A 50.0 mL sample containing Cd 2 + and Mn 2 + was treated with 50.0 mL of 0.0400 M EDTA . Titration of the excess unreacted EDTA required 19.5 mL of 0.0270 M Ca 2 + . The Cd 2 + was displaced from EDTA by the addition of an excess of CN − . Titration of the newly freed EDTA required 17.1 mL of 0.0270 M Ca 2 + . What are the concentrations of Cd 2 + and Mn 2 + in the original solution?

Answer:

Cathode

Cu^2+(aq) + 2e ----> Cu(s)

Zinc is the cathode

Explanation:

The plating of copper is normally done by electrolysis. Electrolysis is generally defined as the chemical decomposition produced by passing an electric current through a liquid or solution containing ions.

There are two electrodes, the anode and the cathode. Recall that electrolysis is not a spontaneous process, hence energy from a battery is required to drive the reaction in the desired direction.

The metal to be plated is normally the cathode while the metal used to plate it is normally the anode. Since copper is to be plated on zinc, zinc must be the cathode while copper will be the anode.

The half-reaction that takes place when pennies are plated with solid copper is :

Cu^2+(aq) + 2e ----> Cu(s)

Zinc is the cathode.

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Answer:

the answer is all of them

Explanation:

it is all of them because organisms in and ecosystem compete for anything and every thing that they need. Hope this helps!

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

T1 = 315 K

V2 = ?

P2 = 463 mmHg

T2 = 777 K

The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

To achieve complete combustion of one mole of butane (C₄H₁₀), 6.5 moles of oxygen (O₂) are necessary, as indicated by the balanced chemical equation for combustion.

To determine the number of moles of oxygen (O₂) required for the complete combustion of one mole of butane (C₄H₁₀), we must first write the balanced chemical equation for the combustion reaction:

C₄H₁₀ + O₂ → CO₂ + H₂O

After balancing the equation, we have:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

This balanced equation tells us that for every 2 moles of butane, we need 13 moles of oxygen for complete combustion, which means that for every mole of butane, we need 6.5 moles of oxygen.

Answer:

c) 107.75

Explanation:

Hot water lost = 61 g * 60C * (4.184 J g¯1 °C¯1) = 3,660

Cold water = 74.6g * 20C  * (4.184 J g¯1 °C¯1) = 1,492

The difference is 3,660 - 1,492 = 2,168

Calorimeter Constant = Heat released by burning / Change in temperature

Calorimeter Constant = 2,168 / 40C * (1.987 J g¯1 °C¯1)

Calorimeter Constant = 107.75

Answer:

Calculate the molar concentration of the NaOH solution that you prepared Number of moles of KHP = Number of moles NaOH = 2.476 x 10 -3 moles Number of moles NaOH = Mb x Vb Mb = 2.476 x 10 -3 moles / 0.0250 L (equivalence point) = 0.0990 M 3

Explanation:

Answer:

The total pressure of the mixture is 1.657 atm

Explanation:

Step 1: Data given

Mol fraction of H2 = 0.35

Pressure of H2 = 0.58 atm

Partial pressure gas = total pressure gas * mol fraction gas

Step 2: Calculate the total pressure

Partial pressure H2 = total pressure * mol fraction

0.58 atm = total pressure * 0.35

Total pressure = 0.58 atm / 0.35

Total pressure = 1.657 atm

The total pressure of the mixture is 1.657 atm

Considering the Dalton's partial pressure, the total pressure in the mixture of gases is 1.657 atm.

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

[tex]P_{T} =P_{1} +P_{2} +... +P_{n}[/tex]

where n is the amount of gases in the mixture.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture.  So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

[tex]P_{A} =x_{A} P_{T}[/tex]

In this case, the partial pressure of gas H₂ can be expressed as:

[tex]P_{H_{2} } =x_{H_{2} } P_{T}[/tex]

You know:

Replacing in the definition of partial pressure of gas H₂:

[tex]0.58 atm=0.35P_{T}[/tex]

Solving:

[tex]P_{T}=\frac{0.58 atm}{0.35}[/tex]

[tex]P_{T}[/tex]= 1.657 atm

In summary, the total pressure in the mixture of gases is 1.657 atm.

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Answer:

The additional information required to solve this problem is the initial volume.

the final pressure P₂ of the gas is 1.108 atm

Explanation:

Given that :

A sample of gas at initial temperature [tex]T_1 = 12.0^0 \ C[/tex] = (12+273)K = 285 K

Pressure (P₁) = 1.06 atm

Initial Volume (V₁) = unknown ???

Final Volume (V₂) = 2.30 L

final temperature [tex]T_2 = 24.9^0 \ C[/tex]  = (24.9 +273)K = 297.9 K

Find the final Pressure (P₂)

The relation between: Pressure, Volume and Temperature can be gotten from the ideal gas equation :

PV = nRT

The Ideal Gas Equation is also reduced to the General Gas Law or the combined Gas Law by assuming that n= 1 .

From ; PV = nRT

[tex]\frac{PV}{T} = R \ \ ( constant) \ if \ n=1[/tex]

∴ [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = \frac{P_3V_3}{T_3}...= \frac{P_nV_n}{T_n} \ \ \ ( n \ constant)[/tex]

The additional information required to solve this problem is the initial volume.

This expression is a combination of Boyle's Law and Charles Law. From the combined Gas Law , it can be deduced that at constant volume, the pressure of a given mass(mole) of gas varies directly with absolute temperature.

∴ [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]  if n & Volume (V) are constant .

[tex]P_2 = \frac{1.06*297.9}{285}[/tex]

P₂ = 1.108 atm

Thus, the final pressure P₂ of the gas is 1.108 atm

Approximately 0.237 milligrams of cesium-137 would remain after 60 years.

The amount of a radioactive substance that remains after a certain amount of time can be calculated using the decay constant. For cesium-137, the decay constant is 0.0871 per year. To determine the amount remaining after 60 years, we can use the formula:
Amount remaining = initial amount * e^(-decay constant * time)
Substituting the values, we get:
Amount remaining = 20mg * e^(-0.0871 * 60) = 20mg * e^(-5.226) ≈ 0.237mg. Therefore, approximately 0.237 milligrams of cesium-137 would remain after 60 years.

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A physical change is when there is an alteration to the material but does not affect at the molecular level. An example of a physical change would be cutting, crushing, freezing, and boiling a material object.

Answer: Amount of Gallium = 0.127g

Explanation:

Electrolysis equation is:

Ga3+   +   3e-    ------> Ga

 To calculate the charge

t = 40.0 min = 40.0 x 60 s = 2400 s

time, t = 2400s

Q = I*t =  

= 0.22A x 2400s

= 528 C

1 mol of Ga requires 3 mol of electron

1 mol of electron = 1 Faraday =96485 C

So,1 mol of Ga requires  96485x 3= 289455 C

mol of Gallium = 528/289455 = 0.00182 mol

Molar mass of Ga = 69.72 g/mol

mass of Ga = number of moles x  molar mass

= 0.00182mol * 69.72

g/mol

= 0.127g

or you can use this direct formula

m=(current*time/Faraday's)*(molar mass/no of electrons transferred)

keeping in mind   Ga3+ + 3e- → Ga

n=3

m=(It/F)*(mew/n)

m =(0.22 x 2400/96485) x (69.72/3)

m=0.127 g

Answer: tap water to ice cube

Explanation:

Answer: tap water to ice cube

Explanation:

Answer:

The Volume of the lungs that would produce 2 mmHg pressure decrease is

         [tex]V_2 = 2.81 \ L[/tex]

Explanation:

From the question we are told that

     The volume of air in the lungs is  [tex]V = 2.8 \ L[/tex]

     The pressure difference for quit normal inspiration is [tex]P = 2 \ mmHg[/tex]

      The temperature of air in the lungs [tex]T = 37^oC[/tex]

      The pressure  after normal  expiration is at  [tex]T = 760 \ mmHg[/tex]

From ideal gas law we have that

         [tex]PV= nRT[/tex]

Now since  nRT is constant we have that

        [tex]P_1 V_1 = P_2 V_2[/tex]

As the pressure decreased by 2 mmHg the volume becomes

        [tex]V_2 = \frac{P_1 V_1}{P_2}[/tex]

        [tex]V_2 = \frac{2.8 * 760}{758}[/tex]

        [tex]V_2 = 2.81 \ L[/tex]

The change in the lung volume during the process of quiet inspiration can be calculated using Boyle's Law by determining the change in pressure and relating it to an equivalent change in volume. Once pressure values are converted to the same units, the Boyle's Law equation (P1V1 = P2V2) is used to solve for the final volume (V2) that represents the expanded lung volume.

The changes in volume and pressure in the lungs during inspiration (breath in) can be described using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. In this scenario, a decrease in pressure by 2 mmHg drives an expansion of the lungs, and we are asked to determine this corresponding change in volume.

Firstly, the pressure change needs to be converted into standard pressure units - the SI unit is Pascal. 1 mm Hg equals 133.322 Pa, so a 2 mm Hg difference equals 266.644 Pascal (Pa).

Using Boyle's Law equation (P1V1=P2V2), where P1=the initial pressure, V1=the initial volume, P2=the final pressure (P1 - 2mm of Hg) and V2=the final volume that we are trying to find, we can solve for V2. P1 is atmospheric pressure, which is approximately 101,325 Pa, and V1 is the initial volume of air in lungs, which is 2.8 Litres.

Solving the equation gives the final volume, V2, after the pressure decrease and resulting lung expansion.

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Answer:

The correct answer is c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm

Explanation:

In order to calculate the partial pressures of the mixture components, we have to first calculate the number of moles:

For N₂O:

Molecular weight (MW): (14 g/mol x 2) + 16 g/mol= 44 g/mol

Number of moles of N₂O (n) = mass/Mw = 6.46g/44 g/mol= 0.1468 mol

For CO:

Molecular weight (MW): 12 g/mol + 16 g/mol= 28 g/mol

Number of moles of CO (n) = mass/Mw = 2.74 g/28 g/mol= 0.0978 mol

For O₂:

Molecular weight (MW): 16 g/mol x 2= 32 g/mol

Number of moles of O₂ (n) = mass/Mw = 5.40 g/32 g/mol= 0.1687 mol

Once calculated the number of moles of each component, we can calculate the total number of moles (nt):

nt = 0.1468 mol + 0.0978 mol + 0.1687 mol = 0.4133 moles

The partial pressure of a gas in a mixture can be calculated from the molar fraction of the gas (X) and the total pressure of the mixture (Pt=4.33 atm):

P(N₂O) = X(N₂O) x Pt

           = (moles N₂O/nt) x Pt

           = 0.1468 moles/0.4133 moles x 4.33 atm

           = 1.538 atm

P(CO) = X(CO) x Pt

           = (moles CO/nt) x Pt

           = 0.0978 moles/0.4133 moles x 4.33 atm

           = 1.0246 atm

P(O₂) = X(O₂) x Pt

           = (moles O₂)/nt x Pt

           = 0.1687 moles/0.4133 moles x 4.33 atm

           = 1.767 atm

Answer:

See explaination

Explanation:

In order to have the detailed and step by step solution of the given problem, check or see the attached files.

Answer:

Methane

Explanation:

Methane is a potent greenhouse gas with the formula CH₄. Hope this helps!

Answer:

-Hybridization does not account for observed bond angles in molecules.

-Single bonds are always pi bonds.

-The length of a bond is determined by where the energy of the system is at its lowest point                                                                

Explanation:

-The very first statement is incorrect because it does account for different bong angles as the hybrid orbitals are responsible for contributing for bond angles in a way that more the hybrid orbitals present the lesser the angles it forms.

-The second statement is incorrect because single bonds are considered as sigma bonds and not a pi bond.

-The third statement is incorrect because hybridization is responsible for deciding the bond length.

The incorrect statements about bonding and hybridization are that single bonds are always pi bonds and pi bonds are always between unhybridized p orbitals.

The incorrect statements about bonding and hybridization are:

Hybridization does not account for observed bond angles in molecules, so this statement is correct. The length of a bond is determined by where the energy of the system is at its lowest point, so this statement is also correct. Multiple bonds can have a combination of sigma and pi bonds, so this statement is correct as well.

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Answer:

911.36mmhg

Explanation:

1 torr is almost equivalent to 1mmhg.but to convert from atm to mmhg,multiply by 760

Final answer:

The total pressure in the tank, after converting all pressures to the same unit (mm Hg), is the sum of the partial pressures of helium, nitrogen, and neon, resulting in a value of 911.36 mm Hg.

Explanation:

To calculate the total pressure in a mixture of gases, you can use Dalton's Law of partial pressures which states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Since the pressure values are given in different units, we first need to convert them all to millimeters of mercury (mm Hg). The pressure of helium gas is already given in mm Hg. The nitrogen gas pressure needs to be converted from atmospheres (ATM) to mm Hg by multiplying by 760 mm Hg/ATM, and the neon gas pressure in torr simply needs to be directly converted to mm Hg because 1 torr equals 1 mm Hg.

Now we can sum these values to find the total pressure in the tank.

The conversion and summation are as follows:

Helium: 191 mm Hg

Nitrogen: 0.261 ATM x 760 mm Hg/ATM = 198.36 mm Hg

Neon: 522 torr = 522 mm Hg

Adding these together, the total pressure in the tank is:

191 mm Hg + 198.36 mm Hg + 522 mm Hg = 911.36 mm Hg

Therefore, the total pressure in the tank in mm Hg is 911.36 mm Hg.

Answer:

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

Answer:

7.5 moles

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2

From the balanced equation above,

3 moles of Cu reacted with 2 moles of H3PO4.

Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e

Xmol of Cu = (3 x 5)/2

Xmol of Cu = 7.5 moles

Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.

Answer:

≈ -57.2 kJ/mol

Explanation:

Total volume of the solution = (50.0 + 45.0) mL = 95.0 mL.

Density of the solution = 0.997 g/mL.

Mass of the solution = (volume of the solution)*(density of the solution)

= (95.0 mL)*(0.997 g/mL)

= 94.715 g

Heat gained by the solution, qsoln = (mass of the solution)*(specific heat capacity)*(change in temperature)

= (94.715 g)*(4.184 J/ºC.g)*(25.65 – 25.00)ºC

= 257.5869 JAccording to the principle of thermochemistry,

qsoln + qrxn = 0

where qrxn denotes the heat change during the neutralization reaction.

Therefore,

(257.5869 J) + qrxn = 0

======> qrxn = -257.5869 J

HNO3 is the limiting reactant.

Moles of limiting reactant = (volume of the limiting reactant in L)*(molarity of limiting reactant)

= (45.0 mL)*(0.100 M)

= (45.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 0.0045 mole.

ΔHneutralization = qrxn/(moles of HNO3)

= (-257.5869 J)/(0.0045 mol)

= -57241.5 J/mol

= (-57241.5 J/mol)*(1 kJ)/(1000 J)

= -57.2415 kJ/mol

≈ -57.2 kJ/mol

Answer:

Explanation:

Given that:

the temperature [tex]T_1[/tex] = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

[tex]\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}[/tex]

where; B = - [tex]152.5 \ cm^3 /mol[/tex]   C = -5800 [tex]cm^6/mol^2[/tex]

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

[tex]\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]

[tex]4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]

Multiplying through with V² ; we have

[tex]4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0[/tex]

[tex]4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0[/tex]

V = 2250.06  cm³ mol⁻¹

Z = [tex]\frac{PV}{RT}[/tex]

Z = [tex]\frac{1800*2250.06}{8.314*10^3*523.15}[/tex]

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

[tex]T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345[/tex]

[tex]T__{\gamma}} = \frac{T}{T_c}[/tex]

[tex]T__{\gamma}} = \frac{523.15}{647.1}[/tex]

[tex]T__{\gamma}} = 0.808[/tex]

[tex]P__{\gamma}} = \frac{P}{P_c}[/tex]

[tex]P__{\gamma}} = \frac{1800}{22055}[/tex]

[tex]P__{\gamma}} = 0.0816[/tex]

[tex]B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}[/tex]

[tex]B_o = 0.083 - \frac{0.422}{0.808^{1.6}}[/tex]

[tex]B_o = 0.51[/tex]

[tex]B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}[/tex]

[tex]B_1 = -0.282[/tex]

The compressibility is calculated as:

[tex]Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}[/tex]

[tex]Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}[/tex]

Z = 0.9386

[tex]V= \frac{ZRT}{P}[/tex]

[tex]V= \frac{0.9386*8.314*10^3*523.15}{1800}[/tex]

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At [tex]T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa[/tex]

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = [tex]\frac{PV}{RT}[/tex]

Z = [tex]\frac{1800*10^3 *0.1249}{729.77*523.15}[/tex]

Z = 0.588

Final answer:

To determine Z and V for steam at 250°C and 1800 kPa, we can use the truncated virial equation with the given experimental values of the virial coefficients, or we can use the generalized Pitzer correlation to obtain the B value and then use the truncated virial equation. Alternatively, we can look up the values in the steam tables.

Explanation:

The question asks us to determine Z and V for steam at 250°C and 1800 kPa using three different methods: (a) The truncated virial equation with given experimental values of the virial coefficients, (b) The truncated virial equation with B value obtained using the generalized Pitzer correlation, and (c) The steam tables.

(a) To determine Z and V using the truncated virial equation with B and C values, we substitute the given temperature and pressure into the equation and solve for Z and V.

(b) To determine Z and V using the truncated virial equation with the B value obtained from the generalized Pitzer correlation, we substitute the given temperature and pressure into the equation and solve for Z and V.

(c) To determine Z and V using the steam tables, we look up the values for Z and V at the given temperature and pressure.

Answer:

2. because water molecules are dipoles and the dipoles orient in an energetically favorable manner to solvate the ions.

Explanation:

For the water to dissociate electrolytes, this one has to orientate its molecules in an energetically favorable way that allows them to interact with ions and dissociate electrolytes. This has to do with the way that intermolecular forces of a solute and a solvent, which is the water, interact to form a solution. The different intermolecular forces that interact in a solution are dipole-dipole force, ion-dipole interactions, Van Der Waals forces, and Hydrogen bonding.

Answer:

3.89 atm

Explanation:

Given data

If we treat air as an ideal gas, we can calculate the final pressure using the Gay-Lussac's law.

[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}\\P_2 = \frac{P_1 \times T_2 }{T_1} = \frac{2.80atm \times 396K }{285K}\\P_2 = 3.89 atm[/tex]

To convert atoms to moles, divide the number of atoms by Avogadro's number of 6.02 x 10²³ atoms per mole.

Your friend is partially correct. In order to convert atoms to moles, you use the constant known as Avogadro's number (6.02 x 10²³ atoms per mole). However, it's important to note that you need to divide the number of atoms by Avogadro's number, not multiply it. Let's give an example:

Example: If we have 2.56 x 10²⁴ atoms of Uranium, we'd use Avogadro's number to convert this to moles like so: (2.56 x 10²⁴ atoms) / (6.02 x 10²³ atoms/mol) = 4.25 moles of Uranium

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No, your friend is not correct. To convert atoms to moles, divide the number of atoms by Avogadro's number, 6.022 × 10²³. This is because 1 mole of any substance contains 6.022 × 10²³ atoms.

To convert the number of atoms to moles, you should divide the number of atoms by Avogadro's number, which is 6.022 × 10²³. This relationship is based on the fact that 1 mole of any substance contains exactly 6.022 × 10²³ atoms, a constant known as Avogadro's number.

Step-by-Step Explanation:

For example, if you have 1.2044 × 10²⁴ atoms of hydrogen:

Complete Question: -

I want to convert atoms to moles. My friend tells my to multiply the number of atoms by 6.02 x 10²³. Is my friend correct?

Answer:

40659.855 J

Explanation:

From the question given above, we obtained the following:

Mass (m) = 18.015g

Heat of vaporisation (ΔHv) = 2257 J/g

Heat (Q) =?

The heat required to vaporise the water can be calculated as follow:

Q = mΔHv

Q = 18.015 x 2257

Q = 40659.855 J

Therefore, the heat required to vaporise the water is 40659.855 J

Answer:

The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.

Explanation:

Because the heat of the plate will be releases warming up the air making it move faster

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Final answer:

To estimate the mixing-cup average concentration of benzoic acid in water, the Schmidt number and fluid flow characteristics are vital. Understanding incompressible fluid flow through constrictions helps in analyzing scenarios like Venturi tubes.

Explanation:

Estimate the mixing-cup average concentration of benzoic acid in the water leaving the tube by considering the overall flow conditions in the cylindrical tube.

Sc (Schmidt number) represents the fluid flow characteristics such as the diffusion rate of momentum and mass transfer in the system.

Understanding concepts like incompressible fluid flow through constrictions can aid in analyzing scenarios like flow in a Venturi tube where diameters change.

Answer:

x= 138.24 g

Explanation:

We use the avogradro's number

6.023 x 10^23 molecules -> 1 mol C2H8

26.02 x 10^23 molecules -> x

x= (26.02 x 10^23 molecules  * 1 mol C2H8 )/6.023 x 10^23 molecules

x= 4.32 mol C2H8

1 mol C2H8     -> 32 g

4.32 mol C2H8 -> x

x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8

x= 138.24 g

The correct answer is 156.69 * 10^46 grams.

How to convert molecules to grams?

learn more about conversion below,

https://brainly.com/question/24513742

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