Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate the amount of Ga ( s ) that can be deposited from a Ga ( III ) solution using a current of 0.220 A that flows for 40.0 min .

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:

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:

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.

Learn more:

Answer:

neither half will have a pole

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:

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:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = [tex]\frac{936}{8}[/tex] = 117

By substituting the value of (y+z) in equation- (1) we get,

                          x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

                           y = [tex]\frac{535-(6\times 32)}{7}[/tex] = 49

By substituting the value of y  into equation- (2) we get,

                           z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Final answer:

To find the amount of sodium in each item, set up a system of equations and solve to get x = 9, y = 55, and z = 85.

Explanation:

To find the amount of sodium in each item, we can set up a system of equations.

Let x represent the amount of sodium in 1 cracker, y represent the amount of sodium in 1 pretzel, and z represent the amount of sodium in 1 cookie.

From the given information, we can create the following equations:

Solving this system of equations, we can find the values of x, y, and z. The amount of sodium in 1 cracker is 9 mg, 1 pretzel is 55 mg, and 1 cookie is 85 mg.

Answer:

790 g/s

Explanation:

The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.

The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;

Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].

Where g = wind direction, y and h are the distance in metres.

Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;

Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.

Emission rate of pollutant,E = 790 g/s.

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:

Carbon monoxide has a bond order of 3

Explanation:

A molecular orbital diagram ( MO diagram ), is a qualitative descriptive tool tha explains chemical bonding in molecules in terms of the molecular orbital theory in general and particularly adopts the linear combination of atomic orbitals (LCAO) method in its approach. A fundamental principle of these theories is that as atoms bond to each other to form molecules, a certain number of atomic orbitals combine to form an equal number of molecular orbitals , although the electrons involved may be redistributed among the orbitals. This tool is very well suited for simple diatomic molecules such as hydrogen molecule , oxygen molecule, and carbon monoxide but becomes more complex when discussing even comparatively simple polyatomic molecules, such as methane. MO diagrams usually explains why some molecules exist and others do not exist. It is used to predict bond strength of molecules, as well as the electronic transitions that can take place.

The MO diagram for carbon monoxide is shown in the image attached. It can clearly be seen that carbon monoxide has a bond order of three. The combination of carbon and oxygen atoms to form molecular orbitals of appropriate energy is also seen in the image.

The molecule SI2 does not exist but if the student meant SiI2, it would have 8 bonding electrons and 14 nonbonding electrons.

The molecule SI2 does not exist. It may have been a typo. If the student meant Si2, this molecule also doesn't exist because silicon, a member of group 14 on the periodic table, needs 4 electrons to achieve a full outer shell, and thus generally forms 4 bonds. If we are talking about SiI2, or silicon diiodide, it would have 8 bonding electrons and 14 nonbonding electrons. This is because all four of the bonding electrons from silicon and four from iodine atoms make up the bonding electrons while the remaining 7 electrons from each iodine atom make up the nonbonding electrons.

https://brainly.com/question/34264394

#SPJ3

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:

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:

c. 7.14

Explanation:

The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.

pH = pKa + log [base]/[acid]

pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211

pH = 7.14

Answer:

B) 125 mL

Explanation:

M1V1=M2V2

(0.120M)(x)=(150.0 mL)(0.100M)

x= 125 mL

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

Answer:

Methane

Explanation:

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

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:

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:

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.

https://brainly.com/question/2862004

#SPJ3

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:

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]

Answer:

a. pH = 4.75

b. pH = 9.20

c. pH = 8.42

d. pH = 13.53

Explanation:

This is a titration between a strong base, the KOH and a weak acid, HCN.

The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN

          HCN + H2O ⇄  H₃O⁺  +  CN⁻

Initial    0.5                      -             -

Eq.      0.5-x                    x             x

Ka =  x² / (0.5-x) = 6.17ₓ10⁻¹⁰

Ka is really small, so we can say that 0.5-x = 0.5. Then,

x² = 6.17ₓ10⁻¹⁰ . 0.5

x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]

pH = - log [H₃O⁺]  →  - log 1.75×10⁻⁵ = 4.75

b. First of all, we determine the moles of base, we are adding.

0.250 mol/L . 0.006 L = 0.0015 moles

In conclussion we have 0.0015 moles of OH⁻

Now, we determine the moles of our acid.

0.500 mol/L . 0.020L = 0.01 moles

The  0.0015 moles of OH⁻ will be neutralized with the acid, so:

      HCN     +    OH⁻         →     H₂O   +    CN⁻

       0.01         0.0015                          0.0085

The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)

Our new volume is 20 mL and 6mL that we added, so, 26mL

This is a buffer with the weak acid, and its conjugate base.

Our concentrations are 0.0085 moles / 0.026 L = 0.327 M

We apply Henderson-Hasselbach

pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)

pH = pKa

c. When we add 40 mL, our volume is 20mL +40mL  = 60 mL

These are the moles, we add:

0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)

 HCN     +    OH⁻         →     H₂O   +    CN⁻

  0.01          0.01                                 0.01

All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.

0.01 moles / 0.060 L = 0.16 M → [CN⁻]

pH at this point will be

       CN⁻  +  H₂O ⇄  HCN + OH⁻             Kb = 1.62ₓ10⁻⁵ (Kw/Ka)

In.   0.16                        -          -

Eq. 0.16-x                     x          x

Kb = x² / (0.16-x)

We can also assume that 0.16-x = 0.16. Then:

[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ .  0.16) = 2.59×10⁻⁶

- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58

pH = 14 - pOH  → 14 - 5.58 = 8.42

This is a basic pH, because the titration is between a weak acid and a strong base.

d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL

We add 0.5 mol/L . 0.062L = 0.031 moles

These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.

0.031 moles - 0.01 moles = 0.021 moles of OH⁻

[OH⁻] = 0.021 moles / 0.062L = 0.34M

- log [OH⁻]  = pOH → - log 0.34 = 0.47

pH = 14-pH → 14 - 0.47 = 13.53

This answer details the titration of HCN with KOH, explaining ways to calculate the pH values at different stages of the titration: the initial state, partway through, at the equivalence point when all acid has been neutralized, and after the equivalence point.

We are considering the titration of a solution of HCN with KOH. First, we calculate the initial pH using the formula pH=-log10[H+], where the H+ concentration can be determined from the Ka expression for HCN.

For question b, when 6.00 mL base KOH is added, we need to determine which ion is in excess, and then use its concentration to calculate the pH. In this case, there are still excess HCN ions leading to a buffer solution. The Henderson-Hasselbalch equation can be used to determine the pH.

For question c, at the equivalence point, 40.00 mL of base KOH has been added. All HCN has been titrated to form CN-, so the pH is determined by the OH- ions hydrolyzed from CN-. We use the Kb of CN- (which can be calculated using Kw/Ka) and the total volume to calculate [OH-], and then pH = 14 - pOH.

Finally, for question d, after 42.00 mL of KOH, OH- is in excess. We can calculate the [OH-] and then find the pH using pH = 14 - pOH.

https://brainly.com/question/31271061

#SPJ6

Answer:

Potassium and Argon (⁴⁰K-⁴⁰Ar)

Explanation:

Potassium-Argon dating is a method used in geology and archeology to date rocks or volcanic ash, generally the oldest.

It is based on the principle of radioactive decay, it is a process that has a half-life of 1248 million years, during which time the gas is concentrated in the rock crystals. Taking advantage of this known rhythm and half-life, the method lends itself to dating samples ranging from 100.000 years to several billion years.

Answer:

Its D to sum it up.

Explanation:

Answer:

Explanation:

check the attached file below for the diagram and better expalnation.

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:

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.

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.

The information needed includes the heat of vaporization and specific heat of H₂O(l).

The information needed to calculate the enthalpy change for converting 1 mol H₂O(g) at 100 °C to H₂O(l) at 80 °C is the (d) heat of vaporization and specific heat of H₂O(l).

The heat of vaporization is required because it accounts for the energy needed to convert water from a liquid to vapor state. The specific heat of H₂O(l) is needed to determine the amount of heat required to change the temperature of liquid water from 100 °C to 80 °C.

Answer:

The volume when the conditions were altered is 0.5109 L or 510.9 mL

Explanation:

Using the general gas equation,

P1 V1 / T1 = P2 V2 / T2

where;

P1 = 756 mmHg

V1 = 475 ml = 0.475 L

T1 = 23.5°C = 23.5 + 273K = 275.5 K

P2 = 722 mm Hg

T2 = 10°C = 10 + 273 K = 283 K

V2 = ?

Rearranging to make V2 the subject of the formula, we obtain:

V2 = P1 V1 T2 / P2 T1

V2 = 756 * 0.475 * 283 / 722 * 275.5

V2 = 101, 625.3 / 198911

V2 = 0.5109 L or  510.9 mL