Who was the first person that was to attend to arrange the element in what year

In summary, the invention of the Haber process led to the production of synthetic fertilizers, which reduced the demand for natural nitrates from mines like Humberstone, ultimately leading to its closure.

The invention of the Haber process, which is a method of synthesizing ammonia from nitrogen and hydrogen gas, allowed for the production of large amounts of fertilizer. This increased the amount of farmland that could be cultivated and substantially increased crop yields. As a result, the demand for natural resources like nitrate-based fertilizers, which were previously extracted from mines like Humberstone in Chile, decreased significantly.

Humberstone was one of the largest nitrate mines in Chile and was known for producing high-quality nitrate fertilizers. However, the increased production of synthetic fertilizers through the Haber process meant that the demand for natural nitrate-based fertilizers gradually declined. The falling demand for natural nitrates eventually led to the closure of many nitrate mines like Humberstone in Chile.

In addition to the Haber process, other factors such as the discovery of cheaper sources of natural nitrates in other parts of the world and the decline in the price of nitrate also contributed to the closure of the Humberstone mines.

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Pathogenic bacteria are the most problematic in the food industry. The term "pathogenic" refers to any bacteria that are capable of causing illness when consumed by humans or animals.

Option A is correct

Pathogenic bacteria are often found in foods that are not cooked or handled correctly. This includes raw meat, unpasteurized dairy products, and foods that are stored at incorrect temperatures for an extended period of time. It is crucial for food industry professionals to follow strict safety protocols to minimize the risk of pathogenic bacteria contaminating their products, leading to foodborne illnesses.Foodborne illnesses can be caused by a variety of microorganisms, including bacteria, viruses, and parasites. Pathogenic bacteria are a significant concern in the food industry, as they can cause serious illness and even death in vulnerable populations such as young children, pregnant women, and the elderly. The most problematic bacteria in the food industry are those that are pathogenic and are capable of multiplying quickly in the food environment. Pathogenic bacteria can cause food poisoning, which is a severe illness that can result in dehydration, fever, vomiting, and diarrhea.

The symptoms can be severe and can lead to hospitalization or even death if left untreated. The most common pathogenic bacteria found in food are Salmonella, Listeria, and E.coli. These bacteria are often found in raw or undercooked meat, unpasteurized dairy products, and fruits and vegetables that are not washed correctly. It is essential for food industry professionals to follow strict safety protocols to prevent the growth and spread of these bacteria. Proper handwashing, storage, and cooking techniques can help minimize the risk of foodborne illnesses caused by pathogenic bacteria. In conclusion, pathogenic bacteria are the most problematic in the food industry. Food safety protocols should be followed to minimize the risk of foodborne illnesses caused by these bacteria.

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Crystallization of sugar in a syrup can be avoided by adding inverted sugar. Invert sugar is a mixture of glucose and fructose, produced from sucrose. It is produced by hydrolysis of sucrose by heating with water and an acid such as lemon juice, vinegar or cream of tartar.

By adding invert sugar to a syrup, it reduces the amount of sucrose present in the solution and promotes the formation of glucose and fructose. This causes the solution to become more saturated and the syrup less likely to crystallize. Invert sugar is often used in the production of confectionery, ice cream, and other sweets.Inverted sugar can also be used to improve the texture and quality of baked goods. It is often used in cake recipes to enhance the flavor and increase the shelf life. It can also be used as a substitute for honey or corn syrup in recipes that require a liquid sweetener.In conclusion, the addition of invert sugar helps to prevent crystallization of sugar in syrup. Invert sugar is a mixture of glucose and fructose produced from sucrose by hydrolysis.

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From the statement of the second law of thermodynamics,  the entropy of the surroundings increases  for spontaneous processes.

According to the second law of thermodynamics, we know that the entropy of surroundings or the universe would always increase in a sponteanous process.

Having said this, it is natural  fall out of the law stated above that the entropy of the surroundings increases  for spontaneous processes.

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Lithium=li

Iron=Fe

Helium=He

I don't say u must have to mark my ans as brainliest but if u think it has really helped u plz don't forget to thank me....

Answer:

lithium symbol=li helium= he and iron =i

Answer:

exosphere thermosphere mesosphere stratosphere troposphere

The ocean where prevailing winds pass through is the Pacific Ocean.

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the mass defect of Li-7 is -0.035279 u and the nuclear binding energy per nucleon is 5.60553 × 10⁻¹² J/nuclide.

Given data:

Atomic mass of Li-7, A = 7.016003

The atomic mass of Li-7 is the sum of the number of protons and neutrons in it. Therefore, the number of neutrons in Li-7 is:

Neutrons = Atomic mass - Protons= 7.016003 - 3= 4.016003The mass of 3 protons and 4.016003 neutrons in Li-7 is: Mass of protons + Mass of neutrons = (3 x 1.007276) + (4.016003 x 1.008665) = 3.021828 + 4.029454 = 7.051282 u

Therefore, the mass defect in Li-7 is:

Mass defect = Actual mass - Calculated mass

= Atomic mass - Mass of protons and neutrons

= 7.016003 - 7.051282

= -0.035279 u

Nuclear Binding Energy per nucleon (BE/A) can be calculated using the formula:

BE/A = [Δm.c² / A]

where Δm is the mass defect and c is the speed of light which is 2.998 × 10⁸ m/s.

Substituting the values in the above formula:

BE/A = [(-0.035279) × (2.998 × 10⁸)² / 7]= 5.60553 × 10⁻¹² J/nuclide

Therefore, the mass defect of Li-7 is -0.035279 u and the nuclear binding energy per nucleon is 5.60553 × 10⁻¹² J/nuclide.

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Answer:

B

Explanation:

The four factors that increase the rate of the reaction:

1.Concentration of the reactants

2.Size of the particles

3.Temperature

4.Catalyst

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O.

Answer:

Explanation:

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Answer:

Since Socrates didn't leave any written works, we can only learn from his disciple Plato. Plato wrote that Socrates believed in an atomic theory, just not like the one we believe now. We can assume that both Plato and Socrates believed that atoms were geometric forms that were all made up of triangles.

They believed in 4 elements, so we can say that their atomic theory was based on a periodic table made up of:

I guess that their major influence was that all elements could be broken down into smaller parts or atoms, in this case some type of triangle.

Answer:

The answer is D..i.e phosphate. it consists of 1p and 4oxygen atoms

\(PO_4^{3-}\)  is a polyatomic ion. Hence, option D is correct.

What is a polyatomic ion?

Polyatomic ions are ions which consist of more than one atom.

The prefix poly- means many, so a polyatomic ion is an ion that contains more than one atom. This differentiates polyatomic ions from monatomic ions, which contain only one atom.

\(PO_4^{3-}\) is a chemical derivative of phosphoric acid with a chemical name Phosphate. Phosphate is also called Phosphate ion or Orthophosphate.

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Answer:

b. stainless steel

Explanation:

b) stainless steel is not good condoctor of heat

a) The sphere emits thermal radiation at a rate of 139.75 Watts.

b) The sphere absorbs thermal radiation at a rate of 37.66 Watts.

c) The sphere's net rate of energy exchange is 102.09 Watts.

To calculate the rates of thermal radiation emission and absorption, we can use the Stefan-Boltzmann law, which states that the rate of thermal radiation emitted or absorbed by an object is proportional to its surface area, temperature, and the Stefan-Boltzmann constant.

a) The rate of thermal radiation emitted by the sphere can be calculated using the formula:

Emitting Rate = emissivity * surface area * Stefan-Boltzmann constant * (\(temperature^4 - environment\ temperature^4\))

Plugging in the given values:

Emitting Rate = \(0.924 * (4\pi * (0.457)^2) * 5.67 \times 10^{-8} * ((32.2 + 273.15)^4 - (82.9 + 273.15)^4)\)

Emitting Rate ≈ 139.75 Watts

b) The rate of thermal radiation absorbed by the sphere can be calculated in a similar way but using the environment temperature as the object's temperature:

Absorbing Rate = emissivity * surface area * Stefan-Boltzmann constant * (\(environment\ temperature^4 - temperature^4\))

Plugging in the given values:

Absorbing Rate = \(0.924 * (4\pi * (0.457)^2) * 5.67 \times 10^{-8} * ((82.9 + 273.15)^4 - (32.2 + 273.15)^4)\)

Absorbing Rate ≈ 37.66 Watts

c) The net rate of energy exchange is the difference between the emitting rate and the absorbing rate:

Net Rate = Emitting Rate - Absorbing Rate

Net Rate = 139.75 Watts - 37.66 Watts

Net Rate ≈ 102.09 Watts

Therefore, the sphere emits thermal radiation at a rate of 139.75 Watts, absorbs thermal radiation at a rate of 37.66 Watts, and has a net rate of energy exchange of 102.09 Watts.

Note: The units for all the rates are Watts.

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The mass of \(Al_{2} S_{3}\) produced is 15.616 grams.

The reaction taking place is as follows:

\(2Al+ 3S\) →\(Al_{2} S_{3} (s)\)

Moles of Al = mass/molar mass = 10.0g/27.0g/mol

= 0.370 mol

Moles of S = mass/molar mass = 15.0g/(32.065g/mol)

= 0.468 mol

Al and S reacts in the molar ratio of 2:3.

2 moles of Al reacts with 3 moles of S

0.370 moles of Al will react with S = (3/2)*0.370mol

= 0.555 mol

Similarly, 0.468 moles of S will react with Al = 2/3 *0.468mol

= 0.312 mol

Thus, Al is in excess and S is the limiting reactant (some of Al will be left over ,S will completely react)

So, moles of \(Al_{2} S_{3}\) produced=1/3*0.312 mol of S

= 0.104 mol

Mass of \(Al_{2} S_{3}\) produced = moles*molar mass of \(Al_{2} S_{3}\)

= 0.104mol*150.158g/mol

= 15.616 grams

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75.1 grams of \(CO_{2}\) would occupy a volume of 8.75 liters at 45.2 degrees Celsius and a pressure of 1.251 atmospheres.

How do you calculate occupied volume?

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Answer: the mole

Explanation:

The mole is the unit of amount in Chemistry.

It provides a bridge between the atom and the macroscopic amounts of material that we work with in the laboratory.

Explanation:

Metal is a better conductor than wood, so a metal spoon could burn

Paul's hand.  So, Sue is incorrect. The metal being good conductor of heat, heats up faster than wood spoon and thus, could burn Paul's hand.

The subscript of the oxygen molecule in the equation 2Mg + O2 → 2MgO is 2.

This means that two oxygen molecules (O2) are involved in the reaction, and they combine with two magnesium atoms (2Mg) to form two magnesium oxide molecules (2MgO). The subscript "2" in front of MgO indicates that two MgO molecules are formed as a result of the reaction. The equation 2Mg + O2 → 2MgO represents a combustion reaction in which magnesium (Mg) reacts with oxygen (O2) to produce magnesium oxide (MgO). This is an example of an exothermic reaction, which releases energy in the form of heat and light.

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Answer:

Change in colour is a chemical change.

Explanation:

Because when two substances combine chemically, they create (a) new substance(s) that have/has different molecular structures from the parent substances, thus mostly likely absorb and reflect light in different ways.

There are 6 oxygen atoms on the left side of the equation and 7 oxygen atoms in the product (CO2 and H2O) formed.

In the given balanced chemical equation:

C2H6O + 3O2 → 2CO2 + 3H2O

We can determine the number of oxygen atoms on the left side of the equation by examining the coefficient in front of the O2 molecule. In this case, there are three O2 molecules, meaning there are a total of 3 × 2 = 6 oxygen atoms on the left side.

Moving on to the right side of the equation, we can see that two CO2 molecules are formed as products. Each CO2 molecule consists of one carbon atom and two oxygen atoms. Therefore, there are a total of 2 × 2 = 4 oxygen atoms in the CO2 molecules.

Additionally, there are three H2O molecules formed as products. Each H2O molecule consists of two hydrogen atoms and one oxygen atom. Therefore, there are a total of 3 × 1 = 3 oxygen atoms in the H2O molecules.

To find the total number of oxygen atoms in the product, we add the number of oxygen atoms from CO2 and H2O: 4 + 3 = 7 oxygen atoms.

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Answer: At 389 K, 6.8% of N2O4 decomposes to NO2.

Explanation: The decomposition of N2O4 is an exothermic reaction, and the forward reaction is favored at lower temperatures. Thus, increasing the temperature will shift the equilibrium towards the reactants, resulting in a decrease in the extent of the reaction.

To solve this problem, we can use the expression for the equilibrium constant (Kp) for the reaction:

Kp = (PNO2)2 / PN2O4

where PNO2 and PN2O4 are the partial pressures of NO2 and N2O4, respectively. At equilibrium, we know that the pressure of N2O4 is 0.100 bar, and the pressure of NO2 is 0.058 x 0.100 = 0.0058 bar. Using these values, we can calculate the equilibrium constant (Kp) at 298 K:

Kp = (0.0058)2 / 0.100 = 0.0003364

Now, we can use this value of Kp to calculate the percentage of N2O4 that decomposes at 389 K. Let x be the fraction of N2O4 that decomposes at 389 K. Then, the partial pressures of N2O4 and NO2 at equilibrium are:

PN2O4 = (1 - x) x 0.100 = 0.100x - 0.100x^2

PNO2 = 2x x 0.100 = 0.200x

Using these expressions, we can calculate the value of Kp at 389 K:

Kp' = (0.200x)2 / (0.100x - 0.100x^2)

At equilibrium, Kp' = Kp, so we can set these expressions equal to each other and solve for x:

0.0003364 = (0.200x)2 / (0.100x - 0.100x^2)

0.00003364(0.100x - 0.100x^2) = 0.040x^2

0.000003364 - 0.000003364x = 0.040x^2

0.040x^2 + 0.000003364x - 0.000003364 = 0

Using the quadratic formula, we get:

x = 0.068 or x = -0.000125

Since x represents a fraction, the solution x = -0.000125 is extraneous, and we can discard it. Therefore, the fraction of N2O4 that decomposes at 389 K is:

x = 0.068

To convert this to a percentage, we multiply by 100:

x = 6.8%

Therefore, at 389 K, 6.8% of N2O4 decomposes to NO2.

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The isotope used to date young, organic material is carbon-14.

Carbon-14 is the isotope used to date young, organic material. It breaks down relatively quickly and has a half-life of about 5,700 years.

Carbon-14 is constantly being formed in the upper atmosphere through the interaction of cosmic rays with nitrogen, and it is taken up by plants during photosynthesis.

Animals then consume plants and incorporate the carbon-14 into their own bodies. When an organism dies, the carbon-14 begins to decay, and by measuring the amount of carbon-14 remaining in a sample, scientists can determine how long ago the organism died.

Carbon-14 dating is most effective for organic materials up to about 50,000 years old.

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"the number of ions produced by one formula unit of the electrolyte," refers to the van't Hoff factor (i) in an ideal solution of a strong electrolyte. It represents the extent of dissociation of the electrolyte into ions.

In an ideal solution of a strong electrolyte, the van't Hoff factor (i) represents the number of ions that are produced when one formula unit of the electrolyte dissociates completely in the solution. It is a measure of the extent of dissociation of the electrolyte.

For example, for a strong electrolyte such as sodium chloride (NaCl), when it dissolves in water, it completely dissociates into sodium ions (Na+) and chloride ions (Cl-). In this case, the van't Hoff factor (i) would be 2 because one formula unit of NaCl produces two ions (Na+ and Cl-).

Similarly, for other strong electrolytes, the van't Hoff factor (i) can be determined based on the number of ions produced per formula unit. It is important to note that for non-electrolytes or weak electrolytes, the van't Hoff factor (i) is typically less than 1, indicating partial dissociation or no dissociation in the solution.

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To determine whether the protein is an enzyme, the scientist should primarily measure the rate of the chemical reaction. Option(b)

Enzymes are specialized proteins that act as catalysts, facilitating chemical reactions by lowering the activation energy required for the reaction to occur. By comparing the rate of the chemical reaction with and without the protein, the scientist can determine if the presence of the protein enhances the reaction rate. If the reaction proceeds at a significantly higher rate in the presence of the protein, it suggests that the protein is acting as an enzyme, catalyzing the reaction. Enzymes typically accelerate reaction rates by providing an alternative reaction pathway with a lower activation energy, allowing the reaction to occur more readily. While measuring the amount of protein consumed may provide information about the protein's involvement in the reaction, it does not directly determine whether the protein is an enzyme. Similarly, measuring the amount of products bound to the active site or the temperature change of the reaction can provide additional insights, but they alone are not sufficient to confirm enzymatic activity. The most reliable indicator to ascertain whether the protein is an enzyme is to compare the rate of the chemical reaction in the presence and absence of the protein. Option(b)

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Answer: The iron equilibrium is the more positive, and so will go in the forward direction. The tin equilibrium is less positive (more negative), and will be driven backwards. So the tin(II) ions will reduce iron(III) ions to iron(II) ions. In the process, of course, the tin(II) ions will be oxidised to tin(IV) ions.

Explanation:

The concentrations of H2, Br2, and HBr are approximately 0.125 M, 0.125 M, and 0.250 M, respectively.

To solve this problem, we'll use the equilibrium constant expression and the given information to calculate the equilibrium concentrations of H2, Br2, and HBr.

The balanced equation for the reaction is:

H2 + Br2 ⇌ 2HBr

Given:

Initial moles of H2 = 0.500 mole

Initial moles of Br2 = 0.500 mole

Volume of the container = 2.00 L

Equilibrium constant (Kc) = 36.0

Let's assume that at equilibrium, the concentration of H2 is [H2], the concentration of Br2 is [Br2], and the concentration of HBr is [HBr]. We'll use the stoichiometric coefficients from the balanced equation to set up the equilibrium expression:

Kc = [HBr]^2 / ([H2] * [Br2])

Since we have the initial moles of H2 and Br2, we can calculate their initial concentrations:

Initial concentration of H2 = (0.500 mole) / (2.00 L) = 0.250 M

Initial concentration of Br2 = (0.500 mole) / (2.00 L) = 0.250 M

Let's assume the equilibrium concentrations of H2 and Br2 are x M, and the equilibrium concentration of HBr is 2x M.

Substituting the concentrations into the equilibrium expression and solving for x:

36.0 = (2x)^2 / (0.250 * 0.250)

36.0 = 4x^2 / 0.0625

144x^2 = 0.0625 * 36.0

144x^2 = 2.25

x^2 = 2.25 / 144

x^2 ≈ 0.015625

x ≈ 0.125

Since [H2] = x and [Br2] = x, the equilibrium concentrations are:

[H2] ≈ 0.125 M

[Br2] ≈ 0.125 M

[HBr] ≈ 2x ≈ 2 * 0.125 M ≈ 0.250 M

Therefore, at equilibrium, the concentrations of H2, Br2, and HBr are approximately 0.125 M, 0.125 M, and 0.250 M, respectively.

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