You want to make a 50 mL SATURATED solution of potassium chloride at 40 degrees Celsius. How many grams of potassium chloride do you need?

Answer: 6.11 L

Explanation:

STP= 1atm, 273.15K

Molar mass of CO2=44.01g/mol so n= (12.0/44.01)

PV=nRT

V=(nRT)/P

V=((12.0/44.01)(0.0821)(273.15))/1

V=6.11L

Answer:

The Correct answer is option 3

Step by Syep Explanation:

4Fe+3O2---->rust

formula for rust----->Fe2O3

4Fe+3O2---->Fe2O3

Balancing the Chemical Equation

both the reactant and product side

we have that;

4Fe+3O2------->2FeO3

the equation is Chemically Balanced

therefore 4Fe+3O2------->2×rust

Answer:

Explanation:

Given that,

Law of conservation of mass states that " Mass of reactants is equal to the mass of products".

Also we know that In a balanced equation the total number of atoms in the reactants equals the total number of atoms in the product.

We are given with 4Fe + 3O₂ i.e the reactant.

First Let's calculate the number of atoms in the reactant.

Now, Let's find the product .

Also, We can see 2Fe₂O₃ (Product)

4Fe + 3O₂ → 2Fe₂O₃

Number of atoms in the reactants = the total number of atoms in the product.

Therefore, 2Fe₂O₃ (Option 3) will the required answer .

Johannes Brsted and Thomas M. Lowry, two chemists, identified the Bromsted-Lowry acids and bases as a particular kind of acid-base reaction in 1923.

Acids are substances that give a base a proton (H+), whereas bases are substances that take a proton from an acid. In a Bromsted-Lowry acid-base reaction, the acid gives the base a proton in order to create the conjugate base and the conjugate acid, two new compounds.

Nitric acid (HNO3), sulfuric acid (H2SO4), and hydrochloric acid (HCl) are a few examples of acids. Sodium hydroxide (NaOH), ammonium hydroxide (NH4OH), and potassium hydroxide (KOH) are a few examples of bases.

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The reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) has an equilibrium constant of 0.3698.

The equilibrium constant (Kb) for the reaction can be calculated using the Ka and Kb values of the reactants and the equation:

Kw = Ka x Kb

where Kw = ion product constant of water (1.0 x 10⁻¹⁴ at 25°C).

Calculate the Kb value for aniline:

Kb = Kw/Ka = (1.0 x 10⁻¹⁴)/(4.3 x 10⁻¹⁰) = 2.33 x 10⁻⁵

Use the Kb value for aniline and the Ka value for benzoic acid to calculate the equilibrium constant (K) for the reaction:

K = Kb/Ka = (2.33 x 10⁻⁵)/(6.3 x 10⁻⁵) = 0.3698

Therefore, the equilibrium constant for the reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) is 0.3698.

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The ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

The relationship between ΔG and K is given by the following equation:

ΔG = -RTln(K)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 °C = 298.15 K), and ln denotes the natural logarithm.

To calculate K, we need to use the equilibrium expression and the given concentrations:

[tex]K = [C]/([A][B])[/tex]

[tex]K = (5.00 * 10^{-5} M)/((1.50 M)(1.00 M))[/tex]

[tex]K = 3.33 x 10^{-5}[/tex]

Now we can substitute the values for R, T, and K into the equation for ΔG:

ΔG = -RTln(K)

ΔG = [tex]-(8.314 J/(mol.K))(298.15 K)ln(3.33 x 10^{-5})[/tex]

ΔG = -8.35 kJ/mol

Therefore, the ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

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To find the pH of the solution, we need to first calculate the concentration of H+ ions in the solution using the following equation:

[H+] = (moles of HCl) / (volume of solution in liters)

First, let's convert the mass of HCl to moles:

moles of HCl = mass / molar mass = 4.0 g / 36.46 g/mol = 0.1096 moles

Next, let's convert the volume to liters:

475 mL = 0.475 L

Now we can calculate the concentration of H+ ions:

[H+] = 0.1096 moles / 0.475 L = 0.2306 M

Finally, we can calculate the pH using the equation:

pH = -log[H+]

pH = -log(0.2306) = 0.637

Therefore, the pH of the solution is approximately 0.637.

a. The mass of Cr2O3 is 0.559 g Cr2O3 is the maximum amount of  Cr2O3 produced.

b.The limiting reactant is  Cr(NO3)3  because it produces less moles of Cr2O3 than Na2O.

c. The percent yield is 84%.

The balanced equation is shown below:

2 Cr(NO3)3 + 3 Na2O → Cr2O3 + 6 NaNO3

moles of Cr(NO3)3 = 1.75 g / 238.01 g/mol = 0.00735 mol

moles of Na2O = 1.75 g / 61.98 g/mol = 0.0282 mol

moles of Cr2O3 = (0.00735 mol Cr(NO3)3) × (1 mol Cr2O3 / 2 mol Cr(NO3)3) = 0.00368 mol Cr2O3 (theoretical yield)

mass of Cr2O3 = (0.00368 mol Cr2O3) × (151.99 g/mol) = 0.559 g Cr2O3

The percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (0.455 g / 0.559 g) × 100% = 81.4%

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We need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

To solve this problem, we need to use the following information:

First, we need to calculate the total amount of CO2 that will be expelled during the mission:

Total CO2 = 6 crew members x 42.0 g CO2/hour x 24 hours/day x 18 days = 136,080 g CO2

Next, we need to calculate the amount of LIOH needed to absorb this CO2. The balanced chemical equation for the reaction between CO2 and LIOH is:

CO2 + 2 LIOH → Li2CO3 + H2O

The molar mass of CO2 is 44.01 g/mol, and the molar mass of LIOH is 23.95 g/mol.

This means that 2 moles of LIOH are needed to absorb 1 mole of CO2.

So, to absorb 136,080 g of CO2, we need:

136,080 g CO2 x (1 mol CO2/44.01 g) x (2 mol LIOH/1 mol CO2) x (23.95 g LIOH/1 mol) = 139,648 g LIOH

Since each CCC contains 750 g of LIOH, we need:

139,648 g LIOH / 750 g CCC = 186.2 CCCs

Therefore, we need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

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The pressure of a gas that occupies 50.5L at 69.0°C is 6.95 atm.

The pressure of an ideal gas can be calculated using Avogadro's equation as follows;

PV = nRT

Where;

According to this question, 12.5 mol of an ideal gas occupies 50.5 L at 69.00°C. The pressure can be calculated as follows:

P × 50.5 = 12.5 × 0.0821 × 342

50.5P = 350.9775

P = 6.95 atm

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Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

Answer:

Atomic number of sodium is 11

Electronic configuration of a sodium atom :

Since sodium has one electron in its outermost shell, Therefore, sodium can easily donate it's one electron. As the result it becomes sodium cation with + 1 charge.

Electronic configuration of a sodium cation,[tex] \: \sf ({Na}^{+1}) [/tex]

In case of sodium cation, it has fully filled electronic configuration.

Cations - Atoms that carry postive charge are called cations. Cations are formed when an atom loses its electron.

For example : [tex]\sf {Na}^{+} [/tex]

Anions - Atoms that carry negative charge are called anions. Anions are formed when an atom gains a electron.

For example : [tex]\sf {Cl}^{-} [/tex]

A mass of 37.0 g of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP.

The molar ratio of S₈ to SO₂ is 1:8.

At STP, one mole of gas occupies 22.4 L. Therefore, 25.7 L of SO₂ gas will contain;

25.7 L / 22.4 L/mol = 1.15 mol of SO₂.

The number of moles of S₈ needed is calculated as;

= 1.15 mol SO₂ / 8 mol S₈ per 1 mol SO₂

= 0.144 mol S₈.

The mass of S₈ needed is calculated as;

0.144 mol S₈ × 256.6 g/mol = 37.0 g of S₈.

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The intital common temperature of copper and water is 9.5°C, under the condition that 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter.

Now to evaluate the initial common temperature of the water and copper calorimeter, we have to apply the formula
m1c1(Tk - Ti) + m2c2(Tk - Ti)
= mcopperccopper(Tk - Ti)

Here,
m1 = mass of water,
c1 =specific heat capacity of water,
m2 = mass of copper calorimeter,
c2 = specific heat capacity of copper calorimeter, mcopper = mass of copper block
ccopper =specific heat capacity of copper.

Here, this equation to evaluate Ti
Ti = (m1c1Tk + m2c2Tk - mcopperccopperTk - m1c1Ti - m2c2Ti) / (m1c1 + m2c2 - mcopperccopper)

Staging the given values into this equation
Ti = (-300.0 g)(4.18 J/g°C)(18.0°C) + (200.0 g)(0.385 J/g°C)(18.0°C) + (10.0 g)(0.385 J/g°C)(18.0°C) / [(300.0 g)(4.18 J/g°C) + (200.0 g)(0.385 J/g°C) - (10.0 g)(0.385 J/g°C)]
Ti = 9.5°C

Hence, the initial common temperature of the water and copper calorimeter was 9.5°C.
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According to the information, we can infer that the difference between photographs 1 and 2 originate from the translation of the Moon around the earth (option C).

To explain the difference between both images we must take into account the movement patterns of the earth and the moon. In the case of the earth, it has 2 main movements, which are rotation on its own axis and translation around the sun.

On the other hand, the moon has a translational movement around the earth, which is what causes the different lunar phases. This motion causes the moon to appear partially shadowed from the earth because the earth blocks the sunlight.

Based on the above, we can infer that the correct answer is option C because this phenomenon is caused by the translation of the moon.

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Answer: As they develop theories, chemists use models to attempt to explain their findings. Chemists assess the model they are using as new evidence becomes available and, if required, continue to refine it by making modifications.

Explanation:

Explanation:

The temperature change of a substance when it absorbs or loses energy can be calculated using the specific heat capacity of the substance. The specific heat capacity of water is approximately 4.18 J/(g°C), which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

To calculate the temperature change of the water sample when 50 joules of energy is added, we need to use the following equation:

q = m * c * ΔT

where q is the amount of energy absorbed by the water, m is the mass of the water sample, c is the specific heat capacity of water, and ΔT is the resulting temperature change.

Rearranging the equation to solve for ΔT, we get:

ΔT = q / (m * c)

Plugging in the values, we get:

ΔT = 50 J / (m * 4.18 J/(g°C))

We need to know the mass of the water sample to calculate the temperature change. Let's assume a mass of 10 grams:

ΔT = 50 J / (10 g * 4.18 J/(g°C))

ΔT = 1.2°C

Therefore, if 50 joules of energy is added to a 10-gram sample of water, the resulting temperature change will be approximately 1.2 degrees Celsius.

The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:

2Ag2O(s) → 4Ag(s) + O2(g)

We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:

0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O

Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:

0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2

1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:

0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL

Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

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The theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

To calculate the theoretical yield of MgO, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO):

2Mg + O₂ → 2MgO

According to the stoichiometry of this equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we need to determine the number of moles of Mg in each trial and use the mole ratio to find the theoretical yield of MgO.

For Trial 1:

The mass of Mg used is: 26.682 g - 27.012 g = 0.330 g

The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:

0.330 g / 24.31 g/mol = 0.0136 mol Mg

According to the balanced equation, 2 moles of Mg produce 2 moles of MgO, so the theoretical yield of MgO is:

0.0136 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.348 g MgO

For Trial 2:

The mass of Mg used is: 26.987 g - 26.695 g = 0.292 g

The number of moles of Mg is:

0.292 g / 24.31 g/mol = 0.0120 mol Mg

The theoretical yield of MgO is:

0.0120 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.307 g MgO

To calculate the percent yield of MgO, we need to use the following formula:

Percent yield = (actual yield / theoretical yield) x 100%

For Trial 1:

The actual yield of MgO is: 27.214 g - 27.012 g = 0.202 g MgO

The percent yield of MgO is:

(0.202 g / 0.348 g) x 100% = 58.0%

For Trial 2:

The actual yield of MgO is: 27.183 g - 26.695 g = 0.488 g MgO

The percent yield of MgO is:

(0.488 g / 0.307 g) x 100% = 159.2%

To calculate the average percent yield of MgO for the two trials, we add the percent yields and divide by 2:

Average percent yield = (58.0% + 159.2%) / 2 = 108.6%

Therefore, the theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

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Answer:

V ≈ 2144.66 m³

Explanation:

Volume of sphere formula is:

V = 4/3 πr³

Radius is half the diameter so we divide the given diameter, 16 by 2 to get 8, the radius. Now we can solve

V = 4/3 π (8)³

V = 4/3 (512π)

V = 2048/3 π

V ≈ 2144.66 m³

Answer:

4/3 x π

Explanation:

The results of the lab activity showed that the larger the mass of the sun, the more likely at least one planet will fall into the habitable zone.

The mass of the sun affects the orbits of planets in a solar system. When the mass of the sun is larger, the gravitational force between the sun and the planets is stronger, causing the planets to move at a slower pace around the sun.

Conversely, when the mass of the sun is smaller, the gravitational force is weaker, causing the planets to move at a faster pace.

Additionally, when Earth is closer to the sun, the gravitational force is stronger, causing its orbit to become faster, while a farther distance from the sun results in a slower orbit.

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To calculate the mass of KOH needed to make a 5 M solution in 185.5 mL, we need to use the formula:

mass = moles × molar mass

where moles is the amount of KOH in moles and molar mass is the mass of one mole of KOH.

We can calculate the moles of KOH as follows:

moles = Molarity × Volume (in liters)

First, we need to convert the volume from milliliters to liters:

185.5 mL = 0.1855 L

Now we can calculate the moles of KOH:

moles = 5 M × 0.1855 L = 0.9275 moles

The molar mass of KOH is 56.11 g/mol. Therefore, the mass of KOH needed is:

mass = 0.9275 moles × 56.11 g/mol = 52.05 g

Therefore, 52.05 grams of KOH are needed to make a 5 M solution in 185.5 mL.

a. We will be able to release 37,827 kJ. b. You can cook a maximum of 19 hamburgers without exceeding the limit of 0.750 kg of carbon dioxide.

a. We need to use the balanced chemical equation and the enthalpy change of the reaction.

Therefore, moles  [tex]CO_2[/tex] produced are[tex]0.750 kg / 44.01 g/mol = 17.03 mol.[/tex]

The enthalpy change of the reaction is -2220.1 kJ/mol. Thus, the maximum number of kilojoules that can be released is:

[tex]\Delta Hrxn * moles of[/tex] [tex]CO_2[/tex] = [tex]-2220.1 kJ/mol * 17.03 mol = -37,827 kJ[/tex]

We need to reverse the sign of the answer, giving us 37,827 kJ.

b. If it requires 1900 kJ to cook one hamburger, we can divide the maximum number of kilojoules that can be released by the energy required to cook one hamburger:

37,827 kJ / 1900 kJ/hamburger = 19.91 hamburgers

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Question #3: 0.8 g of NaHCO3 mass NaCI weight: 0.4 g 0.8 g/84 g/mol, or 0.0095 moles, of NaHCO3 0.4 g/58.5 g/mol = 0.0068 moles of NaCI are the moles.

The reaction's mole to mole ratio and the ratio in the balanced equation (1:1) are in agreement. The highest quantity of NaCI that may be produced from this reaction is 0.0095 moles since NaHCO3 is the limiting reactant.

The theoretical yield of NaCI is 0.0095 moles, which is question #6. The finished product weighs 0.4 g. The percent yield is 0.4 g/0.0095 moles times 100, which is 42.1%.

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The mass of silver oxide needed to decompose in order to produce 120.6 grams of silver is 494.5 grams.

The balanced chemical equation for silver oxide breakdown is:

[tex]Ag_2O[/tex] → [tex]2 Ag[/tex] + [tex]1/2 O_2[/tex]

The equation shows that for every mole of silver oxide that decomposes, two moles of silver are created, and the molar mass of [tex]Ag_2O[/tex] is 231.74 g/mol.

Hence, using stoichiometry, we can calculate the quantity of silver oxide necessary to generate 120.6 grams of silver:

120.6 g Ag × (1 mol Ag / 107.87 g Ag) × (1 mol [tex]Ag_2O[/tex]/ 2 mol Ag) × (231.74 g [tex]Ag_2O[/tex] / 1 mol [tex]Ag_2O[/tex] )

= 494.5 g [tex]Ag_2O[/tex]

As a result, 494.5 grams of silver oxide is needed to decompose in order to produce 120.6 grams of silver.

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Answer:

we need to know the definitions of the two types of mutations:

Looking at the definitions, we can see that a chromosomal mutation can affect many genes at once, while a gene mutation can affect only one or a few nucleotides. Therefore, a chromosomal mutation could have the most drastic effect on a gene, because it could alter or delete an entire gene or multiple genes, resulting in major changes in the phenotype or function of an organism. A gene mutation could also have significant effects on a gene, but it could also be silent or minor depending on the location and type of the mutation. Therefore, the answer is a chromosomal mutation. One possible way to back up this choice is to give an example of a chromosomal mutation that causes a genetic disorder, such as Down syndrome or Turner syndrome.

The energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom is approximately -2.17 × 10⁻¹⁸ Joules.

To calculate the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom, we can use the following equation:

deltaE = -13.6 * (1/n_final^2 - 1/n_initial^2) eV
where n_initial is the initial energy level (6 in this case), n_final is the final energy level (1 in this case), and -13.6 eV is the ionization energy of hydrogen.

Converting eV to Joules, we get:
1 eV = 1.602 x 10^-19 J
Therefore, deltaE in Joules can be calculated as follows:
deltaE = -13.6 * (1/1^2 - 1/6^2) * 1.602 x 10^-19 J/eV
deltaE = -2.179 x 10^-18 J

Therefore, the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom is -2.179 x 10^-18 J.

To calculate the energy difference (ΔE) for an electron's transition from n = 6 to n = 1 in a hydrogen atom, you can use the following formula:
ΔE = -13.6 eV * (1/nf² - 1/ni²)

Where ΔE is the energy difference in electron volts (eV), nf is the final energy level (1 in this case), and ni is the initial energy level (6 in this case).
ΔE = -13.6 eV * (1/1² - 1/6²)
ΔE ≈ -13.56 eV

Now convert electron volts to Joules:
1 eV = 1.6 × 10⁻¹⁹ J
ΔE ≈ -13.56 eV * 1.6 × 10⁻¹⁹ J/eV
ΔE ≈ -2.17 × 10⁻¹⁸ J

So,  approximately -2.17 × 10⁻¹⁸ Joules is  the energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom.

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The number of molecules of BF₃ present in 2 grams of BF₃ is 1.776×10²² molecules (1st option)

We'll begin by calculating the number of mole of 2 grams of BF₃. Details below:

Mole = mass / molar mass

Mole of BF₃ = 2 / 67.81

Mole of BF₃ = 0.02949 mole

Finally, we shall determine the number of molecules of BF₃. This is shown below:

Avogadro's hypothesis suggest that:

1 mole of BF₃ = 6.022×10²³ molecules

Therefore,

0.02949 mole of BF₃ = 0.02949 × 6.02×10²³

0.02949 mole of BF₃ = 1.776×10²² molecules

Thus, the number of molecules of BF₃ is 1.776×10²² molecules (1st option)

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B. To plot the data on a bar chart, draw a horizontal axis for metals and a vertical axis for time to complete the reaction. Then, draw bars for each metal that represent the amount of time required to complete the reaction. The height of the bars must match the time values ​​in the table.

C. No, Emilia was not correct in her forecast. According to the data, aluminum reacted 100 seconds faster than magnesium, which reacted in 50 seconds. Thus aluminum reacts more rapidly with hydrochloric acid than magnesium.

From most reactive to least reactive, the metals are as follows:

This order is consistent with the reactivity series, which is:

We are unable to estimate the reactivity of potassium, sodium, calcium, copper, silver, or gold from this experiment because those variables are not present in the data.

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