Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Answer:

Lies predominantly to the left.

Explanation:

In the reaction:

NH4⁺(aq) + Br-(aq) ⇄ NH3(aq) + HBr(aq)

Conjugate acid + Ion ⇄ weak base + strong acid

HBr is a strong acid whereas NH3/NH4⁺ are the weak base and its conjugate base. A strong acid as HBr dissociates completely in solution as H⁺ and Br⁻. That means in solution you will never have HBr without dissociation doing the reaction:

Answer:

The answer is

Explanation:

The Ka of an acid when given the pH and concentration can be found by

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

Multiply through by - 2

Find antilog of both sides

We have the final answer as

Hope this helps you

Answer:

[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]

Explanation:

Hello,

In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:

[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]

Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:

[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]

Hence, in terms of the molar solubility [tex]x[/tex], we can write:

[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]

In such a way, solving for [tex]x[/tex], we obtain:

[tex]x=0.00238M[/tex]

Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:

[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]

Best regards.

Answer: it is A

Explanation: I am sure

Answer:

0.191 M

Explanation:

i took the test.

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

The given question is incomplete, the complete question is:

An atom of 122In has a mass of 121.910280 amu. Calculate the binding energy in MeV per atom. Use the masses: mass of 1H atom = 1.007825 amu mass of a neutron = 1.008665 amu 1 amu = 931.5 MeV

Answer:

The correct answer is 1029.95 MeV.

Explanation:

Based on the given information,  

The mass of proton is 1.007825, the mass of neutron is 1.008665.  

The atomic number of Indium is 49, therefore, the number of neutrons will be, 122-49 = 73.  

Now the calculated mass will be,  

= 49 * 1.007825 + 73 * 1.008665

= 49.383425 + 73.632545

= 123.01597

The mass defect is calculated by subtracting the actual mas from calculated mass,  

Mass defect = Actual mass - Calculated mass

= 121.910280 - 123.01597

= -1.10569 amu

Now the binding energy will be,  

Binding energy = 1.10569 * 931.5 MeV

= 1029.95 MeV

Answer:

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]

[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Answer:

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Explanation:

Being:

the molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:

Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole

So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?

[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]

amount of moles=0.0289 moles

Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:

[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in [tex]\frac{moles}{liter}[/tex]

So in this case:

Replacing:

[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]

Solving:

[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]

volume=0.2223 liters

Being 1 L= 1,000 mL:

volume=0.222 liters= 222.3 mL

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Answer:

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Explanation:

Consider the reaction:

H2CO(g) + O2(g) --------> CO2(g) + H2O(l)

Using standard thermodynamic data;

Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol

At standard thermodynamic data

[tex]\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}[/tex]

[tex]S^0(CO_2)[/tex] = 213.79 J/mol.K

[tex]S^0(H_2O)=[/tex]  69.95  J/mol.K

[tex]S^0 ({H_2CO}) =[/tex] 218.95 J/mol.K

[tex]S^0 (O_2)[/tex]  = 205.2 J/mol.K

[tex]\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K - (218.95+ 205.2) J/mol.K}[/tex]

[tex]\mathtt{S^0_{rxn} = (283.74) J/mol.K - (424.15) J/mol.K}[/tex]

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Answer:

25.7 mL

Explanation:

Step 1: Given data

Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

V₁ = 25.7 mL

Answer:

8.90 M

Explanation:

Step 1: Given data

Step 2: Calculate the initial concentration

We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₁ = C₂ × V₂  / V₁

C₁ = 5.00 M × 1.13 L  / 0.635 L

C₁ = 8.90 M

Answer:

[tex]\large \boxed{\text{8.90 mol/L}}[/tex]

Explanation:

We can use the dilution formula to calculate the concentration of the original solution.

[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]

Answer:

[tex]1.628 M[/tex]

Explanation:

From the question we were given 0.155 moles of HBr, but Br and H are in ratio 1:1, then there are 0.155 moles of Br- ions.

We were also told that the solution contain NaBr, of 25.9 g. Then it must be converted to moles.

molar mass of NaBr =(22.99g + 79.90 )

= 102.89 g per mol.

the moles of NaBr can be calculated as 25.9 / 102.89

=0.252 moles

But Na and Br are in a ratio 1:1 , then there are 0.252 moles of Br-.

Then to get two Br- mol , we will add the first and second mol of Br- together

= 0.155 + 0.252

=0.407 moles.

The given solution has volume of 250 mL, but we know that there are 1000 ml in a liter, then if we convert to L for unit consistency we have

= 250/1000

= 0.25 L

molarity=0.407 moles/0.25 L

= 1.628 M.

Therefore, Br ion molarity is 1.628 M.

The molarity of the Br ions in the 250 ml solution has been 1.628 M.

Moles can be defined as the mass per unit molecular mass. Moles can be expressed as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of NaBr = [tex]\rm \dfrac{25.9}{102.89}[/tex]

Moles of NaBr = 0.252 mol

Moles of HBr = 0.155 mol.

Since both the compounds have 1:1 ratio of atom: Br, the Br produced has been equal to the concentration of the compound.

Br from NaBr = 0.252 mol

Br from HBr = 0.155 mol.

Total Br ions = 0.407 mol.

Molarity can be expressed as:

Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]

Molarity of Br ions = 0.407 × [tex]\rm \dfrac{1000}{250\;ml}[/tex]

Molarity of Br ions = 1.628 M.

The molarity of the Br ions in the 250 ml solution has been 1.628 M.

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Answer:

0.500 moles of oxygen

Explanation:

Full question says: "Calculate the moles of oxygen needed to produce 0.500 moles of acetic acid.

The reaction of ethanol (C₂H₅OH) with oxygen (O₂) is:

C₂H₅OH + O₂ → CH₃COOH + H₂O

Where 1 mole of ethanol reacts per mole of oxygen to produce 1 mole of acetic acid (CH₃COOH) and 1 mole of water

Based on the chemical equation (1 mole of oxygen produce 1 mole of acetic acid; Ratio 1:1). Thus, if you want to produce 0.500 moles of acetic acid you will need:

Answer:

D

Explanation:

D is the answer

Answer:

thank you...

Answer:

The smallest whole-number coefficient for OH⁻ is 2

Explanation:

Step 1: The equation redox reaction is divided into two half equations

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 5 : addition of the two half equations to yield a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole number coefficient for OH⁻ is 2

A redox reaction is divided into two half equations which are shown below:

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The two half equations are then added and written together to form a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole-number coefficient for OH⁻ is therefore 2.

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Answer:

Following are the answer to this question:

Explanation:

For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:

please find the attachment:

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

Balance the equation and make sure both the reactant and the products are the same

Hope it will be helpful

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex]  is the correct half-reactions.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

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Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:

0.5 mol MgCl₂

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCl → MgCl₂ + H₂

In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.

Step 2: Establish the appropriate molar ratio

The molar ratio of HCl to MgCl₂ is 2:1.

Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl

1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂

Answer:

it is 2.0, the above one is wrong

Explanation:

I did the test :

Answer:

[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]

2. Molecules of B₄H₁₀

[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]

3. Atoms of H

[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]

Answer:

Entropy change of ice changing to water at 0°C is equal to 57.1 J/K

Explanation:

When a substance undergoes a phase change, it occurs at constant temperature.

The entropy change Δs, is given by the formula below;

Δs = q/T

where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur

From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J

Δs = 15600 J / 273.15 K

Δs = 57.111 J/K

Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K

The entropy change of ice changing to water will be "57.1 J/K".

The shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.

According to the question,

Temperature, T = 0°C or,

                          = 273.15 K

Heat, q = 15.6 KJ or,

            = 15600 J

We know the formula,

Entropy change, Δs = [tex]\frac{q}{T}[/tex]

By substituting the values, we get

                                 = [tex]\frac{15600}{273.15}[/tex]

                                 = 57.11 J/K

Thus the above answer is correct.    

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Answer:

The correct answer is 0.25 atm.

Explanation:

As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.  

The formula for calculating mole fraction is,  

Mole fraction = mole of the substance/total moles  

The mole fraction of Kr = x/4x = 0.25

The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,  

Partial pressure = mole fraction * Total pressure

Partial pressure = 0.25 * 1 = 0.25 atm.  

The partial pressure of Kr is 0.25 atm.

We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles  of all the gases will now be; x + x+ x + x = 4x

Let us recall that partial pressure of a gas can be obtained form the formula;

Partial pressure = mole fraction × total pressure

Mole fraction of Kr = x/4x = 1/4

Total pressure = 1 atm

Partial pressure of Kr = 1/4 × 1 = 0.25 atm

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Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

Answer:

1. In the compound, H3C-NH-CH2-CH3, there are no chiral centers present, chiral centers refer to the configuration in which carbon is attached with four different groups. In the molecules, as there are no chiral centers, therefore the molecule is optically inactive, that is, it will not demonstrate pair of an enantiomer is one of the essential characteristics of optically active compounds is the possession of enantiomeric pairs.  

2. On the nitrogen of aniline, the lone pair of electrons can produce hydrogen bonds, play an essential function in basicity, play an essential role in dipole moment or polarity, and wit the increase in solubility there is an increase in the formation of the hydrogen bond, eventually increasing to boiling point. However, the melting point is not affected. As the melting point is the characteristic of the packing efficacy of a molecule and does not rely upon the anilinic nitrogen's lone pairs.  

3. With the increase in the tendency to donate an electron, basicity increases. However, if the electron is taking part in resonance, the donation will not take place easily, and the compound will be the least basic. Apart from cyclohexyl amine, in all the other given compounds, the lone pair of nitrogen takes part in the process of delocalization or conjugation. Thus, cyclohexyl amine will be most basic as the lone pairs are easily available for donation.  

Answer:

In order from lowest to highest:

Methane < Ethane < Chloroethene < Methanol

i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl

Explanation:

Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.

I'm positive this explanation would suffice. Best of luck.

The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.

Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.

The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.

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Answer:

7.50 L

Explanation:

The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 6.25 atm × 1.20 L / 1.00 atm

V₂ = 7.50 L

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

Answer:

a. 3.8856x10⁻³M HCl

b. 1.23x10⁻⁴M OH⁻

c. 1.23x10⁻⁴M Ag⁺

d. Ksp = [Ag⁺] [OH⁻]

Explanation:

a. The reaction that you are studying is:

HCl(aq) + AgOH(aq) → H₂O(l) + AgCl(s)

The HCl solution is diluted from 10.00mL to 250.00mL, that is:

250.00mL / 10.00mL = 25 -The solution is diluted 25 times-

As original concentration of HCl is 0.09714M, the concentration of the diluted solution is:

0.09714M / 25 =

b. 1 mole of HCl reacts per mole of AgOH, moles of HCl that reacts are:

7.93mL = 7.93x10⁻³L × (3.8856x10⁻³mol HCl / L) = 3.0813x10⁻⁵ moles of HCl.

Based on the reaction, you have in solution

3.0813x10⁻⁵ moles of AgOH = Ag⁺ = OH⁻

The AgOH solution was 250.0mL = 0.2500L, its concentration is:

3.0813x10⁻⁵ moles OH⁻ / 0.2500L =

c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:

1.23x10⁻⁴M OH⁻ =

d. The solubility product reaction of AgOH(s) is:

AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)

Where Ksp for this reaction is defined as: