Part C
What is produced at each electrode in the electrolysis of an aqueous solution of both NaBr and AgF?
Drag the appropriate items to their respective bins.
Anode Cathode not produced
Na(s) H2(g) Ag(s) F2(g) O2(g) Br2(l)

A loan is being repaid by \(2n\) level payments, starting one year after the loan. Just after the nth payment the borrower finds that she still owe \((3/4)\)of the original amount. The proportion of the next payment that is interest is \(1 - 2 * (1 - (1 + r)^{(-n)}) / (3 * (1 - (1 + r)^{(-2n)})).\)

Let \(P\) be the original amount of the loan, and let \(x\) be the level payment made at each of the \(2n\) payments. Then the total amount repaid will be \(2nx\). We know that after \(n\) payments, the borrower still owes \((3/4)P\).

Therefore, the amount repaid after \(n\) payments is \((P - (3/4)P) = (1/4)P\). This means that the total amount repaid after the remaining \(n\) payments is \((3/4)P\).

We can set up an equation using the formula for the present value of an annuity:

\(P = x * (1 - (1 + r)^{(-2n)}) / r\)

where \(r\) is the interest rate per payment period (which we will assume is constant), and the first payment is due one year after the loan.

After \(n\) payments, the outstanding balance is \((3/4)P\). We can use the same formula to find the present value of the remaining \(n\) payments, but with \(P\) replaced by \((3/4)P\):

\((3/4)P = x * (1 - (1 + r)^{(-n)}) / r\)

We can rearrange this equation to solve for \(x\):

\(x = (3/4)P * r / (1 - (1 + r)^{(-n)})\)

Now we need to find the proportion of the next payment that is interest. The interest component of each payment is the difference between the total payment and the amount of principal being repaid. The total payment is \(x\), and the amount of principal being repaid is:

\((3/4)P * r / (1 - (1 + r)^{(-n)})\).

So the proportion of the next payment that is interest is:

Interest component / Total payment

\(= (x - (3/4)P * r / (1 - (1 + r)^{(-n)})) / x\\= 1 - (3/4)P * r / (x * (1 - (1 + r)^{(-n)}))\\= 1 - (3/4)P / (2nx * (1 - (1 + r)^{(-n)}))\\\)

We can simplify this expression by using the equation we derived earlier for \(x\):

\(1 - (3/4)P / (2nx * (1 - (1 + r)^{(-n)}))\\= 1 - (3/4)P / ((3/2)P * r / (1 - (1 + r)^{(-n)}) * (1 - (1 + r)^{(-2n)}))\\= 1 - 2 * (1 - (1 + r)^{(-n)}) / (3 * (1 - (1 + r)^{(-2n)}))\)

So the proportion of the next payment that is interest is:

\(1 - 2 * (1 - (1 + r)^{(-n)}) / (3 * (1 - (1 + r)^{(-2n)})).\)

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But there is no context or question provided for me to give a main answer in a single line. Please provide more information or a specific question.

In \(H2C=C=CH2\), the central carbon atom has two double bonds, one with each of the adjacent carbon atoms.

This means that the central carbon atom must use two of its available orbitals to form these double bonds.

The remaining two orbitals must then hybridize to form two new orbitals, which will be perpendicular to each other and parallel to the plane of the molecule. This type of hybridization is called sp hybridization.

Therefore, the second carbon atom (bolded) in \(H2C=C=CH2\) has sp hybridization.

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Answer:

A detailed map which shows every feature of the earth accurately including roads, airports, names of places etc is called a topographical map

Explanation:

Answer:

a) Acidified potassium dichromate solution is used to test for the presence of a primary or secondary alcohol.

b) The orange color of the potassium dichromate solution is reduced to green when it reacts with an alcohol.

c) The oxidation of a primary alcohol produces a carboxylic acid.

d) The oxidation of a secondary alcohol produces a ketone.

e) The dichromate test does not work for tertiary alcohols because they cannot be further oxidized. Methylpropan-2-ol is a tertiary alcohol with the chemical structure:

CH3

|

CH3—C—OH

|

CH3

Since there are no hydrogen atoms attached to the carbon atom bearing the hydroxyl group, it cannot be oxidized.

a) A simple test for the presence of a carboxylic acid group is the addition of sodium hydrogencarbonate solution to the compound. The reagent reacts with the carboxylic acid to produce carbon dioxide gas.

Reagent: Sodium hydrogencarbonate solution

Observation: Effervescence (bubbling) due to the release of carbon dioxide gas.

b) To confirm that the gas produced in the hydrogencarbonate test is carbon dioxide, it can be tested with limewater. Carbon dioxide turns limewater milky/cloudy due to the formation of calcium carbonate.

c) The hydrogencarbonate test is not conclusive proof that a carboxylic acid group is present in a completely unknown compound because some other functional groups such as phenols and alcohols can also react with the reagent and produce carbon dioxide. Therefore, additional tests such as the dichromate test or Tollens' test may be needed to confirm the presence of a carboxylic acid group.

(please could you kindly mark my answer as brainliest and a follow would be really nice)

Answer:

It comes from the sun.

Option c) NH4Cl will hydrolyze when dissolved in water.

Hydrolysis is a chemical reaction that occurs when a salt reacts with water, resulting in the formation of an acidic or basic solution. In this case, when NH4Cl (ammonium chloride) is dissolved in water, it undergoes hydrolysis.

The ammonium ion (NH4+) is the conjugate acid of ammonia (NH3), which is a weak base. When NH4Cl dissociates in water, the ammonium ion reacts with water to form NH3 and H3O+ (hydronium ion). This process is called hydrolysis.

NH4+ (aq) + H2O (l) ↔ NH3 (aq) + H3O+ (aq)

The formation of NH3 leads to an increase in the concentration of hydroxide ions (OH-) in the solution, making it slightly basic. At the same time, the presence of H3O+ ions makes the solution slightly acidic. Therefore, the hydrolysis of NH4Cl results in a slightly acidic and slightly basic solution.

In contrast, salts like NaCl and KCl do not undergo hydrolysis when dissolved in water because they consist of cations (Na+ and K+) and anions (Cl-) that do not react with water to form acidic or basic species.

Na2SO4 (sodium sulfate) is also an example of a salt that does not undergo hydrolysis. The sulfate ion (SO42-) does not react with water to form acidic or basic species, so the solution remains neutral.

Therefore, among the given options, only NH4Cl undergoes hydrolysis when dissolved in water.

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The question is incomplete. Find the full content below:

Which one of the following salts when dissolved in water will hydrolyse?

a) NaCl

b) KCl

c) NH4Cl

d)Na2SO4

Answer:

pretty sure they're called groups

Answer:

The theoretical yield of urea = 120.35kg

The percent yield for the reaction = 72.70%

Explanation:

Lets calculate -

The given reaction is -

\(2NH_3(aq)+CO_2\) →\(CH_4N_2O(aq)+H_2O (l)\)

Molar mass of urea \(CH_4N_2O\)= 60g/mole

Moles of \(NH_3\) = \(\frac{62.8kg/mole}{17g/mole}\) (since \(Moles=\frac{mass of substance}{mass of one mole}\))

                     = 4011.76 moles

Moles of \(CO_2\) = \(\frac{105kg}{44g/mole}\)

                = \(\frac{105000g}{44g/mole}\)

                = 2386.36 moles

Theoritically , moles of \(NH_3\) required = double the moles of \(CO_2\)

    but , \(4011.76<2\times 2386.36\) , the limiting reagent is \(NH_3\)

Theoritical moles of urea obtained = \(\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3\)

                                                      = \(2005.88mole CH_4N_2O\)

Mass of 2005.88 mole of \(CH_4N_2O\) =\(2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}\)

                                                     = 120352.8g

                                                     \(120352.8g\times \frac{1kg}{1000g}\)

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \(\frac{87.5kg}{120.35kg}\times100\)

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

The molar ratio of NaOH:\(KHC_{8}H_{4}O_{4}\) after titration is 1:1.

How to find the molar ratio of compounds during titration?

Potassium hydrogen phthalate (KHP, \(KHC_{8}H_{4}O_{4}\)) is also a good primary standard. 20 mL of NaOH was titrated with 0.600 M \(KHC_{8}H_{4}O_{4}\) solution. The data were graphed and the equivalence point was found when 15.5 mL of the standard 0.600 M KHP solution was added.

The reaction equation is:

NaOH (aq) + \(KHC_{8}H_{4}O_{4}\)(aq) --> \(KNaC_{8}H_{4}O_{4}\)(aq) + \(H_{2}O\)(l).

Let's find the moles of NaOH first.

Molarity = moles/volume

Moles = molarity x volume

Moles of NaOH = 0.6 mol/L × 0.02 L = 0.012 mol \(KHC_{8}H_{4}O_{4}\)

and NaOH are present in equal amounts according to the balanced equation.

i.e, Moles of NaOH = moles of \(KHC_{8}H_{4}O_{4}\)

0.012 mol of \(KHC_{8}H_{4}O_{4}\) was used in the titration since 15.5 mL of a 0.6 mol/L \(KHC_{8}H_{4}O_{4}\) solution was required to reach the equivalence point.

The molar ratio of NaOH: \(KHC_{8}H_{4}O_{4}\) is 1:1 since they are present in the same quantity in the reaction.

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A basic amino acid has an R group that contains ( D) a carboxyl group.

Acid is a substance that has a pH level of lower than 7.0 and is capable of corroding or dissolving other substances. It is usually found in aqueous solutions and is a highly reactive substance. Examples of acid include sulfuric acid, hydrochloric acid, nitric acid and acetic acid. These are used in a variety of industries such as food production, industrial cleaning and chemical engineering. Acid is also used in the laboratory for titrations, pH testing and other experiments. Acids can be dangerous if mishandled and can cause skin, eye and respiratory irritation and even chemical burns.

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The rock layers are described as follows:

Igneous rocks: These rocks form when molten rock (magma or lava) cools and solidifies. They can be either intrusive, meaning that they formed below the earth's surface, or extrusive, meaning that they formed above the earth's surface. Examples of igneous rocks include granite, basalt, and obsidian.

Note that in the above case, It looks like this table is describing the relative ages and characteristics of different rock layers. The oldest rock layer is at the bottom and the youngest rock layer is at the top.

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The change in entropy (ΔS) when 4.4 g of carbon disulfide boils at -78.55°C is approximately 235.29 J/mol·K.

To calculate the change in entropy (ΔS) when 4.4 g of carbon disulfide boils at -78.55°C, we need to use the equation:

ΔS = ΔHv / T

where ΔHv is the heat of vaporization and T is the temperature in Kelvin.

First, let's convert the given temperature from Celsius to Kelvin:

T = -78.55°C + 273.15 = 194.6 K

Next, we calculate the number of moles of carbon disulfide:

moles = mass / molar mass

The molar mass of CS₂ is approximately 76.14 g/mol:

moles = 4.4 g / 76.14 g/mol = 0.0577 mol

Now, we can calculate the change in entropy:

ΔS = ΔHv / T

= 26.74 kJ/mol / 0.0577 mol / 194.6 K

= 235.29 J/mol·K

Therefore, the change in entropy (ΔS) when 4.4 g of carbon disulfide boils at -78.55°C is approximately 235.29 J/mol·K.

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Explanation:

1) 2-methyl prop-1-ene

2) 4-methyl hex-2-ene

3) 2-methyl pent-1-ene

4) 4-propyl hept-3-ene

5) 5-ethyl-4-methyl hept-2-ene

6) 6-methyl-4-propyl oct-1-ene

7) 4,5,6-trimethyl oct-2-ene

Many agricultural practices involve the emission of greater amounts of methane. The two agricultural practices which result in methane emission is manure management techniques and rice cultivation. The correct options are C and D.

The methane is defined as the simplest hydrocarbon with a chemical formula CH₄. It contains four 'H' atoms and one 'C' atom and it is the simplest alkane.

The main sources of methane emissions from agriculture are enteric fermentation, manure management, rice cultivation and residue burning. The production of methane also occurs animal wastes. Methane emissions occurs from paddy rice when soils are flooded.

Thus the correct options are C and D.

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The reason why CaCl2 is used and not NaCl in the preparation of macrocapsules is due to the difference in solubility. Calcium chloride is a salt that is soluble in water, whereas sodium chloride is also soluble in water, but less so than calcium chloride.

A macrocapsule is a type of capsule that is large enough to be seen with the unaided eye. It is also known as a "large capsule." Macrocapsules are usually used in the medical industry to deliver drugs or other substances to specific parts of the body. The substance to be delivered is typically contained within the capsule, which is then implanted into the body.

In order to prepare macro-capsules, a process known as microencapsulation is used. During this process, the substance to be encapsulated is suspended in a solution, and then this solution is mixed with a polymer. The polymer hardens around the substance, creating a capsule that can be implanted into the body.

In the preparation of macro-capsules, CaCl2 is used instead of NaCl because of its solubility. Calcium chloride is highly soluble in water, which makes it ideal for use in the microencapsulation process. The solubility of CaCl2 allows for the formation of a hard, impermeable capsule that is able to protect the substance inside from the surrounding environment. On the other hand, NaCl is less soluble in water than CaCl2, which makes it unsuitable for the microencapsulation process.

Other factors which make CaCl2  suitable for macrocapsule preparation include:

Gel formation: CaCl2 can participate in gel formation reactions with certain polymers or gelling agents. It can crosslink polymers, resulting in the formation of a stable gel structure, which can be useful for encapsulating materials and providing mechanical stability to the macro-capsules.

Compatibility: The specific material being encapsulated or the application of the macrocapsules may require compatibility with CaCl2 rather than NaCl. For example, certain biological or chemical processes may be more compatible with CaCl2 as a component of the encapsulation system.

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Answer:

Explanation:

A neutral titanium atom will have 22 electrons. Therefore, its electron configuration will be 1s2 2s2 2p6 3s2 3p6 4s2 3d2.

Answer: i dont think anyone know because its ghard

Explanation:

The mole ratio between sodium bicarbonate and that of acetic acid is 1:1

A mole ratio can be defined as that ratio in whole numbers between the amounts in moles of any two compounds involved in a balanced chemical reaction.

In conclusion, the mole ratio between sodium bicarbonate and that of acetic acid is 1:1

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Answer:

Volume = 10ml

Density = 1/5 g/ml or 0.20g/ml

Explanation:

The rocks are 10ml since the initial volume went up by 10.

Since density = mass/volume, you divide 2 by 10.

D = 2/10

D = 1/5 g/ml or 0.20g/ml

(Unit is g/ml aka grams/millileter)

Explanation:

The atoms of one element differs from the atoms of other elements in terms of the number of protons they contain. This is often taken as the atomic number of such an atom.

II.

It is possible to change the identity of an atom. This is only possible by altering the atomic number of the atom.

Only nuclear reactions have this capability.

When an atom undergoes nuclear reaction that involves change in number of protons, transmutation occurs and a new atom forms.

The rock containing 3 mg 40K. The half-life of 40K is 1.3 billion years. is approximately 3.0448172338 billion years old.

Half-life is a term used in nuclear physics to describe the time it takes for half of the atoms in a sample of a radioactive substance to decay. It is the amount of time required for the quantity of a substance to decrease by half its initial value.

Each radioactive substance has a unique half-life, which can range from fractions of a second to billions of years. The concept of half-life is important in many fields, including nuclear physics, geology, chemistry, and medicine, as it helps scientists understand how long it takes for radioactive materials to decay and how much radiation they emit over time.

The ratio of the remaining 40K to the initial amount of 40K is given by:

(3 mg) / (12 mg) = 0.25

Taking the natural logarithm of both sides of the equation and using the formula for the half-life, we get:

ln(0.25) = -0.693147 * t / (1.3 billion years)

Solving for t, we get:

t = -(1.3 billion years) * ln(0.25) / 0.693147

t ≈ 3.0448172338 billion years

Therefore, the rock is approximately 3.0448172338 billion years old.

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To calculate the grams of NaCl in a 100 g solution with water, when the solution is 19% NaCl by weight, we can use the formula:

Grams of NaCl = Total weight of solution (in grams) × Percentage of NaCl / 100

In this case, the total weight of the solution is 100 g and the percentage of NaCl is 19%. Plugging in these values:

Grams of NaCl = 100 g × 19 / 100 = 19 grams

Therefore, there are 19 grams of NaCl in the 100 g solution.

Regarding the chemical reaction equation, to balance it, we can use the coefficients to adjust the number of atoms on each side.

The equation is: ___SO2 + ___O2 -> ___SO3

The correct balanced equation is: 2SO2 + O2 -> 2SO3

The coefficients in this balanced equation indicate that we need 2 molecules of SO2, 1 molecule of O2, and 2 molecules of SO3 to balance the reaction.

B. To calculate the density of a substance, we use the formula:

Density = Mass / Volume

In this case, the mass of the gasoline is given as 20.2 kg and the volume is given as 23.7 liters.

Density = 20.2 kg / 23.7 L

Calculating this:

Density = 0.851 Kg/L

Rounding this value to the correct significant figures gives:

Density = 0.85 Kg/L

Therefore, the density of gasoline is approximately 0.85 kg/L.

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Answer:

Lead(IV) iodide

Explanation:

Answer:

Plasmid #50548

Explanation:

If the pH of a solution is one and it is diluted by 1*10^3 times, the pH of the resulting solution will be 4.

The pH of a solution is a measure of its acidity or basicity. It is defined as the negative logarithm of the concentration of hydrogen ions in the solution.

When a solution is diluted by a factor of 10, its pH increases or decreases by 1 depending on whether it is an acidic or basic solution, respectively. In this case, the solution has been diluted by a factor of 1*10^3, which means that its pH will increase by 3 units. Therefore, the pH of the resulting solution will be 4 (pH of the original solution + 3).

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Answer:

a) 3.16 m

b) 0.39 M

Explanation:

a)

The wavelength of a radio wave with a frequency of 95 x 10^6 Hz can be calculated using the formula:

wavelength = speed of light / frequency

The speed of light is approximately 3 x 10^8 m/s. Substituting the given values:

wavelength = (3 x 10^8 m/s) / (95 x 10^6 Hz)

wavelength ≈ 3.16 m (rounded to 2 significant figures)

Therefore, the wavelength of the radio wave is approximately 3.16 meters.

b)

The molarity (M) of a solution can be calculated using the formula:

Molarity = moles of solute / liters of solution

Substituting the given values:

Molarity = 0.44 mol / 1.12 L

Molarity ≈ 0.39 M (rounded to 2 significant figures)

Therefore, the molarity of the solution is approximately 0.39 M.

Answer:

Food chain, in ecology, the sequence of transfers of matter and energy in the form of food from organism to organism. Food chains intertwine locally into a food web because most organisms consume more than one type of animal or plant. ... In a predator chain, a plant-eating animal is eaten by a flesh-eating animal.

Explanation:

O2 will diffuse from alveoli into blood and CO2 will diffuse into alveoli from blood, if the partial pressure of oxygen in the alveoli is 104 mmhg and 40 mmhg in the capillaries surrounding the alveoli.

If a container is filled with multiple gases, each gas will exert pressure, which is known as partial pressure. Its partial pressure is the pressure of any gas contained within the container. The pressure that one gas in a mixture will exert if it takes up the same volume on its own is referred to as partial pressure. The tiniest air sacs in your lungs are called alveoli. The alveoli's job is to transport molecules of oxygen and carbon dioxide (CO2) into and out of your bloodstream.

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