Even if an object is sitting perfectly still and not moving, the atoms and molecules that the object is made of still have kinetic energy due to their constant motion
true or false?

To solve this problem, we'll use the Beer-Lambert Law and the information given.

(a) According to the Beer-Lambert Law, the absorbance (A) is directly proportional to the concentration (c) of the absorbing species:

A = εlc

where A is the absorbance, ε is the molar absorptivity constant, l is the path length (in cm), and c is the concentration (in M).

Given:

Slope of the Beer's Law plot (εl) = 3.5 × [tex]10^3[/tex]

Absorbance (A) = 0.700

Since we have a 1 cm path length, the slope (εl) is equal to the molar absorptivity constant (ε).

Substituting the values into the equation, we can calculate the equilibrium concentrations:

ε = [tex]3.5*10^3 M^{(-1)} cm^{(-1)}[/tex]

A = 0.700

Using the Beer-Lambert Law equation, we can rearrange it to solve for the concentration (c):

c = A / (εl)

For Fe3+:

c(Fe3+) = A / (εl) = 0.700 / (3.5 × [tex]10^3[/tex]) = 2 ×[tex]10^{(-4)}[/tex] M

For SCN-:

c(SCN-) = A / (εl) = 0.700 / (3.5 × [tex]10^3[/tex]) = 2 × 1[tex]0^{(-4)}[/tex] M

For FeSCN2+:

Since FeSCN2+ is the product of the reaction between Fe3+ and SCN-, the concentration of FeSCN2+ at equilibrium will be zero until the reaction reaches equilibrium.

Therefore, the equilibrium concentrations are:

[Fe3+] = 2 × [tex]10^{(-4)}[/tex] M

[SCN-] = 2 × [tex]10^{(-4)}[/tex] M

[FeSCN2+] = 0 M

(b) To calculate the equilibrium constant (Keq), we'll use the equation:

Keq = ([FeSCN2+]) / ([Fe3+][SCN-])

Substituting the given values:

Keq = (0) / ([tex](2 * 10^{(-4)})^2[/tex]) = 0

Therefore, Keq is equal to 0.

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Carbon-14 dating has limitations and loses its accuracy when the age of the sample exceeds 50,000 years. Therefore, option D, 40,000 years, is the closest to the correct answer, as it falls within this range.Carbon-14 (or radiocarbon) dating is a technique used to estimate the age of organic material that is up to 50,000 years old.

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Carbon-14 dating, also known as radiocarbon dating, is a scientific method used to determine the age of an object containing organic material by measuring the amount of carbon-14 remaining in it. Carbon-14 is a radioactive isotope of carbon that is formed naturally in the atmosphere and decays at a known rate over time.

As the carbon-14 decays, it turns into nitrogen-14. By measuring the ratio of carbon-14 to nitrogen-14 in a sample, scientists can determine how long ago the organic material died and ceased to take in carbon-14 from the atmosphere. However, there is a limit to the accuracy of carbon-14 dating.

This is because the amount of carbon-14 in the atmosphere is not constant over time. It has varied in the past due to factors such as solar activity and changes in the Earth's magnetic field. In addition, carbon-14 dating can only be used to date objects that contain organic material, so it cannot be used to date rocks or other inorganic substances. Carbon-14 dating loses its accuracy after about 40,000 years.

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Answer:

22

Explanation:

calcium has 20 proteins

and given the mass number is 42.

mass number = protons + neutrons

42= 20+ n

n= 42-20

n= 22

Answer:

if I'm not mistaken, it is a mixture

It's a mixture my dude

We have that from the Question"After careful analysis, an electromagnetic wave is found to have a frequency of 7.8 x 10^6 Hz. What is the speed of the wave?" it can be said that  the speed of the wave is

From the Question we are told

After careful analysis, an electromagnetic wave is found to have a frequency of 7.8 x 10^6 Hz. What is the speed of the wave?

Generally the equation for speed  is mathematically given as

[tex]V=f\lambda[/tex]

Therefore

The frequency doesn't affect the speed and as v in the equation is a constant

Therefore

the speed of the wave is

v=3*10^{-8}m/s

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The mass (in grams) of CaCO₃ required to react with 5.67 g of HCl as shown by the equation is 7.77 g

The mass of CaCO₃ required to react with 5.67 g of HCl can be obtain as follow:

CaCO₃ + 2HCl → CaCl₂ + H₂O

From the balanced equation above,

73 g of HCl reacted with 100 g of CaCO₃

Therefore,

5.67 g of HCl will react with = (5.67 × 100) / 73 = 7.77 g of CaCO₃

Thus, the mass of CaCO₃ required for the reaction is 7.77 g

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Answer:

An alloy

Explanation:

srry if its wrong but i hope this helps!

Answer:

An alloy

Explanation:

Part A: If the solution is NaCl(aq), the chloride ion concentration is simply equal to the concentration of the NaCl. This is because NaCl dissociates completely in water to produce Na+ and Cl- ions. Therefore, the concentration of chloride ions is equal to the concentration of NaCl.

For example, if the concentration of NaCl is 0.10 M, then the chloride ion concentration is also 0.10 M.Part B: If the solution is FeCl3(aq), we need to consider the fact that FeCl3 dissociates into Fe3+ and Cl- ions. Therefore, we need to calculate the concentration of chloride ions from the concentration of FeCl3.

To do this, we need to use the stoichiometry of the reaction. Each formula unit of FeCl3 produces three chloride ions. Therefore, the concentration of chloride ions is three times the concentration of FeCl3.For example, if the concentration of FeCl3 is 0.10 M, then the chloride ion concentration is 3 × 0.10 M = 0.30 M.

In summary, the chloride ion concentration in a solution depends on the concentration of the salt and the degree of dissociation.

For a salt that dissociates completely, the concentration of chloride ions is equal to the concentration of the salt. For a salt that does not dissociate completely, we need to consider the stoichiometry of the reaction to determine the concentration of chloride ions.

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Boiling point of the solution ≈ 100.9672°C

To determine the concentration of ethylene glycol in a solution of water in molality, we can use the equation:

ΔTf = kf * molality

Given that the freezing point dropped by 2.64°C and kf for water is 1.86°C/m, we can rearrange the equation to solve for molality:

molality = ΔTf / kf

Substituting the values, we get:

molality = 2.64°C / 1.86°C/m

molality ≈ 1.42 m

For the second question, to calculate the boiling point of the solution, we can use the equation:

ΔTb = kb * molality

Given that kb for water is 0.52°C/m, and the solution contains 10.0g of sodium chloride dissolved in 100.0g of water, we need to convert the masses to moles:

moles of sodium chloride = mass / molar mass
moles of sodium chloride = 10.0g / (22.99g/mol + 35.45g/mol) ≈ 0.186 mol

molality = moles of solute / mass of solvent (in kg)
molality = 0.186 mol / 0.1 kg = 1.86 m

Substituting the values into the equation, we get:

ΔTb = 0.52°C/m * 1.86 m

ΔTb ≈ 0.9672°C

The boiling point of the solution is the sum of the boiling point of pure water (100°C) and the ΔTb:

Boiling point of the solution = 100°C + 0.9672°C

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Answer:

C) 3

Explanation:

The compound written is called ;

[tex]NH_4NO_3[/tex]

In this compound , the elements present are;

The molecular formula of all the compounds is C9H18O2. The spectral data helps to differentiate the compounds. The spectral data reveals the carbon environment of each atom of the molecule.

The structure of a molecule can be predicted by analyzing NMR spectra and the type of carbon atom in the molecule by analyzing 13C NMR spectra.The NMR spectrum for Compound B shows a peak at δ 2.2, indicating the presence of an -CH2- group. In the 13C NMR spectrum for Compound B, a peak appears at δ 30.3, indicating the presence of a carbon atom with two hydrogens attached. In the molecule of Compound B, the only carbon atom with two hydrogens attached is the one that is part of the -CH2- group. As a result, Compound B is made up of a straight-chain carbon chain with a terminal carboxyl group and a single methyl group attached to the 3rd carbon atom.The NMR spectrum for Compound A shows peaks at δ 1.6, 2.3, and 2.8, indicating the presence of an -CH3 group, a -CH2- group, and a -CH- group, respectively. In the 13C NMR spectrum for Compound A, peaks appear at δ 15.3, 22.8, and 28.5, indicating the presence of three carbon atoms with one hydrogen each. In the molecule of Compound A, the carbon atoms with one hydrogen each are the three carbon atoms that make up the -CH3 group, the -CH2- group, and the -CH- group. As a result, Compound A is made up of a branched carbon chain with a terminal carboxyl group and a single methyl group attached to the 4th carbon atom.

Compound A is made up of a branched carbon chain with a terminal carboxyl group and a single methyl group attached to the 4th carbon atom. Compound B is made up of a straight-chain carbon chain with a terminal carboxyl group and a single methyl group attached to the 3rd carbon atom. Therefore, the NMR and 13C NMR spectra are consistent with Compounds A and B, respectively, and their molecular formulas are C9H18O2.

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Answer:

Temperature

Explanation:

The given compound has the empirical formula NO2, and the molecular molar mass is 92 g/mol. The empirical formula of a compound is the simplest formula that represents the ratio of elements in the compound. To find the molecular formula, we need to determine the number of empirical formula units in the compound.

For this purpose, we need to calculate the molecular formula mass by multiplying the empirical formula mass by a factor "x." The value of 'x' can be determined by dividing the molecular molar mass by the empirical formula mass. Let's solve this problem.

Molecular formula mass = (empirical formula mass) x n

Where n is a whole number, and the empirical formula of the compound is NO2

Empirical formula mass of NO2 = (14.01 + (2 x 16.00))

= 46.01 g/mol

Number of empirical formula units in the compound = Molecular molar mass / Empirical formula mass

x = Molecular formula mass / Empirical formula mass

x = 92 / 46.01

x = 2Molecular formula = Empirical formula x n

x = (NO2)2

= N2O4

Therefore, the molecular formula of the given compound is N2O4.

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At temperature of 100 °C, the vapor pressure of the aqueous solution of Ca(NO3)2 is approximately 1.002 atm.

To find the vapor pressure of an aqueous solution of Ca(NO3)2 at a given temperature, we can use the concept of boiling point elevation and the equation for the Clausius-Clapeyron equation. The boiling point elevation is related to the molality of the solution.

The equation for the boiling point elevation is:

ΔTb = Kbm

where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for the solvent (water), b is the molality of the solution, and m is the number of dissociated particles.

For Ca(NO3)2, it dissociates into three ions: Ca2+ and two NO3-. So, m = 3.

Given that the molality of the solution is 0.465 mol/kg and the boiling point elevation constant for water is approximately 0.512 °C/m, we can calculate the boiling point elevation as follows:

ΔTb = (0.512 °C/m) * (0.465 mol/kg) * (3)

ΔTb ≈ 0.710 °C

Now, we need to determine the new boiling point by adding the boiling point elevation to the normal boiling point of water, which is 100 °C:

New boiling point = 100 °C + 0.710 °C = 100.710 °C

Finally, using the Clausius-Clapeyron equation, we can calculate the vapor pressure of water at the new boiling point:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

Where P1 is the vapor pressure of water at the normal boiling point (1 atm), P2 is the vapor pressure of water at the new boiling point, ΔHvap is the enthalpy of vaporization of water, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T1 is the absolute temperature at the normal boiling point, and T2 is the absolute temperature at the new boiling point.

Assuming ΔHvap is constant, we can simplify the equation to:

ln(P2/1) = ΔHvap/R * (1/373 K - 1/373.71 K)

Solving for P2:

P2/1 = e^(ΔHvap/R * (1/373 K - 1/373.71 K))

P2 ≈ 1.002 atm

Therefore, at a temperature of 100 °C, the vapor pressure of the aqueous solution of Ca(NO3)2 is approximately 1.002 atm.

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Answer:

gas is this: a substance (such as oxygen or hydrogen) that is like air and has no fixed shape. : a gas or mixture of gases that is burned as a fuel.

Explanation:

Gas is one of the four fundamental states of matter. A pure gas may be made up of individual atoms, elemental molecules made from one type of atom, or compound molecules made from a variety of atoms. A gas mixture, such as air, contains a variety of pure gases.

Answer:

Answer is 'B' I think

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

Answer:

I think chemical reaction

Explanation:

Answer:

Explanation:

X-ray

The higher the pH value, the lower the acidity of the solution. This is observed when comparing the pH values of two solutions produced by the same concentration of ammonia (NH3) and ethylamine (C2H7N). C2H7N has a stronger basicity than NH3, leading to a higher pH value.

The higher the pH value, the lower the acidity of the solution. In other words, the solution's base is stronger. This will be observed when comparing the pH values of the two solutions produced by the same concentration of ammonia (NH3) and ethylamine (C2H7N). The Kb of NH3 is 1.8 x 10-5 m and the Kb of C2H7N is 5.1 x 10-4 m, respectively. It's worth noting that NH3 is weaker than C2H7N. This implies that the base of C2H7N is stronger than that of NH3. This will make the pH of the solution produced by C2H7N greater than that of the solution produced by NH3 at the same concentration.  C2H7N has a stronger basicity than NH3. As a result, a solution with a higher pH would be formed by the former. As we compare the pH values of the solutions produced by the two compounds, the solution produced by C2H7N would have a higher pH value than the solution produced by NH3 at the same concentration.

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Answer:

It is 15N in that direction. When it's in the exact same direction, you combine them.

Explanation:

(i) The initial reaction rate is [tex]2*10^{-4} mol litre^{-1} sec^{-1.[/tex]

(ii) The half-life period of the reaction is 1386 seconds.

(iii) The time taken to complete 75% of the reaction is approximately 2772 seconds.

We can use the first-order rate equation:

Rate = k[N2O5]

Where:

Rate is the reaction rate,

k is the rate constant,

[N2O5] is the concentration of N2O5.

Given:

Rate constant (k) = [tex]5*10^{-4} sec^{-1}[/tex]

Initial concentration of N2O5 =[tex]0.4 mol litre^{-1}[/tex]

(i) To find the initial reaction rate:

Substitute the given values into the rate equation:

Rate = k[N2O5]

Rate = [tex](5*10^{-4} sec^{-1})(0.4 mol litre^{-1})[/tex]

Rate = [tex]2*10^{-4} mol litre^{-1} sec^{-1}[/tex]

The initial reaction rate is [tex]2*10^{-4} mol litre^{-1} sec^{-1}[/tex].

(ii) To find the half-life period:

The half-life of a first-order reaction is given by the equation:

t(1/2) = (0.693 / k)

Substitute the given value of k into the equation:

t(1/2) = [tex](0.693 / 5*10^{-4} sec^{-1})[/tex]

t(1/2) = 1386 sec

The half-life period of this reaction is 1386 seconds.

(iii) To find the time taken to complete 75% of the reaction:

The time required to complete a certain percentage of a reaction can be found using the equation:

t = (ln(1 / (1 - x)) / k)

Where x is the fraction of the reaction completed (in this case, 75%).

Substitute the given values into the equation:

t =[tex](ln(1 / (1 - 0.75)) / 5*10^{-4} sec^{-1})[/tex]

t = 2772 sec

The time taken to complete 75% of the reaction is approximately 2772 seconds.

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Answer:

Volume of gas  is 24.7×10⁴ L

Explanation:

Given data:

Moles of neon gas = 1.1×10⁴ mol

Temperature of gas = standard = 273 K

Pressure of gas = standard = 1 atm

Volume of gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × V  = 1.1×10⁴ mol ×0.0821 atm.L/ mol.K  × 273 K

V = 24.7×10⁴ atm.L / 1 atm

V = 24.7×10⁴ L

2-chloro-3-methyl-2-butene exhibits three different ¹H NMR signals.

To determine the number of ¹H NMR signals exhibited by 2-chloro-3-methyl-2-butene, we need to analyze the different types of hydrogen (proton) environments in the molecule.

The structure of 2-chloro-3-methyl-2-butene can be represented as follows:

CH3-CH(Cl)-C(CH3)=CH2

Let's examine the different proton environments:

The methyl group (CH3) adjacent to the chlorine atom is chemically equivalent to the other methyl group attached to the double bond. These two methyl groups are part of the same functional group and should produce only one NMR signal. Therefore, we have one signal for these protons.

The hydrogen attached to the carbon adjacent to the chlorine atom is different from the other hydrogens in the molecule. This hydrogen is allylic to the double bond and exhibits different chemical shifts due to its proximity to the pi electrons. Thus, we have a separate signal for this hydrogen.

The hydrogens on the carbon atoms in the double bond are also chemically different from the rest of the hydrogens in the molecule. However, they are part of the same functional group and should produce only one NMR signal.

Taking all these proton environments into account, 2-chloro-3-methyl-2-butene exhibits a total of three ¹H NMR signals.

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Answer:

Let us assume the balanced chemical equation of H2 and I2 forming HI is as follows: H2(g) + I2(g)  2HI(g) in the chemical reaction at equilibrium is said to be constant.

Hence, we can write the expression for the equilibrium constant (Kc) of the above chemical reaction as follows:

Kc = [HI]2 / [H2][I2] We know the concentration of HI is 6M, and we have to find out the molarity of H2 present. Now that we know the concentration of I2, we can use the expression above to solve for the molarity of H2. However, we don't have a concentration of I2.

Therefore, we must use the balanced chemical reaction to find the molarity of H2 as follows:

3H2(g) + 3I2(g) → 6HI(g) 2H2(g) + I2(g) → 2HI(g)

Since 2 mol of HI are formed from 1 mol of I2, if we use up all the I2, we will produce 4 mol of HI using 2 mol of H2. Therefore, we have:

2H2(g) + I2(g) → 4HI(g)At equilibrium, we have a total of 6 M of HI and 0.5 M of I2. Since I2 is a limiting reagent, we must use 0.5 M to solve for the molarity of H2 as follows:

2H2(g) + I2(g) → 4HI(g) 2x            0.5        4x x = 0.5 / 2 = 0.25 M

Therefore, the molarity of H2 is 0.25 M.

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Answer:

there should be around 41 moles

Answer: 39.99711 moles

The amount of excess reactant remaining when 59.3 g of Na2SO4 reacts with 75.0 g of AgNO3 is 29.49 grams of Na2SO4.

To determine the amount of excess reactant remaining in the given reaction between Na2SO4 and AgNO3, we need to follow these steps:

Step 1: Convert the given masses of Na2SO4 and AgNO3 to moles.

The molar mass of Na2SO4 is:

2(Na) + 1(S) + 4(O) = 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.04 g/mol

Number of moles of Na2SO4 = 59.3 g / 142.04 g/mol = 0.417 moles

The molar mass of AgNO3 is:

1(Ag) + 1(N) + 3(O) = 107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 169.87 g/mol

Number of moles of AgNO3 = 75.0 g / 169.87 g/mol = 0.442 moles

Step 2: Determine the stoichiometric ratio between Na2SO4 and AgNO3.

From the balanced equation:

Na2SO4 + 2AgNO3 ---> Ag2SO4 + 2NaNO3

The stoichiometric ratio between Na2SO4 and AgNO3 is 1:2. This means that for every 1 mole of Na2SO4, we need 2 moles of AgNO3 to react completely.

Step 3: Determine the limiting reactant.

To determine the limiting reactant, we compare the number of moles of each reactant to their stoichiometric ratio.

For Na2SO4: 0.417 moles * (2 moles AgNO3 / 1 mole Na2SO4) = 0.834 moles AgNO3 required

Since we have 0.442 moles of AgNO3, which is greater than 0.834 moles required, AgNO3 is the limiting reactant.

Step 4: Calculate the excess reactant.

To determine the excess reactant, subtract the moles required for the limiting reactant from the actual moles of the excess reactant.

Excess moles of Na2SO4 = 0.417 moles - (0.417 moles * (1 mole Na2SO4 / 2 moles AgNO3)) = 0.208 moles

Step 5: Convert the excess moles of Na2SO4 to grams.

Mass of excess Na2SO4 = Excess moles of Na2SO4 * Molar mass of Na2SO4

Mass of excess Na2SO4 = 0.208 moles * 142.04 g/mol = 29.49 g

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The amount of solute (osmoles) in each compartment is as follows: 11.4 osmoles in total body water, 3.705 osmoles in extracellular fluid (ECF), 7.56 osmoles in intracellular fluid (ICF), and 0.798 osmoles in plasma.

To calculate the osmolarity of the patient's body fluids, we can use the freezing point depression equation:

ΔT = Kf × m × i

Where ΔT is the freezing point depression, Kf is the cryoscopic constant (1.86°C for an ideal solution), m is the molality of the solution, and i is the van't Hoff factor (the number of particles per molecule of solute).

Given that the patient's plasma shows a freezing point depression of 0.53°C, we can rearrange the equation to solve for the molality:

m = ΔT / (Kf × i) = 0.53°C / (1.86°C/molal × 1)

Converting °C to K (Kelvin), we get:

m = 0.53 K / (1.86 K/molal) = 0.285 mol/kg

Since osmolarity is expressed in osmoles per liter (mol/L), we can convert the molality to osmolarity by multiplying by the density of water (1 kg/L):

Osmolarity = 0.285 mol/kg × 1 kg/L = 0.285 osmol/L = 285 mOsM

Therefore, the osmolarity of the patient's body fluids is 285 mOsM.

To determine the amount of solute (osmoles) in each compartment, we multiply the osmolarity by the volume of distribution for each compartment.

For total body water:

Osmoles in total body water = Osmolarity × Volume of distribution of total body water = 285 mOsM × 40 L = 11.4 osmoles

For extracellular fluid (ECF):

Osmoles in ECF = Osmolarity × Volume of distribution of ECF = 285 mOsM × 13 L = 3.705 osmoles

For intracellular fluid (ICF):

Osmoles in ICF = Osmolarity × Volume of distribution of ICF = 285 mOsM × (40 L - 13 L) = 7.56 osmoles

For plasma:

Osmoles in plasma = Osmolarity × Volume of distribution of plasma = 285 mOsM × 2.8 L = 0.798 osmoles

Therefore, the amount of solute (osmoles) in each compartment is as follows: 11.4 osmoles in total body water, 3.705 osmoles in ECF, 7.56 osmoles in ICF, and 0.798 osmoles in plasma.

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Answer:

Cerium

Explanation:

The equation to determine the number of hours required for plating 10.5 kg of copper onto the cathode, assuming that copper in the electrolytic solution is present as Cu²⁺ is given by;t = (m * z) / (I * F)

Where; t is the time taken for plating in hours, m is the mass of the substance to be plated in grams, z is the number of electrons transferred per molecule of the substance,

I is the current in amperes and F is the Faraday constant  F = 96500 C mol⁻¹Cu²⁺ + 2e⁻ → Cu(s)From the balanced equation, we can see that z = 2

.As such, substituting the values given in the question into the formula gives ; t = (m * z) / (I * F) = (10500 * 2) / (44.5 * 96500)= 0.051 hours (to 3 significant figures)

Therefore, it will take approximately 0.051 hours to plate 10.5 kg of copper onto the cathode.

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Answer:

Since electromagnetic forces are responsible for regular chemical bonds, every chemical compound would dissolve. All life and all ordinary objects would cease to exist. I think your question would call on electrons and protons to cease to exist, since these particles are associated with the electromagnetic force.

Explanation: