chemical compounds. In this method, the chemical compound along with the impurities is dissolved in an appropriate hot solvent, to separate the chemical compound and the impurity.

The pure chemical compound that separates after the recrystallization process exists in the form of crystals.

Generally, the impure chemical compound is fully dissolved in the minimum amount of the solvent, thus forming a saturated solution. This is because if large amount of solvent is used, then a large amount of the chemical compound will remain dissolved in the solvent, even after cooling.

Therefore, all of the chemical compound will not get recrystallized.

Answer:

A)Boiling point constant of benzene = 2.63°C/m

B) 242.77 g/mol is the molar mass of the solute.

Explanation:

[tex]\Delta T_b=T_b-T[/tex]

[tex]\Delta T_b=K_b\times m[/tex]

[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]

where,

[tex]\Delta T_f[/tex] =Elevation in boiling point

[tex]K_b[/tex] = boiling point constant od solvent= 3.63 °C/m

1 - van't Hoff factor

m = molality

A) Mas of solvent = 500 g = 0.500 kg

T = 80.1°C ,[tex]T_b[/tex] =82.73°C

[tex]\Delta T_b=T_b-T[/tex]

[tex]\Delta T_b[/tex]= 82.73°C - 80.1°C = 2.63°C

[tex]2.63^oC=K_b\times \frac{36 g}{72 g/mol\times 0.500 kg}[/tex]

[tex]K_b=\frac{2.63^oC\times 72 g/mol\times 0.500 kg}{36 g}=2.63 ^oC/m[/tex]

Boiling point constant of benzene = 2.63°C/m

B) Mass of solute = 1.2 g

Molar mas of solute = M

Mass of solvent = 50 g= 0.050 kg

i = 1

T = 80.1°C ,[tex]T_b[/tex] =80.36°C

[tex]\Delta T_b=T_b-T[/tex]=80.36°C -  80.1°C = 0.26°C

[tex]0.26^oC=1\times K_b\times \frac{1.2 g}{M\times 0.050 kg}[/tex]

[tex]M=1\times 2.63^oC/m\times \frac{1.2 g}{0.26^oC\times 0.050 kg}[/tex]

M = 242.77 g/mol

242.77 g/mol is the molar mass of the solute.

Final answer:

The boiling point constant (Kb) of benzene was found to be 2.63°C/m using the data from part A. Using this constant and the boiling point elevation data in part B, the molar mass of the unknown solute was calculated to be 240 g/mol.

Explanation:

The subject of this question is Chemistry, focusing specifically on the boiling point elevation of solutions and the calculation of molar mass from boiling point data. The problem pertains to colligative properties of solutions, which are properties that depend on the number of particles in a solution but not their identity.

Part A: Calculating Boiling Point Elevation

To find the boiling point constant (Kb) of benzene, we use the formula for boiling point elevation, ΔTb = Kb × m, where ΔTb is the change in boiling point and m is the molality of the solution. Given that the boiling point of benzene increases from 80.1°C to 82.73°C, the elevation in boiling point is 2.63°C. Molality (m) is calculated by the number of moles of pentane divided by the mass of benzene in kilograms. The molar mass of pentane (C5H12) is 72.15 g/mol, and so 36 g corresponds to 0.5 moles. The mass of benzene is 500 g, which is 0.5 kg. Thus, m = 0.5 moles / 0.5 kg = 1 mol/kg. Using these values, we can calculate Kb: 2.63°C = Kb × 1 mol/kg, so Kb = 2.63°C/m.

Part B: Determining Molar Mass of an Unknown Solute

For part B, to find the molar mass of the solute, we first calculate the change in boiling point of benzene, which is 80.36°C - 80.1°C = 0.26°C. With the boiling point constant of benzene from Part A (Kb = 2.63°C/m), and assuming i (the van 't Hoff factor) is 1 for a non-electrolyte, we establish the molality of the solution as m = ΔTb / Kb = 0.26°C / 2.63°C/m = 0.1 mol/kg. Knowing the molality and the mass of solvent (benzene), we can now calculate the molar mass (M) of the unknown solute using the formula M = mass of solute / (molality × mass of solvent in kg). With the mass of solute as 1.2 g and the mass of solvent as 0.05 kg, M = 1.2 g / (0.1 mol/kg × 0.05 kg)= 240 g/mol.

Answer:

The rate is [tex] 4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

Explanation:

Stoichiometry

[tex]CH_{3}Cl+NaOH \rightarrow CH_{3}OH+NaCl [/tex]

Kinetics

[tex]-r_{A}=k \times [CH_{3}Cl] \times [NaOH] [/tex]

The rate constant K can be calculated by replacing with the initial data

[tex] 1 \times 10^{-4}\frac{mole}{Ls}=k \times [0,2M] \times [1,0M]  =5 \times 10^{-4}\frac{L}{mole s}[/tex]

Taking as a base of calculus 1L, when half of the [tex] CH_{3}Cl [/tex] is consumed the mixture is composed by

[tex] 0,1 mole CH_{3}Cl [/tex] (half is consumed)

[tex] 0,9 mole NaOH [/tex] (by stoicheometry)

[tex] 0,1 mole CH_{3}OH [/tex]  

[tex] 0,1 mole NaCl [/tex]

Then, the rate is

[tex]-r_{A}=5 \times 10^{-4} \frac{L}{mole s}\times 0,1M \times 0,9 M=4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

The reaction rate decreases because there’s a smaller concentration of reactives.

Answer:

150 × 10⁻⁹ μL

Explanation:

Data provided in the question:

Molarilty of the stock solution, M₁ = 200 M

Final Volume of the solution, V₂ = 100 μL = 100 × 10⁻⁶ L

Final concentration, M₂ = 300 nM = 300 × 10⁻⁹ M

Now,

M₁V₁ = M₂V₂

where,

V₁ is volume of the stock solution

Thus,

200 × V₁ = 100 × 10⁻⁶ × 300 × 10⁻⁹

or

V₁ = 150 × 10⁻⁹ μL

Answer:

Measuring with a ruler and using final volume minus initial volume

Explanation:

You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be

[tex]volume=length*width*height[/tex]

You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.

[tex]Volume of the object= volume of water and object - volume of water[/tex]

The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.

Answer: The volume of 2.0 M solution of glucose required is 11.3 mL

Explanation:

To calculate the volume of concentrated solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution

We are given:

Conversion factor: 1 L = 1000 mL

[tex]M_1=2.0M\\V_1=?mL\\M_2=0.15M\\V_2=0.15L=150mL[/tex]

Putting values in above equation, we get:

[tex]2.0\times V_1=0.15\times 150\\\\V_1=11.3mL[/tex]

Hence, the volume of 2.0 M solution of glucose required is 11.3 mL

Answer:

[tex]K_{f}[/tex] for solvent is [tex]17^{0}\textrm{C}.kg.mol^{-1}[/tex]

Explanation:

[tex]\Delta T_{f}=K_{f}.m[/tex], where [tex]\Delta T_{f}[/tex] is depression in freezing point and m is molality of solution

Molality of solution (m) = (moles of solute/mass of solvent in kg)

                                      = [tex]\frac{\frac{0.980}{178.2}}{0.01387}mol/kg[/tex]

                                      = 0.396 mol/kg

[tex]\Delta T_{f}=(12.0-5.1)^{0}\textrm{C}=6.9^{0}\textrm{C}[/tex]

So, [tex]K_{f}=\frac{\Delta T_{f}}{m}=\frac{6.9}{0.396}^{0}\textrm{C}.kg.mol^{-1}=17^{0}\textrm{C}.kg.mol^{-1}[/tex]

Answer:

0.400 mL

Explanation:

Hello, the dilution factor (in folds) is given by:

[tex]folds=\frac{total_{volume}}{sample_{volume}}[/tex]

Thus, the sample volume with three significant figures is given by:

[tex]sample_{volume}=\frac{2mL}{5"folds"}=0.400 mL[/tex]

Best regards.

Answer:

v=3.92*10¹⁷ m/s, and l=2.93*10⁻³⁰m

Explanation:

The velocity of the electrons in the x-ray tube is related to the mass of electrons and the energy applied, the equation is E=0.5*m*v², where E is the energy in kV, m is the mass of the electrons (9.11x10-31kg), and v the velocity. Substituting, 70000 = 0.5*(9.11x10-31kg)*v², and rearranging the terms, sqrt((70000)/(0.5)(9.11x10-31)) = v = 3.92*10¹⁷ m/s. Then the energy is related to the constant h (6.62607015×10−34 Js), the speed of light 299792458 m/s, and "l" that is the wavelength. So the equation is l=(h*c)/E, and l = 2.93*10⁻³⁰m.

Answer:

hypochlorite   -1  others -2

Explanation:

Hypochlorite, polyatomic ion has the lowest charge when compared with other given options.

Answer: hypochlorite

Explanation:

Ions that possess more than single atom is always referred as polyatomic or molecular ions. It is the charged chemical compound consisting more than one covalently bonded atoms or metal complex that can be taken as the single unit.

[tex]\text { Hypochlorite is } C l O^{-1}[/tex] is the charge for Hypochlorite and it is the polyatomic ion with the lowest charge.

                  [tex]\text { Hypochlorite is } C l O^{-1}[/tex]

                  [tex]\text { carbonate is } C O_{3}^{-2}[/tex]

                  [tex]\text { sulfite is } S O_{3}^{-2}[/tex]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = [tex]K_a=6.8\times 10^{-4}[/tex]

Concentration of HF (weak acid)= 0.0310 M

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

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

[tex]pK_a=-\log (K_a)[/tex]

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

[tex]pK_a=-\log (6.8\times 10^{-4})[/tex]

[tex]pK_a=4-\log (6.8)[/tex]

[tex]pK_a=3.17[/tex]

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[NaF]}{[HF]}[/tex]

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

[tex]2.60=3.17+\log (\frac{[NaF]}{0.0310})[/tex]

[tex][NaF]=0.00834M[/tex]

Now we have to calculate the moles of NaF.

[tex]\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole[/tex]

Now we have to calculate the mass of NaF.

[tex]\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g[/tex]

Therefore, the mass of sodium fluoride added should be 0.105 grams.

Answer:

Part 1

(a) 0.0113

(b) 11300 ppm

(c) 1.82 *10⁻³

(d) 0.100 M

(e) 0.101 m

Part 2

(a) 9.45 *10⁻³

(b) mole fraction = 2.45 *10⁻³

(c) 11.3 ppm

Explanation:

Chlorobenzene formula is C₆H₅Cl

Part 1: We are given a concentration of chlorobenzene in water of 100 mol/m³, and a density of the solution of 1.00 g/cm³.

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / mass solution

We know there are 100 moles of C₆H₅Cl per m³ of solution.

To get the mass of C₆H₅Cl we'll convert the moles to mass by using the molar mass:

Molar mass C₆H₅Cl = 6*12.011 + 5*1.00794 + 35.4527 = 112.558 g/mol

mass C₆H₅Cl = moles C₆H₅Cl * molar mass C₆H₅Cl

mass C₆H₅Cl = 100 moles * 112.558 g/mol = 11255.8 g

11255.8 g of C₆H₅Cl are in 1 m³ of solution.

Next we'll convert 1 m³ of solution to mass by using the density

mass solution = volume solution * density of solution

[tex]mass solution = 1m^{3} *\frac{(100cm)^{3} }{ 1m^{3}} * \frac{1.00 g}{cm^{3} } = 1.00 *10^{6} g[/tex]

weight fraction C₆H₅Cl = 11256 g / 1.00 *10⁶ g = 0.0113

(b) ppm stands for "parts per million" and it is usually expressed as mg per Liter of solution

We already calculated that there are 11256 g or more exactly 11300 g of C₆H₅Cl in 1 m³ of solution, so lets convert to mg/L:

[tex]\frac{11300 g}{1 m^{3} } * \frac{1000 mg}{1 g} * \frac{1 m^{3} }{1000 L} = 11300 mg/L[/tex]

So the solution is 11300 ppm

(c)  mole fraction = moles of C₆H₅Cl / total moles in solution

total moles = moles C₆H₅Cl + moles water

moles water = mass water / molar mass water

mass water = mass solution - mass C₆H₅Cl

moles of C₆H₅Cl = 100 moles

mass water = 1.00 *10⁶ g of solution - 11256 g = 988744 g of water

moles water = 988744 g / 18.0153 g/mol = 54884 moles water

total moles = 100 + 54884 = 54984 moles

mole fraction = 100 moles of C₆H₅Cl / 54984 moles = 1.82 *10⁻³

(d) Molarity = moles C₆H₅Cl / Liters of solution

We know the solution is 100 mol / m³ so we just have to convert the m³ to L:

[tex]\frac{100 mol}{m^{3} } * \frac{1 m^{3}}{1000 L} = 0.100 mol / L = 0.100 M[/tex]

(e) Molality = moles C₆H₅Cl / kg water

We know that there are 100 moles per 988744 g of water, so we need to convert the grams of water to kilograms.

[tex]Molality = \frac{100 moles}{988744 g} *\frac{1000 g}{1 kg} = 0.101 m[/tex]

____________________________________

Part 2: Concentration of C₆H₅Cl in air is 0.100 mol/m³, at 25 °C and 1 atm.

Molar mass air = 28.84 g/mol

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / total mass

mass C₆H₅Cl = 0.100 mol * 112.558 g/mol = 11.26 g

total mass = mass C₆H₅Cl + mass air

mass air = moles air * molar mass air

moles air = total moles - moles C₆H₅Cl

We can calculate the total moles by using the ideal gas law:

P V = n R T

where P is pressure in atm, V is volume in L, n is the number of moles, R is the gas constant and T is temperature in Kelvin.

n = P V / R T

P = 1 atm

V = 1 m³ = 1000 L

R = 0.08206 L atm K⁻¹ mol⁻¹

T = 25 + 273.15 = 298 K

n = (1 atm * 1000 L) / (0.08206 L atm K⁻¹ mol⁻¹ * 298 K) = 40.89 moles

moles air = 40.89 - 0.100 = 40.79 moles air

mass air = 40.79 mol * 28.84 g/mol = 1176.4 g

total mass = 1176.4 g + 11.26 g = 1188 g

weight fraction = 11.26 g / 1188 g = 9.45 *10⁻³

(b) mole fraction = moles C₆H₅Cl / total moles

mole fraction = 0.100 / 40.89 = 2.45 *10⁻³

(c) ppm = mg C₆H₅Cl / Liters

We already know there are 11.26 g C₆H₅Cl in 1 m³, which is the same as 1000 L, so:

[tex]\frac{11.26 g}{1000 L} *\frac{1000 mg}{1 g} = 11.3 mg/L[/tex]

The concentration is 11.3 ppm

Final answer:

a) The weight fraction of chlorobenzene is 11.26. b) The chlorobenzene concentration in ppm is 100,000 ppm. c) The mole fraction of chlorobenzene is 0.1. d) The molarity of chlorobenzene is 0.1 M. e) The molality of chlorobenzene is 0.1 m.

Explanation:

a) The weight fraction of chlorobenzene can be calculated by dividing the mass of chlorobenzene by the total mass of the solution. Since the density of the solution is 1.00 g/cm3, the mass of chlorobenzene can be calculated as 100 mol/m3 * 112.6 g/mol / 1000 cm3 = 11.26 g/cm3. Therefore, the weight fraction of chlorobenzene is 11.26 g/cm3 / (1.00 g/cm3) = 11.26.

b) To convert the concentration of chlorobenzene from mol/m3 to ppm (parts per million), we need to multiply by 10^6. Therefore, the chlorobenzene concentration in ppm is 100 mol/m3 * 10^6 ppm/mol = 100,000 ppm.

c) The mole fraction of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the total number of moles in the solution. Since the concentration of chlorobenzene is given in mol/m3, the number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. The total number of moles in the solution is 1 L * (1 mol/m3) = 1 mol. Therefore, the mole fraction of chlorobenzene is 0.1 mol / 1 mol = 0.1.

d) The molarity of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the volume of the solution in liters. Since the concentration of chlorobenzene is given in mol/m3, the number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. The volume of the solution is 1 L. Therefore, the molarity of chlorobenzene is 0.1 mol / 1 L = 0.1 M.

e) The molality of chlorobenzene can be calculated by dividing the number of moles of chlorobenzene by the mass of the solvent in kilograms. Since the density of the solution is 1.00 g/cm3, the mass of the solvent (water) can be calculated as 1 L * (1000 cm3 / 1 L) * (1.00 g/cm3) = 1000 g. The number of moles of chlorobenzene can be calculated as 100 mol/m3 * (1 L / 1000 cm3) = 0.1 mol. Therefore, the molality of chlorobenzene is 0.1 mol / 1 kg = 0.1 m.

Answer:

D. The particles move up and down without changing their position

Explanation:

These are gases. They are typically known for their randomness and no fixed arrangement of their atoms.

Gases generally assume the volumes of the containers they fill. They spread easily and readily to fill the volume where they occupy. Also, gases are readily compressible as they lack intermolecular attraction between their molecules.

Answer: Option (A) is the correct answer.

Explanation:

In a solid substance, the atoms are held together by strong intermolecular forces of attraction. Hence, the atoms are not able t move freely but they are able to vibrate at their mean position.

As a result, solids have a definite shape and volume and also, they are incompressible in nature.

In liquids, the atoms are held together by less strong intermolecular forces of attraction as compared to solids. Therefore, atoms of a liquid are able to slide past each other. So, they are partially compressible in nature.

Also, liquids acquire the shape of container in which they are placed.

But in gases, the molecules are held together by weak intermolecular forces of attraction. Hence, atoms are able to move far away from each other as they have high kinetic energy.

Gases are highly compressible in nature.

Therefore, in the given situation as substance took different shapes in each container.

Thus, we can conclude that it is partially compressible could be another characteristic of the substance.

Answer:

e. 18

Explanation:

A neutral P atom has an atomic number of 15, which means there are 15 protons in the atom. In order to be neutral, the P atom must also have 15 electrons.

The P³⁻ anion has 3 electrons more than the neutral P atom since it has a charge of -3.

Thus, the total number of electrons are 15 + 3 = 18 electrons.

Answer:

36 micrograms of [tex]Cu^{2+}[/tex] ion  are therein one spot.

Explanation:

Amount of copper(II) ions in 1 liter solution = 6 g

Volume of solution used in spotting = 6μL = [tex]6\times 10^{-6} L[/tex]

[tex]1 \mu L = 10^{-6} L[/tex]

Amount of copper (II) ion in [tex]6\times 10^{-6} L[/tex]:

[tex]6\times 6\times 10^{-6} L=3.6\times 10^{-5} g[/tex]

[tex]1 g = 10^{6} \mu g[/tex]

[tex]3.6\times 10^{-5} g=3.6\times 10^{-5}\times 10^{6} \mu g=36 \mu g[/tex]

36 micrograms of [tex]Cu^{2+}[/tex] ion  are therein one spot.

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x)  has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

Answer:

185.5156 g

Explanation:

The reaction of copper sulfate with zinc is shown below as:

[tex]CuSO_4+Zn\rightarrow ZnSO_4+Cu[/tex]

Given that :

Amount of copper sulfate = 454 g

Molar mass of copper sulfate = 160 g/mol

The formula for the calculation of moles is shown below:

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

Thus, moles are:

[tex]moles= \frac{454\ g}{160\ g/mol}[/tex]

[tex]moles\ of\ copper\ sulfate= 2.8375\ mol[/tex]

From the reaction,  

1 mole of copper sulfate reacts with 1 mole of zinc

Thus,

2.8375 moles of copper sulfate reacts with 2.8375 moles of zinc

Moles of Zinc that should react = 2.8375 moles

Mass of zinc= moles×Molar mass

Molar mass of zinc = 65.38 g/mol

Mass of zinc = 2.8375 ×65.38 g = 185.5156 g

Final answer:

185.595 grams of zinc would react with 454 g of copper sulfate, based on the one-to-one mole ratio between copper sulfate and zinc and the molar mass of zinc (65.38 g/mol).

Explanation:

The student is asking how many grams of zinc would react with 454 g of copper sulfate, which has a molar mass of 160 g/mol. According to the provided reaction, CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq), there is a one-to-one mole ratio between copper sulfate and zinc. Given the molar mass of copper sulfate (160 g/mol), we first find the number of moles of copper sulfate in 454 g.

Number of moles of CuSO4 = (454 g) / (160 g/mol) = 2.8375 mol

Since the mole ratio between copper sulfate and zinc is 1:1, an equal number of moles of zinc will react with the copper sulfate. We then need to find the molar mass of zinc to convert moles of zinc to grams. Zinc has a molar mass of approximately 65.38 g/mol.

Mass of zinc = Number of moles of Zn × molar mass of Zn = 2.8375 mol × 65.38 g/mol = 185.595 g

Therefore, 185.595 grams of zinc would react with 454 g of copper sulfate.

Answer:

make thermometers smaller using mercury

Explanation:

Daniel Fahrenheit invented first accurate thermometer which used mercury instead of the alcohol and the water mixtures. In laboratory, he used this invention of him to develop first temperature scale which was enough precise to become the worldwide standard.

The key to the Fahrenheit's thermometer was that the mercury is able to rise and fall within tube without sticking to sides. It was ideal substance for the reading temperatures since mercury expanded at more constant rate than the alcohol and is also able to be read the temperature at much higher and also lower temperatures.

Answer: Option (b) is the correct answer.

Explanation:

Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.

As the given reaction is as follows.

      [tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]

Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.

Thus, we can conclude that in the given reaction carbon atoms are reduced.

Explanation:

The given data is as follows.

          [tex]\Delta H[/tex] = 286 kJ = [tex]286 kJ \times \frac{1000 J}{1 kJ}[/tex]

                            = 286000 J

 [tex]S_{H_{2}O} = 70 J/^{o}K[/tex],      [tex]S_{H_{2}} = 131 J/^{o}K[/tex]

 [tex]S_{O_{2}} = 205 J/^{o}K[/tex]

Hence, formula to calculate entropy change of the reaction is as follows.

          [tex]\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)[/tex]

                     = [tex][(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}][/tex]

                    = [tex][(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)][/tex]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             [tex]\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}[/tex]

                            = [tex]286000 J - (163.5 J/K \times 298 K)[/tex]

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

Final answer:

To find the electrical work required for the electrolysis of water to produce 1 mole of hydrogen, calculate the Gibbs Free Energy (ΔG) for the reaction. Using the given thermodynamic data, ΔG at 298 K is 236.4 kJ, representing the minimum electrical work needed.

Explanation:

The student has asked how to determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298 K and 1 atm. The question involves understanding the thermodynamics of the reaction: H2O(l) → H2(g) + 0.5O2(g), with given data points including ΔH and standard entropies (S°) for the reactants and products. To find the electrical work required, you first calculate the ΔG (Gibbs Free Energy) of the reaction using the formula ΔG = ΔH - TΔS. Knowing ΔG allows you to determine the maximum work that can be extracted from the reaction, which, in the case of electrolysis, corresponds to the minimum work required to drive the reaction in reverse.

ΔS for the reaction can be calculated using the given entropies: ΔS = ⅛∑S°(products) - ⅛∑S°(reactants) = (131 + 0.5×205) - 70 = 166.5 J/K. Therefore, ΔG at 298 K can be calculated as ΔG = 286,000 J - (298K × 166.5 J/K) = 286,000 J - 49,617 J = 236,383 J or 236.4 kJ. This value represents the minimum electrical work required to produce one mole of hydrogen gas via electrolysis of water under the specified conditions.

Answer:

The temperature difference across the stainless steel is 18°C

Explanation:

The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.

The heat flux q/A can be expressed as:

[tex]q/A = h*\Delta T[/tex]

It can also be expressed as

[tex]q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)[/tex]

being Δx/k*(T1-T2) the conductive heat flux through the steel.

If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:

[tex](k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^{\circ} C[/tex]

Answer:

2486 mmHg

Explanation:

Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:

P₁/T₁ = P₂/T₂

Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.

The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.

°C must be converted to absolute temperature (K), thus:

25°C + 273,15 = 298,15 K

755°C + 273,15 = 1028,15 K

Thus, pressure P₂ is:

(T₂·P₁) / T₁ = P₂

1028,15K · 721mmHg / 298,15 K =  2486 mmHg

I hope it helps!

Answer:

The hydroxide ion concentration is 0.003981 M.

The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].

Explanation:

The pOH of an aqueous solution at 25°C =  2.40

[tex]pOH=-\log[OH^-][/tex]

[tex]2.40=-\log[OH^-][/tex]

[tex][OH^-]=0.003981 M[/tex]

The hydroxide ion concentration is 0.003981 M.

The pH of an aqueous solution at 25°C =  ?

The relationship between pH and pOH :

pH + pOH = 14

[tex]pH=14- pOh = 14-2.40 =11.6 [/tex]

[tex]pH=-\log[H_3O^+][/tex]

[tex]11.6=-\log[H_3O^+][/tex]

[tex][H_3O+]=2.5119\times 10^{-12} M[/tex]

The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].

Answer:

If you want to calculate the number of moles of any compund, you should look for the molar mass. For LiF, its molar mass is 25,9 g/m So if 25,9 g are in one mol, the 317 in how many?. Then 12,2 is the number of moles of lithium fluoride produced.

Explanation:

To get the molar mass of a compound, you should know the mass at the Periodic Table, for Li 7,00 , for F 18,9

Answer:

[tex]\boxed{\text{12.2 mol}}[/tex]

Explanation:

[tex]\text{M$_{r} $ of LiF} = 6.94 + 19.00 = 25.94\\\text{Moles} = \text{317 g } \times \dfrac{\text{1 mol}}{\text{25.94 g}} = \text{12.2 mol}\\\\\text{The sample contains $\boxed{\textbf{12.2 mol}}$ of LiF}[/tex]

Answer: The given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]

Explanation:

Scientific notation is defined as the notation in which a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 0.0251089

Converting this into scientific notation, we get:

[tex]\Rightarrow 0.0251089=2.511\times 10^{-2}[/tex]

Hence, the given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]

The average atomic weight of Hafnium is calculated by multiplying the atomic weight of each isotope by its relative abundance and summing the products. The result is an average atomic weight of 178.433 amu.

To calculate the average atomic weight of Hafnium (Hf), we multiply the atomic weight of each isotope by its natural abundance, expressed as a fraction, and then sum the results. Here are the calculations:

Adding these products together we get the average atomic weight of Hafnium:

(0.0016 × 173.940) + (0.0526 × 175.941) + (0.1860 × 176.943) + (0.2728 × 177.944) + (0.1362 × 178.946) + (0.3508 × 179.947) = 0.278304 + 9.255366 + 32.922978 + 48.523072 + 24.373172 + 63.079956 = 178.432848 amu

The average atomic weight of Hafnium truncated to three decimal places is 178.433 amu.

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Explanation:

Boiling point is defined as the point at which liquid state and vapor state of a substance are existing in equilibrium.

Equilibrium is defined as the state in which rate of forward and rate of backward reaction are equal to each other.

For example, [tex]Br(l) \rightleftharpoons Br(g)[/tex]

So, when we boil bromine which is present in liquid state then at the boiling point its vapors will exist in equilibrium. And unless all the liquid state of bromine will not convert into vapors its temperature will not change.

Therefore, we can conclude that at boiling point the liquid and the vapur of Bromine are in equilibrium.

Answer:

Pyruvic acid: conjugate base

Lactic acid: conjugate base

Explanation:

The ratio of conjugate base to conjugate acid can be found using the Henderson-Hasselbalch equation when the pH and pKa are known.

pH = pKa + log([A⁻]/[HA])

The equation can be rearranged to solve for the ratio:

pH - pKa = log([A⁻]/[HA])

[A⁻]/[HA] = 10^(pH-pKa)

Now we can calculate the ratio for the pyruvic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 2.50) = 79433

[A⁻] = 79433[HA]

There is a much higher concentration of the conjugate base.

Similarly for lactic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 3.86) = 3467

[A⁻] = 3467[HA]

For lactic acid the conjugate base also dominates at pH 7.4