NH4NO3 is ammonium nitrate is used for supplies ammonium and nitrate ions.
a. Calculate the percentage nitrogen, hydrogen and oxygen by mass in this fertilizer.
b. Calculate the mass of nitrogen in 500kg of ammonium nitrate
Relative atomic masses: H = 1; N = 14 ; O= 16​

The balanced chemical equation for the reaction between Fe2+ and MnO4- in acidic solution is :

5 Fe2+ + MnO4- + 8 H+ -> 5 Fe3+ + Mn2+ + 4 H2O

From the equation, we can see that 1 mole of MnO4- reacts with 5 moles of Fe2+.

The initial moles of Fe2+ in the solution are :

0.500 mol/L x 0.100 L = 0.050 mol

Since the stoichiometric ratio between MnO4- and Fe2+ is 1:5, the moles of MnO4- required to react with all the Fe2+ are :

5 x 0.050 mol = 0.250 mol

Therefore, the quantity of Fe2+ ion reacting, in moles, is 0.050 mol.

Answer: 10.51 kg of iron is produced in the reaction.

Explanation: The balanced equation for the reaction between iron III oxide and iron is:

2Fe2O3 + 3C → 4Fe + 3CO2

From the equation, we can see that 2 moles of Fe2O3 react to produce 4 moles of Fe. We can use the molar mass of Fe2O3 to convert the given mass to moles, and then use the mole ratio to calculate the moles of Fe produced. Finally, we can convert the moles of Fe to mass using the molar mass of Fe.

The molar mass of Fe2O3 is:

2(55.85 g/mol Fe) + 3(16.00 g/mol O) = 159.69 g/mol Fe2O3

So, 15 kg (or 15000 g) of Fe2O3 is equal to:

15000 g / 159.69 g/mol = 94.03 mol Fe2O3

According to the balanced equation, 2 moles of Fe2O3 produce 4 moles of Fe. So, 94.03 moles of Fe2O3 will produce:

4/2 x 94.03 = 188.06 moles of Fe

The molar mass of Fe is 55.85 g/mol, so the mass of Fe produced is:

188.06 mol x 55.85 g/mol = 10507.8 g or 10.51 kg

The specific heat capacity of the metal can be calculated to be 2.03 J/g･ °C. This result can be explained by the fact that the heat lost by the metal is equal to the heat gained by the ethylene glycol, since no other source of heat is present.

Specific heat capacity is the amount of heat energy needed to raise the temperature of a unit mass of a substance by 1 degree Celsius. It is measured in joules per kilogram Kelvin (J/kg K).

The specific heat capacity (c) of the metal can be calculated using the following equation:

Q = mc∆T

Where Q is the heat lost by the metal, m is the mass of the metal, and ∆T is the change in temperature of the metal.

Q = (115.0 g)(c)(165.0 °C - 25.8 °C)

On the other hand, the heat gained by the ethylene glycol can be calculated using the following equation:

Q = (265.0 g)(2.43 J/g･ °C)(41.5 °C - 25.8 °C)

By equating the two equations, we can solve for the specific heat capacity of the metal:

(115.0 g)(c)(165.0 °C - 25.8 °C) = (265.0 g)(2.43 J/g･ °C)(41.5 °C - 25.8 °C)

c = 2.03 J/g･ °C

Therefore, the specific heat capacity of the metal can be calculated to be 2.03 J/g･ °C.

This result can be explained by the fact that the heat lost by the metal is equal to the heat gained by the ethylene glycol, since no other source of heat is present.

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Answer:

Pressure of the gas =766.4 mmHg.

Explanation:

Gas pressure greater than atmospheric pressure = Atmospheric pressure + Height

745.0 mm + 26.4 cm = 766.4 mmHg.

The enthalpy of vaporization of ethanol is 38.56 kJ/mol at its boiling point (78°C). Calculate the value of entropy of  surrounding when 1.00 mole of ethanol is vaporized at 78°C and 1.00 atm is -110.2 kJ/mol K. The correct option is option A.

Entropy is the measurement of the amount of thermal energy for each degree of temperature in a system that cannot be used for productive work. Entropy is a measure of a system's molecular disorder or unpredictability since work is produced by organised molecular motion.

Entropy theory offers profound understanding of the pattern of impulsive shifts for numerous commonplace events. A standout of 19th-century physics is its invention by the German scientist Rudolf Clausius in 1850.

C[tex]_2[/tex]H[tex]_5[/tex]OH(l)   →  C[tex]_2[/tex]H[tex]_5[/tex]OH(g)    

ΔSsystem= 38.56 /351

                  =110.2 kJ/mol K

ΔSsurrounding=-110.2 kJ/mol K

ΔSuniverse=0

Therefore, the correct option is option A.

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The balanced equation for the given chemical reaction can be written as 3CsCl + K[tex]_3[/tex]PO[tex]_4[/tex]→ Cs[tex]_3[/tex] (PO[tex]_4[/tex]) + 3KCl.

An equation per a chemical reaction is said to be balanced if both the reactants plus the products have the same number of atoms and total charge for each component of the reaction. In other words, each component of the reaction have an equal balance of mass and charge.

The components and outcomes of a chemical reaction are listed in an imbalanced chemical equation, but the amounts necessary to meet the conservation of mass are not specified. The balanced equation for the given chemical reaction can be written as 3CsCl + K[tex]_3[/tex]PO[tex]_4[/tex]→ Cs[tex]_3[/tex] (PO[tex]_4[/tex]) + 3KCl.

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The reaction is endothermic since the energy level of the products have are higher than that of the reactants.

The activation energy (AE) is the energy difference between the reactants and the transition state.

The change in energy E and the energy difference between the reactants and the products

The data given is as follows:

Reactants: 0 kJ/mol

AE forward 125 kJ/mol

AE reverse: 86 kJ/mol

Products: 39 kJ/mol

The values of ΔE forward and ΔE reverse are as follows:

ΔE forward = (39 - 0) kJ/mol

ΔE forward = +39 kJ/mol

ΔE reverse = (0 - 39) kJ/mol

ΔE reverse =  -39 kJ/mol

3. Given that Ea = 154 kJ/mol and ΔE = 136 kJ/mol

AE reverse = ΔE - AE forward

E = 136 kJ/mol - 154 kJ/mol

E = -18 kJ/mol

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In the year 2000, the population of Arctic foxes reached its peak.

A region's population is the total number of people living there. Typically, it is expressed as the number of persons in a given volume or region. A census is typically performed to estimate population in order to ascertain its size, density, and distribution. The term "population" can also refer to the variety of species present in a region. Population growth is the gradual rise in a species' number of individuals in a particular area.

At that time, the total number of Arctic foxes in their habitat, which stretches from northern Europe and Asia over the tundra of the Arctic Circle, was believed to be 745,000. Based on the International Union for Conservation of Nature's most recent estimations, this population figure.

The Arctic fox is a very tough species that can endure some of the world's harshest environments. They can also adjust to their surroundings and eat a range of foods, such as grass, eggs, carrion, small mammals, and carrion.

Unfortunately, a variety of causes are causing the Arctic fox population to decline at the moment. Additionally, human actions like hunting, habitat loss, and fur trapping are lowering their numbers.

The Arctic fox is not now facing extinction, according to the IUCN, which has classified it as a species of Least Concern. However, the species is threatened by a number of factors, and unless conservation efforts are made, its population is likely to keep declining.

Reducing human influences on the Arctic fox's environment is crucial for its protection. This entails restricting hunting and trapping in addition to minimising habitat loss and pollution. The ability of the species to adapt to climate change should also be improved, and their population numbers and trends should be tracked. If these steps are adopted, the number of Arctic foxes may eventually return to its original high.

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The complete question is,

Which year saw the highest arctic fox population? How many people attended?

Explanation:

The molar mass of octane (C8H18) is approximately 114 g/mol.

To find the energy required to increase 1 mole of substance by 10°C, we can use the formula:

ΔE = n * C * ΔT

where ΔE is the energy required, n is the number of moles, C is the specific heat, and ΔT is the change in temperature.

For water, n = 1 mol, C = 4.184 J/gK, and ΔT = 10°C.

So, the energy required to increase 1 mole of water by 10°C is:

ΔE_water = n * C_water * ΔT

= 1 mol * 4.184 J/gK * 10 K

= 41.84 J

For octane, n = 1 mol, C = 2.22 J/gK, and ΔT = 10°C.

So, the energy required to increase 1 mole of octane by 10°C is:

ΔE_octane = n * C_octane * ΔT

= 1 mol * 2.22 J/gK * 10 K

= 22.2 J

Therefore, it takes more energy to increase 1 mole of water by 10°C than 1 mole of octane.

125.4 kJ of heat is absorbed in the process of making 0.887 mol of CF₄.

The given reaction produces one mole of CF₄, and its enthalpy change is 141.3 kJ/mol. Therefore, the enthalpy change for the production of 0.887 mol of CF₄ can be calculated as follows:

Enthalpy change for 1 mol of CF₄ = 141.3 kJ/mol

Enthalpy change for 0.887 mol of CF₄ = (0.887 mol) x (141.3 kJ/mol)

Enthalpy change for 0.887 mol of CF₄ = 125.39 kJ

Therefore, the quantity of heat absorbed in the process of making 0.887 mol of CF₄ is 125.39 kJ.

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C4H10 + O2 --> CO2 + H2O

The mass must be balanced:

2C4H10 + 13O2 --> 8CO2 + 10H2O

The molar mass of butane is 12 × 4 + 10 = 58

The molar mass of carbon dioxide is 16 × 2 + 12 = 44

First, we calculate the moles of carbon dioxide produced

[tex]n_{CO2} = \frac{50,0 g}{44 g/mol} = 1,14 mol[/tex]

If 2 moles of butane are needed for produce 8 moles of CO2 then x moles of butane are needed to produce 1,14 moles of CO2, therefore

[tex]\frac{2}{8} = \frac{x}{1,14} \\ \\ x = \frac{1,14}{4} = 0,285 mol[/tex]

Then the moles can be multiplied by the molar mass of butane in order to get the total mass of butane burned.

[tex]m_{C4H10} = 0,285 mol × 58 g/mol = 1,65 g[/tex]

Answer:

Explanation:

The total mark for this paper is 60.

t The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question.

t Questions labelled with an asterisk (*) are ones where the quality of your

written communication will be assessed

– you should take particular care with your spelling, punctuation and grammar, as

well as the clarity of expression, on these questions.

Advice

t Read each question carefully before you start to answer it.

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Explanation:

Humidity levels are high, averaging around 82% due to the warm moist trade winds that flow through the archipelago, as well as sea surface temperatures, a rich and vibrant vegetative cover and abundant rainfall.

Weight percentage of KClO₃ is 117% indicates that there could have been a mistake in the experiment.

First, calculate the partial pressure of O₂ in the flask, which is equal to the total pressure minus the vapor pressure of water:

Partial pressure of O₂ = 749 torr - 16.5 torr = 732.5 torr

Next, calculate the number of moles of O₂ produced from the decomposition of KClO₃. We can use the ideal gas law to do this:

PV = nRT

n = PV/RT

where P is the partial pressure of O₂, V is the volume of O₂ collected over water (in liters), R is the gas constant (0.08206 L atm mol⁻¹ K⁻¹), and T is the temperature in Kelvin (273.15 + 19 = 292.15 K).

n = (732.5 torr / 760 torr/atm) × (0.260 L / 1000 L) × (0.08206 L atm mol⁻¹ K⁻¹) × (292.15 K)⁻¹

n = 0.00966 mol O₂

From the balanced chemical equation, 2 moles of KClO₃ produces 3 moles of O₂. Therefore, the number of moles of KClO₃ in the sample is:

n(KClO₃) = (3/2) × n(O₂)

n(KClO₃) = (3/2) × 0.00966 mol

n(KClO₃) = 0.0145 mol

Finally, use the formula for weight percentage:

Weight percentage of KClO₃ = (mass of KClO₃ / mass of sample) × 100%

Calculate the mass of KClO₃ from the number of moles using the formula:

mass = n × formula weight

mass(KClO₃) = 0.0145 mol × 122.55 g/mol

mass(KClO₃) = 1.77 g

The mass of the sample is given as 1.515 g, so the weight percentage of KClO₃ is:

Weight percentage of KClO₃ = (1.77 g / 1.515 g) × 100%

Weight percentage of KClO₃ = 117% (This result is not physically meaningful, indicating that there may have been an error in the experiment.)

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Due to changes in stream temperature, stream flow due to shortages or increased storms, and other stressors which can affect ecosystem health, these aquatic ecosystems are at risk from climate change.

The long-term trend of the weather in a place is called the climate. Hour by hour, day by day, month by month, or even year by year, the weather might change. Due to changes in stream temperature (which results in a corresponding drop in oxygen levels).

Stream flow due to shortages or increased storms, and other stressors (such as increased storm runoff such as nutrients, pollutants, along with sediment) which can affect ecosystem health, these aquatic ecosystems are at risk from climate change.

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Breaking tension describes this circumstance. Tensiometers function up to 0.5 bar. The vacuum gauge of the majority of commercial tensiometers has a suction scale that ranges from 0 to 100 cb.

A tube filled with water and fitted with a specific porous tip and vacuum gauge is a tensiometer. This device monitors soil water suction, which is akin to the method a plant root employs to get water from the ground. upkeep of the field as seen to the right. Tensiometers are reusable and not very expensive.

In comparison to ECH2O-5TE and Watermark without site-specific calibration, Tensiometer gave relatively reliable soil moisture measurements, according to statistical analysis of the results. Watermark overestimated the water content of the soil, while ECH2O-5TE did not.

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Dipolar molecules have a separation of charge with a positive center and a negative center located at two different points, resulting in a net dipole moment. The correct option is true

Dipolar molecules are molecules that have a separation of charge, resulting in a positive center and a negative center located at two different points in the molecule.

This separation of charge creates a net dipole moment, which is a measure of the molecule's polarity. The magnitude and direction of the dipole moment depend on the difference in electronegativity between the atoms in the molecule and the molecule's geometry.

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The standard reduction potential of Y₃⁺ is +1.97 V.

The cell potential of a galvanic cell is given by the difference between the standard reduction potentials of the two half-cells. In this case, the cell consists of two half-cells:

Y₂O₃(s) + 6H⁺(aq) + 3e⁻ → 2Y³⁺(aq) + 3H₂O(l) E° = +1.97 V

AgBr(s) + e⁻ → Ag(s) + Br⁻(aq) E° = +0.65 V

The cell potential is given as 1.32 V. We know the reduction potential of one half-cell (AgBr/Ag), so we can use it to determine the reduction potential of the other half-cell (Y₂O₃/Y³⁺):

E°cell = E°reduction (Y³⁺) - E°reduction (AgBr)

1.32 V = E°reduction (Y³⁺) - 0.65 V

E°reduction (Y³⁺) = 1.32 V + 0.65 V

E°reduction (Y³⁺) = 1.97 V

Therefore, Y₃⁺ has a typical reduction potential of +1.97 V, which is the reduction potential of Y₃⁺ under standard conditions (1 M concentration, 1 atm pressure, 25°C temperature).

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Answer:

1. Ionic Compound

2.Lithium Oxide

3. Li2O

Explanation:

This is an Ionic Compound because (Lithium) the metal or cation is giving away electrons to the oxygen a nonmental or anion so both can have a full valence electron shell or have all the electrons on the outer most shell.

This is also Lithium Oxide because the metal goes first then the nonmetal and the ending of the non metal is ending with an ide.

The chemical formula is the periodic table symbols (usually the metal first) and then the charges are switched. E.g  Sample A has charge of 3 and sample B has a charge of 2. The formula would be A2B3. If the charge is 1 then you can leave it blank because it is understood. E.g is sample B had a charge of 1 then it would be AB3

The enthalpy change for the reaction is 443.5 kJ/mol, this indicates that the reaction is exothermic, meaning that heat is released during the reaction.

The enthalpy change of a reaction can be calculated using the following equation:

ΔH = ΣΔH(products) - ΣΔH(reactants)

where ΔH is the enthalpy change, Σ is the sum, and the subscripts refer to the enthalpy of the products and reactants.

Using the data provided, we can look up the standard enthalpy of formation (ΔHf°) for each substance involved in the reaction:

ΔHf°(CO(g)) = -110.5 kJ/mol

ΔHf°(H₂(g)) = -285.8 kJ/mol

ΔHf°(CH₃OH(l)) = -238.6 kJ/mol

To calculate the enthalpy change of the reaction, we can plug in these values into the equation:

ΔH = [ΔHf°(CH₃OH(l))] - [ΔHf°(CO(g)) + 2ΔHf°(H₂(g))]

ΔH = [-238.6 kJ/mol] - [-110.5 kJ/mol + 2(-285.8 kJ/mol)]

ΔH = [-238.6 kJ/mol] - [-682.1 kJ/mol]

ΔH = 443.5 kJ/mol

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The reaction in the experiment is exothermic reaction and 158.72 kilo joules at 34.51 degrees celsius is released.

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An increase in the internal energy of a system can be caused by several factors, such as heating, compression, and work done on the system.

In general, any process that adds energy to a system will result in an increase in its internal energy. For example, if heat is added to a gas in a closed container, the gas molecules will start moving faster and colliding more frequently, resulting in an increase in the internal energy of the gas.

Similarly, if work is done on a system, such as by compressing a gas, the internal energy of the gas will increase due to the increased kinetic energy of its molecules.

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Answer:

The change in cell potential under non-standard conditions can be predicted using the Nernst equation. The Nernst equation relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).

In this case, the concentrations of Al³+ and Ni²+ are not standard. According to the Nernst equation, an increase in the concentration of the oxidizing agent ion in the cathodic side of the system (reduction reaction) should result in an increase in cell potential. Conversely, an increase in the concentration of the reducing agent in the anodic side of the system (oxidation reaction) should result in a decrease in cell potential.

Explanation:

The Nernst equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).

The Nernst equation is given by:

Ecell = E°cell - (RT/nF)lnQ

where Ecell is the cell potential under non-standard conditions, E°cell is the standard cell potential, R is the universal gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred in the balanced equation for the cell reaction, F is Faraday’s constant, and Q is the reaction quotient.

At room temperature (25°C or 298K), this equation can be simplified to:

Ecell = E°cell - (0.0592/n)logQ

where logQ is the base-10 logarithm of the reaction quotient.

To calculate the change in cell potential using this equation, you need to know the standard cell potential (E°cell), the number of moles of electrons transferred in the balanced equation for the cell reaction (n), and the reaction quotient (Q) under non-standard conditions.

We can see that an atom with 14 protons, 14 neutrons, and 16 electrons is unstable, and has a -2 charge.

So the correct option is the third one.

An atom with 14 protons, 14 neutrons, and 16 electrons is not stable. The number of protons in an atom, also known as its atomic number, determines its element and its chemical properties. In this case, the atom has 14 protons, which corresponds to the element silicon (Si) on the periodic table.

For an atom to be stable, it should have a balanced number of protons and electrons. Electrons are negatively charged particles that orbit the nucleus of an atom in energy levels or electron shells. The number of electrons in a stable atom should be equal to the number of protons, resulting in a neutral charge overall.

In this case, the atom has 14 protons and 16 electrons, which means it has two more electrons than protons, resulting in a net charge of -2. This is an example of an ion.

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The electrode indicated is the cathode

The contents of the beaker are the electrolyte

Electrons flow from the cathode  to the anode

Dilute sulfuric acid will give colorless gases at both electrodes

The electrodes used could be made from platinum since they are inert.

Electrolysis is a process of using an electric current to drive a non-spontaneous chemical reaction.

This involves the use of an electrolytic cell, where an external power source is used to drive a chemical reaction that would otherwise not occur.

During electrolysis, positive and negative ions move toward the electrodes (positively charged anode and negatively charged cathode) where they undergo oxidation and reduction reactions respectively.

The process is commonly used in the production of metals, such as aluminum, and in the purification of copper and other metals. It is also used in various industrial and laboratory applications, including electroplating, electrorefining, and the production of hydrogen and oxygen gas.

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1) New plants start to grow and sprout: As temperatures warm up in the spring, seeds that have been dormant over the winter start to germinate and new plants start to grow.

Sprout is a type of plant that is grown from a seed. It is usually a young plant that has just started to grow, and is usually eaten raw or lightly cooked. Some of the most common sprouts are alfalfa, broccoli, and mung bean. Sprouts are highly nutritious, as they contain a large amount of vitamins, minerals, protein, and antioxidants. They are a great addition to salads, sandwiches, and stir-fries, and can also be used as a topping for soups or curries. Sprouts are easy to grow at home, and can be enjoyed in a variety of dishes.

2) The leaves on trees start to change colors: As temperatures warm up in the spring, deciduous trees start to produce new leaves that are lighter in color than the leaves they had in the fall.

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To measure reaction heat under isochoric conditions, a bomb calorimeter is needed.

This device is designed to maintain a constant volume (isochoric) during the reaction, allowing for accurate measurement of reaction heat. To improve the accuracy of the coffee cup calorimeter for measuring reaction heat under isobaric conditions, a modification that could be made is to use a stirring device to ensure uniform mixing of the reactants and to minimize heat loss to the surroundings.

Additionally, a lid with a small hole could be placed over the top of the calorimeter to prevent heat loss while still allowing for pressure equalization. These modifications would help to minimize errors in heat measurement and improve the accuracy of the results obtained.

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The absorbance of this solution at this wavelength would be 0.287.

We can use the relationship between percent transmittance (%T) and absorbance (A) :

% T = 10 ^ ( - A )

Rearranging this equation, we can solve for A:

A = - log (%T / 100 )

Substituting the given value, we get:

A = - log ( 51. 6 / 100) = - log (0. 516) = 0. 287

Therefore, the absorbance of this solution at 550 nm is 0.287 .

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Answer:

B. It is hard to change the temperature of the substance