What is the osmotic pressure, in atm, of 2.2 L of an aqueous solution that contains 261 g of calcium acetate Ca(C2H3O2)2, at 37 °C? Calcium acetate is an electrolyte with a molar mass of 158.17 g/mol.

Please enter your answer with three sig figs.

The statement that correctly describes residual radiation from the patient's body after this type of treatment is the patient's body will be radioactive for a very short period after the treatment.

The correct option is A.

Residual radiation refers to the lingering radiation that remains after a nuclear or radioactive event has occurred.

It is the radiation that persists in the environment or materials even after the primary source of radiation has been removed or decayed.

Residual radiation can arise from various sources, including nuclear accidents, radioactive waste disposal, and from radiotherapy.

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PART A: Approximately 2,887.23 grams of hydrochloric acid would be required to produce 80 grams of hydrogen gas.

PART B: Approximately 7.65 x 10^24 water molecules would be produced.

PART A: To determine the grams of hydrochloric acid required to produce 80g of hydrogen gas, we need to consider the balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl):

2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that for every 3 moles of hydrogen gas produced, 6 moles of hydrochloric acid are required. To calculate the grams of hydrochloric acid needed, we need to use the molar mass of HCl, which is approximately 36.46 g/mol. First, we convert the grams of hydrogen gas to moles using the molar mass of hydrogen (H₂), which is approximately 2.02 g/mol: 80 g H₂ * (1 mol H₂ / 2.02 g H₂) = 39.60 mol H₂

Since the mole ratio of HCl to H₂ is 6:3, we multiply the moles of hydrogen by the ratio: 39.60 mol H₂ * (6 mol HCl / 3 mol H₂) = 79.20 mol HCl

Finally, we convert the moles of HCl to grams using the molar mass of HCl: 79.20 mol HCl * (36.46 g HCl / 1 mol HCl) ≈ 2,887.23 g HCl

Therefore, approximately 2,887.23 grams of hydrochloric acid would be required to produce 80 grams of hydrogen gas.

PART B: To determine the number of water molecules produced in the reaction of 741 grams of magnesium hydroxide (Mg(OH)₂), we need to consider the balanced chemical equation for the reaction:

Mg(OH)₂ → MgO + H₂O

From the equation, we can see that for every one mole of magnesium hydroxide, one mole of water is produced. To calculate the moles of magnesium hydroxide, we divide the given grams by the molar mass of Mg(OH)₂: 741 g Mg(OH)₂ * (1 mol Mg(OH)₂ / molar mass of Mg(OH)₂) = X mol Mg(OH)₂

The molar mass of Mg(OH)₂ can be calculated by adding the atomic masses of magnesium (Mg), oxygen (O), and hydrogen (H) together:

Mg: 24.31 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Molar mass of Mg(OH)₂ = (24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 58.33 g/mol

Substituting the values: 741 g Mg(OH)₂ * (1 mol Mg(OH)₂ / 58.33 g Mg(OH)₂) ≈ 12.70 mol Mg(OH)₂

Since the mole ratio of water to Mg(OH)₂ is 1:1, the number of water molecules produced is the same as the moles of Mg(OH)₂: 12.70 mol H₂O

To calculate the number of water molecules, we multiply the moles of H₂O by Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol: 12.70 mol H₂O * (6.022 x 10^23 molecules/mol) ≈ 7.65 x 10^24 molecules of water .Therefore, approximately 7.65 x 10^24 water molecules would be produced.

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To calculate the amount of heat released when cooling water, we can use the formula:

Q = m × c × ΔT

where:

Q is the heat energy released or absorbed (in Joules),

m is the mass of the water (in grams),

c is the specific heat capacity of water (in J/g°C),

ΔT is the change in temperature (in °C).

Given:

m = 100.0 g (mass of water)

c = 4.186 J/g°C (specific heat capacity of water)

ΔT = 10.0°C (change in temperature, from 75.0°C to 10.0°C)

Using the formula:

Q = 100.0 g × 4.186 J/g°C × (10.0°C - 75.0°C)

Q = 100.0 g × 4.186 J/g°C × (-65.0°C)

Q = -27187 J

The negative sign indicates that heat is released during the cooling process.

Therefore, the correct answer is:

d) -27,209 J

Answer:

D

Explanation:

The correct answer is D. In ethylene (CH2=CH2), the carbon atoms in the double bond have two unhybridized p-orbitals.

The freezing point of the solution is 6.213 °C. This is less than the freezing point of ether.

Freezing point of the solution can be determined by using the concept of freezing point depression, which is a colligative property.

The freezing point depression is given by the equation:

ΔTf = Kf * m

Where:

ΔTf = Freezing point depression

Kf = Cryoscopic constant (a constant specific to the solvent)

m = Molality of the solution

First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

We have 1.727 mol of solute and 483 g of ether (solvent).

First, we convert the mass of ether to kilograms:

Mass of ether = 483 g = 483/1000 kg = 0.483 kg

Now we can calculate the molality:

m = moles of solute / mass of solvent (in kg)

m = 1.727 / 0.483 ≈ 3.575 mol/kg

Next, we need the cryoscopic constant (Kf) for ether. The cryoscopic constant is specific to the solvent. For ether, the cryoscopic constant is approximately 1.74 °C kg/mol.

Finally, we can calculate the freezing point depression (ΔTf):

ΔTf = Kf * m

ΔTf = 1.74 °C kg/mol * 3.575 mol/kg

= 6.213 °C

Therefore, the freezing point of the solution is decreased by approximately 6.213 °C compared to the freezing point of pure ether.

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0.0956 grams of helium will occupy a volume of 575 mL at 760 mmHg and 20°C

To calculate the mass of helium that will occupy a given volume at a specific temperature and pressure, we need to use the ideal gas law equation: PV = nRT. Here's how you can solve the problem:

Convert the volume to liters: 575 mL = 575/1000 = 0.575 L.

Convert the pressure to atmospheres: 760 mmHg = 760/760 = 1 atm.

Convert the temperature to Kelvin: 20°C = 20 + 273.15 = 293.15 K.

Plug the values into the ideal gas law equation: (1 atm) * (0.575 L) = n * (0.0821 L·atm/(mol·K)) * (293.15 K).

Rearrange the equation to solve for n (the number of moles): n = (1 atm * 0.575 L) / (0.0821 L·atm/(mol·K) * 293.15 K).

Calculate the value of n: n = 0.0239 moles.

To find the mass, we need to know the molar mass of helium, which is approximately 4 grams per mole.

Multiply the molar mass by the number of moles: 4 g/mol * 0.0239 moles = 0.0956 grams.

Therefore, approximately 0.0956 grams of helium will occupy a volume of 575 mL at 760 mmHg and 20°C.

Note: The ideal gas law assumes ideal gas behavior, and the calculated result may not be accurate under extreme conditions or for gases that deviate significantly from ideal behavior.

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For a solution of sodium hydroxide:

1. To make a buffer with pH = 5.00, have a ratio of

[tex]\frac{[A-]}{[HA]} = 10^{-5.50}[/tex] = 0.316. The concentration of acetic acid is 0.200 M, so add enough sodium hydroxide to make the concentration of acetate 0.316 M.

The volume of sodium hydroxide needed is calculated as follows:

V(NaOH) = (0.316 M - 0.200 M) / 4.50 M = 0.0316 L = 31.6 mL

Therefore, 31.6 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid solution to make a buffer with pH = 5.00.

2. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.76 + log(0.2/0.1)

= 4.86

Therefore, the pH of the buffer is 4.86.

3. The mass of salt that must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87 is calculated as follows:

[tex]\frac{[A-]}{[HA]} = 10^{-4.87}[/tex] = 0.0114

The concentration of acetate is 0.0114 M, and the concentration of propanoic acid is 1 mol dm³. Therefore, the mass of acetate that must be dissolved is calculated as follows:

Mass of acetate = (0.0114 mol dm³)(0.25 dm³) = 0.00285 g

Therefore, 0.00285 g of sodium propanoate must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87.

4. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.74 + log(0.1/0.1)

= 4.74

Therefore, the pH of the buffer is 4.74.

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Answer:

There are 3.13 x 10^24 water molecules in 5.2 moles of water.

The type of compound represented by the graph at right is a strong acid (option B).

An acid is generally any compound capable of dissociating into its respective constituent ions when in an aqueous solution.

An acid is categorised as strong or weak depending on whether it can dissociate completely or partially. A strong acid dissociates completely in water.

According to this question, HA, when added to water, dissociates into H+ and A- ions, hence, is a strong acid.

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The original volume of the concentrated solution was 182.7 mL.

To solve this problem, we can use the formula for dilution:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

We are given that the initial concentration (C1) is 2.3 M, the final concentration (C2) is 2.1 M, and the final volume (V2) is 191.8 mL. We want to find the initial volume (V1).

Plugging in the values we know into the dilution formula, we get:

(2.3 M) V1 = (2.1 M) (191.8 mL)

Simplifying this expression, we can solve for V1:

V1 = (2.1 M) (191.8 mL) / (2.3 M)

V1 = 182.7 mL

It's important to note that the units of concentration and volume must be consistent in this formula. In this case, the concentrations are given in units of M (moles per liter), and the volumes are given in units of mL (milliliters).

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If the temperature is 273 K, the partial pressure of O₂ is 1.40 atm.

According to the given information:

Volume of bottle  = 2.00 L

Total pressure = 5.00 atm

Mixture of  N₂, O₂ and CO₂

Moles of N₂ = 0.29 moles

Partial pressure of CO₂ = 0.350 atm

Temperature = 273 K

It is known that

PV = nRT

P = pressure

V = volume

n = no of moles

R = universal gas constant

= 0.08206 L atm/mol K

T = temperature

so

To find total no of moles

PV = nRT

n = PV/RT

= (5.00 atm × 2.00 L) ÷ ( 0.08206 L atm/mol K × 273 K)

= 0.446 moles

Number of moles of CO₂

n = PV/RT

= (0.350 atm × 2.00 L) ÷ ( 0.08206 L atm/mol K × 273 K)

= 0.0312 moles CO₂

Total no of moles = moles of O₂ +  moles of CO₂ + moles of N₂

0.446 moles = moles of O₂ + 0.29  + 0.0312

Moles of O₂ = 0.125

Partial pressure of O₂ = (no of moles of O2) × R × T/V

= (0.125 moles × 0.08206 L atm/mol K × 273 K)/ 2.00 L

= 1.40 atm

The partial pressure of O₂ is 1.40 atm.

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Answer:

I'm sorry, but I'm not able to view the chemical structure of glycine.

To determine the molar solubility of Cd(IO₃)₂, it is required to use the solubility product constant (Ksp) and the stoichiometry of the compound.

Given information,

Ksp = 2.5 × 10⁻⁸

Temperature = 25°C

The balanced equation for the dissolution of Cd(IO₃)₂ is:

Cd(IO₃)₂(s) ⇌ Cd₂+(aq) + 2 IO₃⁻(aq)

The expression for the solubility product constant is given by:

Ksp = [Cd²⁺][IO₃⁻]²

The equilibrium concentrations can be expressed as:

[Cd²⁺] = x mol/L

[IO₃⁻] = 2x mol/L

Substituting these values into the Ksp expression:

Ksp = (x)(2x)² = 4x³

Now,

2.5 × 10⁻⁸ = 4x³

Solving for x:

x³ = (2.5 × 10⁻⁸)/4

x = (2.5 × 10⁻⁸)/[tex]4^{1/3}[/tex]

x = (2.5 ×  10⁻⁸)/1.58

x= 1.58 × 10⁻⁸

Therefore, the molar solubility is 1.58 × 10⁻⁸.

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The type of energy transfer between the frozen ice cream and the tongue is convection. The frozen ice cream will gradually melt because of the temperature difference.

Convection is the method of heat transfer that involves the transmission of heat in a fluid (liquid or gas) by the circulation of currents.

According to this question, a spoonful of frozen ice cream is placed on the tongue. The heat transfer will occur between the tongue with warmer temperature and the ice cream with colder temperature.

The heat energy will flow from the tongue to the ice cream, hence, making it defreeze or melt with time. This type of heat transfer is convection because it involves a fluid.

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Answer:

Explanation:

h 7

Answer:

[tex] \huge{ \boxed{3.48 \: g/ml}}[/tex]

Explanation:

The density of the aluminum can be found by using the formula;

[tex]density( \rho) = \dfrac{mass}{volume} [/tex]

From the question

mass = 83.6 g

volume = 23.99 ml

[tex] \rho = \dfrac{83.6}{23.99} = 3.4847[/tex]

We have the final answer as

At a specific temperature, we can use the following equation to determine the root mean square (rms) speed of molecules or atoms:

rms speed = √(3RT/M)

where

R is the gas constant (8.314 J/(mol·K)),

T is the temperature in Kelvin, and

M is the molar mass in kilograms per mole.

For [tex]O_2[/tex] molecules at 427 K:

Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol = 0.032 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.032 kg/mol)

rms speed ≈ √(28165.3 J/kg)

rms speed ≈ 167.9 m/s

The rms speed of [tex]O_2[/tex] molecules at 427 K is approximately 167.9 m/s.

For He atoms at 427 K:

Molar mass of He = 4.00 g/mol = 0.004 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.004 kg/mol)

rms speed ≈ √(7061.925 J/kg)

rms speed ≈ 83.9 m/s

The rms speed of He atoms at 427 K is approximately 83.9 m/s.

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Answer:

Copper Copper Copper Copper

Hello !

The action that would most likely reduce water pollution is A) initiating efforts to remove trash from the ocean.

Trash and plastic waste in the ocean pose a significant threat to marine life and ecosystems. Initiating efforts to remove trash from the ocean can help mitigate this problem and reduce water pollution. When plastic waste accumulates in oceans, it can break down into microplastics, which are ingested by marine organisms and can enter the food chain, causing harm to both marine life and humans.

By actively removing trash from the ocean, we can prevent it from further degrading and releasing pollutants. This can help protect marine habitats, reduce the risk of entanglement or ingestion by marine animals, and preserve the overall health of ocean ecosystems.

On the other hand, options B, C, and D would likely contribute to increased water pollution:

B) Eliminating laws and regulations on industrial waste would remove important safeguards and oversight, potentially allowing industries to discharge pollutants directly into water bodies, leading to increased water pollution.

C) Promoting the use of oil and gas can lead to oil spills, leakage, and contamination of water sources. This can have severe consequences for aquatic ecosystems and human health.

D) Removing the ban on chlorofluorocarbons (CFCs) would result in the release of these harmful substances, which deplete the ozone layer and can contaminate water sources when disposed of improperly.

In conclusion, initiating efforts to remove trash from the ocean is the most effective action to reduce water pollution, while the other options would likely exacerbate the problem. Therefore, Option A is correct.

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The event of the fan stopping when it is switched off in a classroom provides an opportunity to discuss scientific concepts and learning. Here are the various steps involved in this event.

Observation: The students observe that when the switch of the fan is turned off, the fan stops rotating.

Questioning: Encourage students to ask questions about why the fan stops when the switch is turned off, leading to discussions about energy transfer, electrical circuits, and mechanical motion.

Explanation: Explain to the students that when the switch is turned off, it breaks the electrical circuit, interrupting the flow of current to the fan motor. Without the electrical energy from the current, the fan motor stops rotating.

Discussion: Engage students in a discussion about the conversion of electrical energy to mechanical energy in the fan motor. Discuss concepts such as electromagnetic induction, electric motors, and the role of magnetic fields in generating rotational motion.

Application: Relate this event to everyday life, highlighting the importance of understanding electrical circuits and energy conversions in various appliances.

By discussing these concepts, students can deepen their understanding of the principles behind the event and foster scientific learning.

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Answer:

Chemical changes involve the formation of a new substance. 

Explanation:

The main difference between chemical and physical changes is the fact that chemical changes involve the formation of a new substance, results in a new substance. When a physical change occurs, this does not lead to bonds being broken or formed. The properties of the elements involve do not change. This includes changes such as changes of state. Physical change will result in a different state of matter.

The relative atomic mass of atom X is 9.

The relative atomic mass of an element is the average mass of its atoms relative to 1/12th the mass of a carbon-12 atom.

Given that atom X is 9 times heavier than 1/12th the mass of a carbon-12 atom, we can calculate the relative atomic mass of X as follows:

Relative atomic mass of X = (9 × 1/12) × relative atomic mass of carbon-12

The relative atomic mass of carbon-12 is defined as exactly 12. Therefore, we have:

Relative atomic mass of X = (9 × 1/12) × 12

= 9

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At an altitude of 6.5 km where the pressure is 0.5 atm, the volume of the gas in the balloon would be 1000 mL.

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Mathematically, Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

In this case, the initial pressure and volume are given as 1 atm and 500 mL, respectively. The final pressure is 0.5 atm, and we need to find the final volume.

Plugging the given values into Boyle's Law, we have:

(1 atm)(500 mL) = (0.5 atm)(V₂)

Simplifying the equation, we get:

500 mL atm = 0.5 V₂ atm

To solve for V₂, we divide both sides of the equation by 0.5 atm:

V₂ = (500 mL atm) / (0.5 atm)

V₂ = 1000 mL

Therefore, at an altitude of 6.5 km where the pressure is 0.5 atm, the volume of the gas in the balloon would be 1000 mL.

It's important to note that this calculation assumes that the temperature remains constant throughout the process. Additionally, in reality, the temperature may change with altitude, which could affect the behavior of the gas.

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Pressure and volume can be regarded as the entity that is inversely proportional.

It should be noted that this explanatin an be done using the law in chemistry which is the Boyle's law which states that, for a given amount of gas and constant temperature, the volume is inversely proportional to the pressure.

However the Equal quantities of all gases can be seen to have same number of molecules when subjected to the same temperature and pressure (Avogadro's law).

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The volume that the sample of tin occupies in its liquid state is approximately 2.974 cm³.

To find the volume of the sample of tin in its liquid state, we can use the concept of density and the given information.

Given:

Initial volume of the sample (solid tin) = 3.60 cm³

Density of solid tin = 5.770 g/cm³

Density of liquid tin = 6.990 g/cm³

To calculate the volume of the sample in its liquid state, we can use the following formula:

Volume = Mass / Density

First, let's calculate the mass of the sample of tin in its solid state using the density and volume information:

Mass of solid tin = Density of solid tin * Volume of solid tin

= 5.770 g/cm³ * 3.60 cm³

Next, we can calculate the volume of the sample in its liquid state using the mass of the sample and the density of liquid tin:

Volume of liquid tin = Mass of solid tin / Density of liquid tin

Now, let's perform the calculations:

Mass of solid tin = 5.770 g/cm³ * 3.60 cm³

= 20.772 g

Volume of liquid tin = Mass of solid tin / Density of liquid tin

= 20.772 g / 6.990 g/cm³

Finally, calculate the volume:

Volume of liquid tin = 2.974 cm³ (approximately)

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The volume of oxygen will be 0.404 dm³ at 2.02 x 10³ Nm² and 400 K.

To solve this problem, we can use the ideal gas law equation, which states:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

Since we are dealing with the same sample of oxygen, the number of moles (n) and the ideal gas constant (R) will remain constant. Therefore, we can rewrite the equation as:

P₁V₁ / T₁ = P₂V₂ / T₂

Where the subscripts 1 and 2 represent the initial and final conditions, respectively.

Given:

V₁ = 1 dm³ = 1 L = 0.001 m³

T₁ = 500 K

P₁ = 1.01 x 10³ Nm²

P₂ = 2.02 x 10³ Nm²

T₂ = 400 K

Let's substitute the values into the equation and solve for V₂:

(1.01 x 10³ Nm²)(0.001 m³) / 500 K = (2.02 x 10³ Nm²)(V₂) / 400 K

(1.01 x 10³ Nm²)(0.001 m³)(400 K) = (2.02 x 10³ Nm²)(V₂)(500 K)

V₂ = [(1.01 x 10³ Nm²)(0.001 m³)(400 K)] / [(2.02 x 10³ Nm²)(500 K)]

Now let's calculate the value of V₂:

V₂ = (0.404 Nm)(0.001 m³) / (1.01 Nm)

V₂ = 0.000404 m³ = 0.404 dm³

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77.7K

Ideal gas laws allow us to calculate the characteristics of gases through their changes.

Gay-Lussac's Law

One of the ideal gas laws is Gay-Lussac's Law. This law describes the relationship between temperature and pressure. It states that temperature and pressure are directly proportional. This means that as temperature increases, pressure also increases. In equation form, this law is shown as:

Finding Temperature
To find the temperature, plug in the known information and solve for T₂.

Now, to find T₂, multiply both sides by 210.0 kPa.

Since this question is based on measured values, we should round according to significant figure rules. We should round to 3 sig figs because 208K has 3. This means the temperature of the gas tank must be 77.7 Kelvin.

The relative atomic mass of the sample of rubidium is approximately 48.6.

To calculate the relative atomic mass of the sample of rubidium using the given information, we can use the following steps:

Calculate the contribution of each isotope to the average atomic mass based on their abundance:

Isotope 1: Relative abundance = 72%, Mass number = 48

Isotope 2: Relative abundance = 28%, Mass number = 50

Contribution of isotope 1 = (72/100) * 48

Contribution of isotope 2 = (28/100) * 50

Sum the contributions of each isotope to obtain the average atomic mass:

Average atomic mass = Contribution of isotope 1 + Contribution of isotope 2

Calculate the values for the contributions and sum them up.

Round the average atomic mass to the appropriate number of significant figures.

Let's perform the calculations:

Contribution of isotope 1 = (72/100) * 48 = 34.56

Contribution of isotope 2 = (28/100) * 50 = 14.00

Average atomic mass = Contribution of isotope 1 + Contribution of isotope 2

Average atomic mass = 34.56 + 14.00 = 48.56

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