2. If 13. 5 L of nitrogen gas reacts with 17. 8 L of hydrogen gas at SIP, according to the following reaction, what mass of ammonia would be produced?
N2
*
3 H2 - 2 NH3

In order to produce the impact of future droughts that may occur in the area, the scientist's plan would most likely include several key elements.

First and foremost, the plan would likely involve extensive research and data analysis to better understand the climate patterns and environmental factors that contribute to drought in the region.

This could involve collecting and analyzing data on rainfall, temperature, humidity, and other key indicators, as well as examining the impact of human activity on the local ecosystem.

Based on this research, the scientist may develop a range of strategies aimed at mitigating the effects of drought, such as water conservation measures, alternative irrigation techniques, and improved crop management practices.

Additionally, the plan may involve community outreach and education initiatives to raise awareness about the importance of water conservation and sustainable resource management.

Overall, the scientist's plan would likely be a comprehensive and multi-faceted approach aimed at preparing the city for future droughts and promoting long-term resilience and sustainability.

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There are 1.18x10²³ grams of nitrogen in 5.6x10²³ liters of nitrous oxide at STP.

To solve this problem, we need to use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP (standard temperature and pressure), P = 1 atm and T = 273.15 K.

We can rearrange the equation to solve for n:

n = PV/RT

We can then convert the number of moles to grams using the molar mass of nitrogen (N₂), which is 28.02 g/mol.

n(N₂) = n(N₂O) x 2 moles of N per mole of N₂O

n(N₂) = (PV/RT) x 2

n(N₂) = (1 atm x 5.6x10²³ L) / (0.08206 L·atm/mol·K x 273.15 K) x 2

n(N₂) = 4.22x10²¹ mol

mass(N) = n(N₂) x MM(N₂)

mass(N) = 4.22x10²¹ mol x 28.02 g/mol

mass(N) = 1.18x10²³ g

As a result,  1.18x10²³ grammes of nitrogen are present in 5.6x10²³ liters of nitrous oxide at STP.

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The concentration of the stock solution is 0.183 M.

To solve this problem, we can use the equation:

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the NaOH titrant, and V2 is the volume of the titrant used.

First, we need to calculate the number of moles of NaOH used:

Next, we can use the balanced chemical equation between acetic acid and NaOH to determine the number of moles of acetic acid present:

Now we can calculate the concentration of the stock solution:

Therefore, the concentration of the stock solution is 0.183 M.

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The size of the Kf indicate regarding the stability of the transition metal complexes is the large values of the Kf indicate that transition metal complexes are often very stable.

The larger the value of the Kf of the complex ion, the more stable will be the transition metal complexes. Due to the how large formation constants are often is not uncommon to listed as the logarithms in the form of the log Kf.

The Kf values that are the very large in the magnitude for the complex ion formation that indicate that the reaction is heavily favors the products. The complex ions that are the poorly formed and this value will be the very small.

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The third quantum number of a 2p³ electron in phosphorus is M = -1. Option A is the answer.

The electronic configuration of phosphorus is 1s²2s²2p⁶3s²3p³. The 2p subshell has three orbitals, which can hold up to six electrons. The three orbitals are labeled as 2p_x, 2p_y, and 2p_z, where each orbital can hold a maximum of two electrons with opposite spins.

The three quantum numbers that define the state of an electron in an atom are n, l, and m. Here, n represents the principal quantum number, l represents the azimuthal quantum number, and m represents the magnetic quantum number.

The values of l for the 2p subshell are 1, and the possible values of m for l = 1 are -1, 0, and 1. The electron in question is in the 2p subshell, so its value of l is 1. Since the possible values of m for this electron are -1, 0, and 1, we can rule out options B, C, and D. Therefore, the correct answer is A, M = -1. Hence, option A is the answer.

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Concentration of the diluted HCl solution : 0.00389 M

To find the concentration of the diluted HCl solution, we can use the equation:

C1V1 = C2V2

Where C1 is the initial concentration of the HCl solution (0.09714 M), V1 is the initial volume of the solution (10.00 mL), C2 is the final concentration of the diluted HCl solution, and V2 is the final volume of the diluted HCl solution (250.00 mL).

Plugging in the values, we get:

(0.09714 M)(10.00 mL) = C2(250.00 mL)

Solving for C2, we get:

C2 = (0.09714 M)(10.00 mL) / (250.00 mL)

C2 = 0.00389 M

Therefore, the concentration of the diluted HCl solution is 0.00389 M.

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The concentration of the H₂CO₃ solution is 0.162 M.

To solve this problem, we can use the balanced chemical equation for the neutralization reaction between H₂CO₃ and Fe(OH)₃:

2 Fe(OH)₃ + 3 H₂CO₃ → Fe₂(CO₃)+ 6 H₂O

From the balanced equation, we can see that the mole ratio between Fe(OH)3 and H₂CO₃ is 2:3. We can use this information along with the volume and concentration of the Fe(OH)₃ solution to calculate the number of moles of H₂CO₃:

Moles of Fe(OH)₃  = volume x concentration = 0.0880 L x 1.22 M = 0.10776 moles

Moles of H₂CO₃= (2/3) x moles of Fe(OH)₃ = (2/3) x 0.10776 moles = 0.07184 moles

Now, we can calculate the concentration of the H₂CO₃ solution using the volume of the solution provided: concentration = moles / volume = 0.07184 moles / 0.225 L = 0.162 M

Therefore, the molarity of the H₂CO₃ solution has been determined to be 0.162 M.

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The volume changes to 1.36 L, under the condition pressure of a balloon begins at 2. 45 atm and a volume 2. 00.

For this problem we have to apply  Boyle's law that states  at constant temperature, the pressure and volume of a gas are inversely proportional to each other.

Then, pressure increases, volume decreases and vice versa. The formula for Boyle's law is

P1V1 = P2V2

Here

P1 and V1 = initial pressure and volume

P2 and V2 = final pressure and volume

Applying this formula, we can evaluate the final volume of the balloon

P1V1 = P2V2

(2.45 atm)(2.00 L) = (3.60 atm)(V2)

V2 = (2.45 atm)(2.00 L) / (3.60 atm)

V2 = 1.36 L

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When the crest and trough of two waves meet, they undergo destructive interference, causing the amplitude of the resulting wave to be smaller than that of either individual wave.

In this scenario, the two students shaking the rope create waves that travel toward each other. One student creates a crest, which is a point of maximum positive displacement, while the other creates a trough, which is a point of maximum negative displacement. When these two points meet, they interfere destructively, resulting in a wave with a smaller amplitude than either individual wave.

This phenomenon of destructive interference is a result of the superposition principle of waves, which states that the displacement of two waves at any point in space and time is the algebraic sum of the individual displacements of the waves.

When two waves of equal amplitude and opposite phase meet, they cancel each other out, resulting in a wave with a smaller amplitude or even no wave at all.

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To determine the number of moles of chlorine gas required for the formation of 320.5 grams of aluminum chloride, we need to use the balanced chemical equation for the reaction. The equation for the reaction between aluminum and chlorine gas to form aluminum chloride is:

2Al(s) + 3Cl2(g) → 2AlCl3(s)

From the equation, we can see that for every two moles of aluminum, three moles of chlorine gas are required to form two moles of aluminum chloride. Therefore, we can set up a proportion:

2 moles of AlCl3 : 3 moles of Cl2 = 320.5 g of AlCl3 : x

Where x is the number of moles of Cl2 required.

We can use the molar mass of aluminum chloride (133.34 g/mol) to convert the mass of AlCl3 to moles:

320.5 g AlCl3 ÷ 133.34 g/mol = 2.403 moles AlCl3

Substituting the values into the proportion, we get:

2 moles of AlCl3 : 3 moles of Cl2 = 2.403 moles of AlCl3 : x

Solving for x, we get:

x = 3.605 moles of Cl2

Therefore, 3.605 moles of chlorine gas are required to react with 320.5 grams of aluminum to form aluminum chloride.

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The reaction of sulfur dioxide (SO₂) with oxygen and water to form sulfuric acid (H₂SO₄) is responsible for acid rain. The reaction is: SO₂(g) + O₂(g) + H₂O(l) -> H₂SO₄(aq).

When flue gas from a coal-burning furnace is discharged into the atmosphere, it contains sulfur dioxide (SO₂) as one of its components. SO₂ can react with oxygen and water in the atmosphere to form sulfuric acid (H₂SO₄), which is a strong acid that can cause harm to the environment. Sulfuric acid is one of the main components of acid rain, which can damage crops, forests, and bodies of water, as well as erode buildings and other structures.

Acid rain can also be harmful to human health, as it can cause respiratory problems and other illnesses. Therefore, it is important to control the emissions of SO₂ from coal-burning furnaces and other sources to reduce the formation of sulfuric acid and the occurrence of acid rain.

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The letter in response to david li's letter is-

Dear David Li,

Thank you for your letter regarding the groundwater system at Riverdale School. I am glad to hear that you are interested in this innovative system.

To answer your question, the groundwater system at Riverdale School could heat the air by utilizing the stable temperature of the groundwater. Groundwater has a relatively constant temperature throughout the year, which can be warmer than the outside air temperature during the winter. The system could pump the groundwater through a heat exchanger, which transfers the heat to the air and distributes it throughout the school.

If the groundwater system were used, the air temperature at Riverdale School would become more stable because the system would provide a constant source of heat.

This stability in temperature would be beneficial for the comfort and well-being of the students and staff. The air temperature would also change compared to the current heating system, as the groundwater system would provide a more consistent and efficient source of heat.

I hope this answers your questions about the groundwater system at Riverdale School. Please let me know if you have any further inquiries.

Sincerely,

[Your Name]

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Answer: I believe it's A

Source: Trust me bro

The heat of a reaction may be found with the equation q=mcΔT. A 56. 8g sample of aluminum is heated from 79. 5°C to 143. 7°C. The specific heat capacity of aluminum is 0. 900 J/(g*K).  The heat absorbed is C) 6220J.

The heat absorbed can be calculated using the formula q=mcΔT, where q is the heat absorbed, m is the mass of the sample, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Substituting the given values, we get:

q = (56.8 g) x (0.900 J/(g*K)) x (143.7°C - 79.5°C)

q = 6220 J

Therefore, the heat absorbed is 6220 J, and the answer is option C. This means that 6220 Joules of energy is required to heat a 56.8 gram sample of aluminum from 79.5°C to 143.7°C, assuming a specific heat capacity of 0.900 J/(g*K).

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There are approximately 205 marbles in the large pile that weighs 592.45 g.

To determine the number of marbles in the large pile, we need to use a proportion. We know that 15 marbles weigh 43.35 g, so we can set up the following proportion:

15 marbles / 43.35 g = x marbles / 592.45 g

To solve for x, we can cross-multiply and simplify:

15 marbles x  592.45 g = 43.35 g x   x marbles

8886.75 g = 43.35 g x   x marbles

x marbles = 8886.75 g / 43.35 g ≈ 205

Therefore, there are approximately 205 marbles in the large pile that weighs 592.45 g. It's worth noting that this answer is an approximation since we rounded the final result to the nearest whole number. Also, the actual weight of each marble may vary slightly, which could affect the exact number of marbles in the pile.

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A potted plant is placed under a grow lamp, which provides 6,200. J of energy to the plant and the soil over the course of an hour. The specific heat capacity of the soil is about 0. 840 J/g°C and the temperature goes up by 8. 75°C of soil.  800 grams of soil are there

We can use the formula:

Q = m * c * ΔT

where Q is the amount of energy transferred, m is the mass of the material, c is the specific heat capacity of the material, and ΔT is the change in temperature.

We know that Q = 6,200 J, c = 0.840 J/g°C, and ΔT = 8.75°C. We can rearrange the formula to solve for m:

m = Q / (c * ΔT)

Plugging in the values, we get:

m = 6,200 J / (0.840 J/g°C * 8.75°C)

m = 800 grams

Therefore, there are 800 grams of soil.

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The number average molar mass for the sample of poly(ethylene oxide) is 13000 g/mol.

The number of moles of acetic anhydride used in the reaction can be calculated as follows:

Moles of acetic anhydride = (mass of acetic anhydride) / (molar mass of acetic anhydride)

Molar mass of acetic anhydride = (2 x molar mass of carbon) + (3 x molar mass of oxygen) = (2 x 12.011) + (3 x 15.999) = 102.09 g/mol

Moles of acetic anhydride = (2.5 × 10⁻³) / 102.09 = 2.45 × 10⁻⁵ mol

Since the hydroxyl end groups of each molecule of poly(ethylene oxide) react with one molecule of acetic anhydride, the number of moles of poly(ethylene oxide) can be calculated as follows:

Moles of poly(ethylene oxide) = moles of acetic anhydride / 2 = 1.23 x 10⁻⁵ mol

The mass of the sample of poly(ethylene oxide) is given as 2.00 g, so the number average molar mass can be calculated as follows:

Number average molar mass = (mass of sample) / (moles of sample)

Number average molar mass = 2.00 / 1.23 x 10⁻⁵ = 1.626 x 10⁸ g/mol

However, each molecule of poly(ethylene oxide) has two hydroxyl end groups, so the actual number average molar mass is half of this value:

Number average molar mass = 1.626 x 10⁸ / 2 = 13000 g/mol.

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In general, a system tends to favor a reaction that results in an increase in entropy, which is a measure of the number of possible arrangements of the system's particles. The answer is A) lower energy and higher entropy.

This is due to the fact that the increase in the number of particles in the system or the increase in the number of ways the particles can be arranged leads to an increase in entropy. On the other hand, a system also tends to favor a reaction that results in a decrease in energy, which is a measure of the system's ability to do work.

Therefore, when a system undergoes a reaction that decreases its energy while increasing its entropy, it is moving towards a more stable and disordered state.

This is because a lower energy state means that the system is releasing energy, while a higher entropy state means that the system is becoming more disordered and spread out. This tendency towards lower energy and higher entropy is known as the second law of thermodynamics, which governs the behavior of all physical systems.

The answer is A) lower energy and higher entropy.

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Answer:

B

Explanation:

[OH⁻] = 1.1 x 10⁻¹⁰ M, pH = 9.96; No, [HF] is not equal to [NaF] based on stoichiometry as NaF dissociates completely to form Na⁺ and F⁻ ions, whereas HF dissociates partially.

The dissociation of NaF in water can be represented as follows:

Since NaF is a salt of a strong base (NaOH) and a weak acid (HF), the F⁻ ion will hydrolyze in water to produce OH⁻ ions.

The hydrolysis reaction is as follows:

Firstly, we can use the equilibrium expression for the reaction of HF with water to calculate the [H⁺] ion concentration:

Since the initial concentration of HF is negligible, we can assume that the concentration of F- ion at equilibrium is equal to the initial concentration of NaF.

Using Kw = [H⁺][OH⁻], we can calculate the [OH⁻] ion concentration:

Since NaF dissociates completely in water, [F⁻] = 0.105 M. Therefore, [HF] = Ka*[NaF]/[F⁻] = 6.4 x 10⁻⁴ * 0.105/1 = 6.72 x 10⁻⁵ M.

Hence, [HF] is not equal to [NaF] based on stoichiometry.

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The object absorbed 356.4 J of heat energy.

To calculate the amount of heat energy absorbed by an object, we can use the formula:

Q = mcΔT

where Q is the heat energy absorbed, m is the mass of the object, c is the specific heat capacity of the object, and ΔT is the change in temperature of the object.

Let's put in the given values:

m = 18 g (mass of object)

c = 0.900 J/(g*C) (specific heat of object)

ΔT = 40°C - 18°C = 22°C (change in temperature of object)

Now we can calculate the heat energy absorbed:

Q = mcΔT

Q = (18 g)(0.900 J/(g*C))(22°C)

Q = 356.4 J

Therefore, the object absorbed 356.4 J of heat energy.

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6. The mass (in grams) present is 9.72×10⁸ grams

7. The number of atoms is 8.50×10²⁴ atoms

8. The mass (in grams) present is 3.73×10¹⁰ grams

First, we shall determine the mole of LiNO₃. Details below:

6.022×10²³ atoms = 1 mole of LiNO₃

Therefore,

8.48×10³⁰ atoms = 8.48×10³⁰ / 6.022×10²³

8.48×10³⁰ atoms = 1.41×10⁷ moles of LiNO₃

Finally, we shall determine the mass of LiNO₃. Details below:

Mass = Mole × molar mass

Mass of LiNO₃ = 1.41×10⁷ × 68.95

Mass of LiNO₃ = 9.72×10⁸ grams

First, we shall determine the mole in 2105 g of (NH₄)₃PO₃. Details below:

Mole = mass / molar mass

Mole of  (NH₄)₃PO₃ = 2105 / 149.09

Mole of  (NH₄)₃PO₃ = 14.12 moles

Finally, we shall determine the number of atoms. Details below:

From Avogadro's hypothesis,

1 mole of (NH₄)₃PO₃ = 6.022×10²³ atoms

Therefore,

14.12 moles of (NH₄)₃PO₃ = 14.12 × 6.022×10²³

14.12 moles of (NH₄)₃PO₃ = 8.50×10²⁴ atoms

Thus, the number of atoms is 8.50×10²⁴ atoms

First, we shall determine the mole of (NH₄)₂SO₄. Details below:

6.022×10²³ atoms = 1 mole of (NH₄)₂SO₄

Therefore,

1.7×10³² atoms = 1.7×10³² / 6.022×10²³

1.7×10³² atoms = 2.82×10⁸ moles of (NH₄)₂SO₄

Finally, we shall determine the mass of (NH₄)₂SO₄. Details below:

Mass = Mole × molar mass

Mass of (NH₄)₂SO₄ = 2.82×10⁸ × 132.14

Mass of (NH₄)₂SO₄ = 3.73×10¹⁰ grams

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When 142 g Cl₂ combines with lots of aluminium, 1.33 moles of AlCl₃ are formed.

To determine the number of moles of AlCl₃ formed when 142 g Cl₂ reacts with plenty of aluminum, we first need to write a balanced chemical equation for the reaction:

2 Al + 3 Cl₂ → 2 AlCl₃

From the balanced equation, we can see that 3 moles of Cl₂ react with 2 moles of Al to form 2 moles of AlCl₃.

Next, we need to calculate the number of moles of Cl₂ present in 142 g:

n(Cl₂) = m/M

n(Cl₂) = 142 g / 70.9 g/mol

n(Cl₂) = 2.00 moles

Since the reaction consumes 3 moles of Cl₂ for every 2 moles of AlCl₃ formed, we can determine the number of moles of AlCl₃ formed as:

n(AlCl₃) = (2/3) x n(Cl₂)

n(AlCl₃) = (2/3) x 2.00 moles

n(AlCl₃) = 1.33 moles

Therefore, 1.33 moles of AlCl₃ form when 142 g Cl₂ reacts with plenty of aluminum.

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According to the balanced chemical equation, 1 mole of SiC is produced from 1 mole of C. Therefore, the number of moles of SiC produced from 9.3 moles of C is also 9.3 moles.

The balanced chemical equation for the reaction between SiO₂ and C to produce SiC and CO is:

SiO₂ + C ⇒ SiC + CO

The stoichiometric coefficients of C and SiC are both 1. This means that for every 1 mole of C reacted, 1 mole of SiC is produced. Therefore, if we have 9.3 moles of C, we can expect to produce 9.3 moles of SiC.

It is important to note that the balanced chemical equation assumes that the reaction goes to completion, meaning that all of the reactants are consumed and converted into products. In reality, some of the reactants may not be fully consumed, leading to a lower yield of the desired product.

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The concentration of [OH⁻] in the solution is 5.0×10⁻⁹ M.

To find the [OH⁻] of the solution with [H3O⁺] = 2.0×10⁻⁶ M, you can use the ion product constant of water, Kw = [H₃O⁺][OH⁻].

Step 1: Write down the known values and the ion product constant of water (Kw = 1.0×10⁻¹⁴ at 25°C).
[H₃O⁺] = 2.0×10⁻⁶ M
Kw = 1.0×10⁻¹⁴

Step 2: Use the formula Kw = [H₃O⁺][OH⁻] to solve for [OH⁻].
1.0×10⁻¹⁴ = (2.0×10⁻⁶ M) × [OH⁻]

Step 3: Divide both sides by [H₃O⁺] to isolate [OH⁻].
[OH⁻] = (1.0×10⁻¹⁴) / (2.0×10⁻⁶ M)

Step 4: Calculate the concentration of [OH⁻].
[OH⁻] = 5.0×10⁻⁹ M
So, the concentration of [OH⁻] in the solution is 5.0×10⁻⁹ M.

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Approximately 88.67 grams of [tex]Fe_2O_3[/tex] are formed when [tex]1.67*10^{23}[/tex] atoms of Fe react completely with oxygen.

The balanced chemical equation for reaction between iron and oxygen to form iron (III) oxide can be written as:

4 Fe + 3 O2 → 2 [tex]Fe_2O_3[/tex]

To find the number of moles  [tex]Fe_2O_3[/tex] formed when [tex]1.67*10^{23[/tex] atoms of Fe react, we first need to convert the given number of atoms of Fe to moles:

1.67x[tex]10^{23}[/tex] atoms of Fe × (1 mol/6.022 x [tex]10^{23}[/tex] atoms) = 0.2777 mol of Fe

The number of moles of [tex]Fe_2O_3[/tex] formed :

0.2777 mol of Fe × (1 mol of [tex]Fe_2O_3[/tex]/0.5 mol of Fe) = 0.5554 mol of[tex]Fe_2O_3[/tex]

We can calculate the mass of [tex]Fe_2O_3[/tex] :

0.5554 mol of [tex]Fe_2O_3[/tex] × 159.69 g/mol = 88.67 g of [tex]Fe_2O_3[/tex]

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Approximately [tex]6.13 x 10-19[/tex] liters of oxygen gas will be needed to react with 3.3 x 104 molecules of sulfur dioxide to produce sulfur trioxide.

The balanced chemical equation for the reaction between sulfur dioxide and oxygen gas to form sulfur trioxide is:
[tex]SO2 + 1/2O2 → SO3[/tex]
From this equation, we can see that one mole of sulfur dioxide reacts with 1/2 mole of oxygen gas to produce one mole of sulfur trioxide.
To find the amount of oxygen gas required to react with 3.3 x 104 molecules of sulfur dioxide, we need to convert the number of molecules of SO2 to moles. The molar mass of SO2 is 64 g/mol, so 3.3 x 104 molecules of SO2 is equivalent to:

(3.3 x 104 molecules) / (6.022 x 1023 molecules/mol) = 5.48 x 10-20 moles of SO2
Since one mole of SO2 reacts with 1/2 mole of O2, we need half as many moles of oxygen gas as we have moles of SO2. Therefore, the amount of oxygen gas required is:
1/2 x 5.48 x 10-20 moles = 2.74 x 10-20 moles
Finally, we can convert this amount to volume using the ideal gas law, assuming standard temperature and pressure (STP) of 0°C and 1 atm. The volume of one mole of any gas at STP is 22.4 L, so the volume of oxygen gas required is:
2.74 x 10-20 moles x 22.4 L/mol = 6.13 x 10-19 L

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The probability of a random sample of n=9 sections of PVC pipe having a mean diameter between 1.009 inches and 1.012 inches is approximately 0.8185 or 81.85%.

The mean diameter of PVC pipe is 1.01 inches, and the standard deviation is 0.003 inches. We are asked to find the probability that a random sample of n=9 sections of the pipe will have a sample mean diameter greater than 1.009 inches and less than 1.012 inches.

First, we need to find the standard error of the mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is 0.003/√9 = 0.001.

Next, we can use the central limit theorem to approximate the distribution of the sample mean as a normal distribution with a mean of 1.01 inches and a standard deviation of 0.001 inches (the standard error we just calculated).

We can then calculate the z-scores for the lower and upper limits of the sample mean:

z1 = (1.009 - 1.01)/0.001 = -1

z2 = (1.012 - 1.01)/0.001 = 2

Using a z-table or calculator, we can find the probability of the sample mean falling within this range:

P(-1 < Z < 2) = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 = 0.8185

Therefore, the probability of a random sample of n=9 sections of PVC pipe having a mean diameter between 1.009 inches and 1.012 inches is approximately 0.8185 or 81.85%.

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Galena and Pyrite are mineral ores.

Ore is a deposit of one or more precious minerals in the Earth's crust.

Galena is lead ore with formula PbS while pyrite is iron ore having formula FeS₂. In Other words, Galena is sulfide of lead and pyrite is sulfide of iron.

Both Galena and Pyrite are sulfide ores with different specific gravities.

Pyrite shows magnetic property on heating which galena is nonmagnetic component and doesn’t bear any magnetic properties.

Both are semi conductors but they are used for different purpose.

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