what does the nitrogen cycle formed​

Step 1

The reaction is written and balanced:

4 Rb + O2 =>2 Rb2O

-----------

Step 2

Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100

The actual yield is provided by the exercise = 39.7 g

----------

Step 3

Determine the limiting reactant. The molar masses are needed to solve this:

For Rb) 85.4 g/mol

For O2) 32 g/mol

Procedure:

4 Rb + O2 =>2 Rb2O

4 x 85.4 g Rb ----- 32 g O2

82.4 g Rb ----- X = 7.72 g O2 are needed

For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.

--------

Step 4

Determine the theoretical yield from the limiting reactant:

The molar mass Rb2O) 187 g/mol

Procedure:

4 x 85.4 g Rb ------ 2 x 187 g Rb2O

82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield

---------

Step 5

% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.

Answer: % yield = 44 %

In an ecosystem in the hilly areas the should built its factory away from the river so that pollution does not reach the river.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Calculating this expression will give you the pressure of propane in the tank at 24.0 °C.

To determine the pressure (in psi) of propane in a tank of liquid propane at 24.0 °C, you can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization (∆Hvap) and the gas constant (R).

The Clausius-Clapeyron equation can be written as:

ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)

Where:

P1 is the known pressure at temperature T1 (-42.0 °C in this case)

P2 is the unknown pressure at temperature T2 (24.0 °C in this case)

∆Hvap is the enthalpy of vaporization (18.8 kJ/mol)

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

First, convert the given temperatures to Kelvin:

T1 = -42.0 °C + 273.15 = 231.15 K

T2 = 24.0 °C + 273.15 = 297.15 K

Now, we can rearrange the equation to solve for P2:

P2 = P1 * exp((∆Hvap/R) * (1/T1 - 1/T2))

P1 is given as 1 atm, which is equivalent to 14.7 psi.

Substituting the values into the equation:

P2 = 14.7 psi * exp((18.8 kJ/mol / (8.314 J/(mol·K))) * (1/231.15 K - 1/297.15 K))

Calculating this expression will give you the pressure of propane in the tank at 24.0 °C.

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Answer:

Sulfuric acid has a higher density than water, which causes the acid formed at the plates during charging to flow downward and collect at the bottom of the battery.

Answer:

The concentration of H₃O⁺ ions increase whereas the concentration of the OH⁻ decreases.

Explanation:

The water, H₂O, has its own equilibrium thus:

2H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

Where the water equilibrium constant, Kw, is defined as:

Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

The addition of a base represents the increasing of [OH⁻]. that means the concentration of OH⁻ ions.

Based on LeCh principle, in an equilibrium, the addition of a product produce the decreasing in concentration of the other products trying to counteract the effect, that means the concentration of H₃O⁺ ions decreases.

The blue arrow is pointing to an air bubble that is trapped in the column.

The other answer choices do not match with what the blue arrow is pointing to. A dry patch caused by improper mixing of the slurry when pouring the column would not be visible as a single point in the column. The stationary phase in column chromatography is usually a solid or a gel-like material and would not appear as an air bubble. A desiccant bead is typically added to the column to absorb moisture and would not be visible as a single point in the column. Therefore, the correct answer is that the blue arrow is pointing to an air bubble that is trapped in the column.

Components in the mixture interact differently with the stationary phase, causing them to move at different rates through the column and ultimately become separated.

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When an organism dies, bacteria degrade it. During the decomposition process, carbon dioxide is emitted into the environment or water. Hence, it has no impact on the decomposers.

When animals breathe or decay (decompose), carbonate rocks are worn, forest fires happen, and volcanoes erupt, carbon dioxide is naturally released into the atmosphere.

Thus, even during process of cellular respiration, carbon dioxide is expelled into the atmosphere.  Hence, it has no impact on the decomposers.

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To determine the stability of the given closed-loop transfer functions, we need to analyze the locations of the poles in the complex plane. If all the poles have negative real parts, the system is stable.

If any pole has a positive real part, the system is unstable. If there are poles on the imaginary axis or have zero real parts, further analysis is required.

a) Transfer function: 8Kc / (5S + 1)

The pole of this transfer function is at S = -1/5. Since the real part of the pole is negative, the system is stable.

b) Transfer function: 8Kc / (S^2 + 3S + 2)

The poles of this transfer function are at S = -1 and S = -2. Both poles have negative real parts, so the system is stable.

c) Transfer function: 8Kc / (S^3 + 6S^2 + 12S + 8 + 8Kc)

To determine the stability of this transfer function, we need to find the roots of the polynomial in the denominator. The characteristic equation is S^3 + 6S^2 + 12S + 8 + 8Kc = 0.

By analyzing the roots of the characteristic equation for different values of Kc, we can determine the stability. If all the roots have negative real parts for any value of Kc, the system is stable.

However, without a specific value for Kc provided, we cannot conclusively determine the stability of the system. The stability of the system will depend on the specific value of Kc and the locations of the roots of the characteristic equation. Further analysis is required to determine the stability.

In summary:

a) The system is stable.

b) The system is stable.

c) The stability requires further analysis as it depends on the specific value of Kc and the locations of the roots of the characteristic equation.

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Answer:

The equinoxes are the only time when both the Northern and Southern Hemispheres experience roughly equal amounts of daytime and nighttime. On Earth, there are two equinoxes every year: one around March 21 and another around September 22.

Explanation:

Katja will use different colors of paper in her experiment to test her hypothesis and determine which color of paper will increase the most in temperature when exposed to sunlight.

By using a variety of colors, Katja can compare the results and determine if her hypothesis is correct or if another color of paper increases the most in temperature.

This experiment will provide valuable information about the effects of different colors on temperature and can be useful in a variety of applications, such as in the development of materials that are resistant to heat or for designing energy-efficient buildings that reflect sunlight.

Ultimately, the use of different colors of paper in this experiment allows for a more thorough and accurate analysis of the relationship between color and temperature.

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Answer:

Explanation:

A covalent bond consists of the mutual sharing of one or more pairs of electrons between two atoms. These electrons are simultaneously attracted by the two atomic nuclei. A covalent bond forms when the difference between the electronegativities of two atoms is too small for an electron transfer to occur to form ions.

Hey!!!

The answer of your question is Gas.

Explanations:

Solid have a fixed shape and volume.

Liquid have a definite volume but no definite shape

Gas neither have fixed shape nor a fixed volume.

Hope it helps...

Good luck on your assignment

Answer:

Two factors that contribute to the efficiency of a chromatographic separation are the amount of sample being separated and the separation time. A larger sample size or a longer separation time will generally result in a more efficient separation. Other factors that can impact the efficiency of a chromatographic separation include the nature of the sample, the type of chromatographic technique being used, and the characteristics of the separation column.

Answer:

c. Some of the atoms in the paper became gases.

Explanation:

The statement that correctly explains why the ashes of burnt paper have less mass than the original paper is C. Some of the atoms in the paper became gases.

It should be noted that the mass of an object simply has to do with the amount of matter or substance that makes up that particular object. The mass of an object is measured in kilograms.

In this case, the mass of the ashes of burnt paper will be less than the original paper. This is due to the fact that when a substance is burnt, there will be a reduction in the mass just as it is depicted in this case.

Therefore, some of the atoms in the paper became gases.

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To determine if MgCO3 will precipitate from the solution, we need to compare the ion product (Q) with the solubility product (Ksp) of MgCO3. The ion product (Q) is calculated by multiplying the concentrations of the ions involved in the dissociation of MgCO3.

The balanced equation for the dissociation of MgCO3 is:

MgCO3(s) ⇌ Mg2+(aq) + CO32-(aq) Given that the concentration of Mg2+ is 1.5×10^−3 M, we can calculate the concentration of CO32- using stoichiometry. Since 1 mole of MgCO3 dissociates to give 1 mole of Mg2+ and 1 mole of CO32-, the concentration of CO32- is also 1.5×10^−3 M.

The ion product (Q) is then calculated as:

Q = [Mg2+][CO32-] = (1.5×10^−3 M)(1.5×10^−3 M) = 2.25×10^−6

Comparing Q with the solubility product (Ksp) of MgCO3 (6.82×10^−6), we find that Q < Ksp. This means that the ion product is smaller than the solubility product, indicating that no MgCO3 will precipitate from the solution. Therefore, based on the given concentrations and the solubility product of MgCO3, no MgCO3 will precipitate from the solution when 50.0 mg of Na2CO3 is added to 150.0 ml of the solution.

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Answer:

d

Explanation:

i think it's d, but I'm not sure. my answer is a little different but maybe i used a different number for the density of lead (11.35g/cc)

Answer: If you double the force you would have to double the Acceleration.

Explanation:Since the mass does not change as the acceleration increases, we can say that force is equal to acceleration. Therefore, if you double the force you double the acceleration.

The organic product formed when cyclopentene reacts with iodobenzene in the presence of triethylamine and a Pd(0) catalyst is 1-iodocyclopentene.

The reaction you are describing is a cross-coupling reaction known as a Heck reaction. In this reaction, cyclopentene acts as the alkene component, while iodobenzene acts as the aryl halide component. Triethylamine serves as a base to remove the acidic proton from cyclopentene and facilitate the reaction.

The Pd(0) catalyst helps in the formation of the carbon-carbon bond between cyclopentene and iodobenzene. The reaction proceeds by the Pd(0) catalyst coordinating with the alkene and the aryl halide, followed by insertion of the alkene into the Pd-aryl bond. This generates a Pd-alkyl intermediate, which then undergoes reductive elimination to form the final product, 1-iodocyclopentene.

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The reaction of cyclopentene with iodobenzene in the presence of triethylamine and a Pd(0) catalyst leads to the formation of substituted cyclopentene products.

When cyclopentene reacts with iodobenzene in the presence of triethylamine and a Pd(0) catalyst, the organic product formed is a substituted cyclopentene. The reaction is an example of a palladium-catalyzed cross-coupling reaction known as the Heck reaction.

In this reaction, the Pd(0) catalyst facilitates the insertion of the aryl group from iodobenzene into the double bond of cyclopentene. Triethylamine acts as a base to deprotonate the cyclopentene, making it more reactive towards the Pd(0) catalyst.

The specific product formed depends on the regioselectivity and stereoselectivity of the reaction. In the case of cyclopentene, the reaction can occur on either side of the double bond, resulting in two possible products.

One possible product is the 3-substituted cyclopentene, where the iodobenzene adds to the carbon adjacent to the double bond. Another possible product is the 4-substituted cyclopentene, where the iodobenzene adds to the carbon opposite the double bond.

Both products can be obtained in varying amounts depending on the reaction conditions and the electronic and steric factors involved.

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The characteristic of Byzantine art is Maniera greca.

Byzantine art is known for its distinctive style influenced by the Eastern Roman Empire. One of its notable characteristics is the use of "Maniera greca," which refers to the incorporation of elements from Greek art. This influence can be seen in the stylized figures, flat and elongated proportions, and intricate patterns used in Byzantine artwork. Maniera greca also includes the use of gold backgrounds, rich colors, and detailed ornamentation.

These features are commonly found in Byzantine mosaics, icons, and frescoes, reflecting the religious and spiritual nature of the art. This is the correct answer. Maniera greca, or "Greek style," refers to the influence of Greek art and aesthetics on Byzantine art. It is characterized by stylized figures, elongated proportions, intricate patterns, and the use of gold backgrounds. This style was prominent in Byzantine mosaics, icons, and frescoes.

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Answer:

Yes, A  toothpick can be a good model for showing the strength and brittleness of salt as It will break and splinter just like salt.

What are the properties of  salt ?

Salts are ionic in nature due to the presence of ions.

They are brittle, hard and crystalline solids.

Salt is white, odorless and it has a salty taste.

Toothpick will break and splinter just like salt, exclamation, because it breaks easy and if we use a hammer on a tooth pick its gonna break and splinter just like salt.

Therefore , A  toothpick can be a good model for showing the strength and brittleness of salt as It will break and splinter just like salt.

Explanation: hope this helps

c. decreases, increases. this is correct answer.

When a nonvolatile (covalent) solute is added to a solvent, the freezing point of the solution decreases.

This phenomenon is known as freezing point depression. The presence of the solute disrupts the regular arrangement of solvent molecules, making it more difficult for them to form the solid lattice structure during freezing. As a result, the solution requires a lower temperature to freeze compared to the pure solvent.

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The ideal gas law is the state equation of a hypothetical ideal gas. The volume of the nitrogen gas is 51 I.

The general gas equation, commonly known as the ideal gas law, is the state equation of a hypothetical ideal gas. The ideal gas equation is formulated as:

PV = nRT

In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the total amount of ideal gas that is measured in terms of moles, R is the universal gas constant, and T is the temperature.

As it is given that the pressure on the nitrogen is 720 torr, the temperature of the gas is 20°C(293 k), and the moles of nitrogen are 2.0 moles. Therefore, using the ideal gas equation the volume of the nitrogen can be written as,

\(pv=nRT\\\\v = \dfrac{nRT}{p}\\\\v = \dfrac{2 \times 62.396 \times 293}{720}\\\\v = 50.7834 \approx 51\)

Thus, the volume of the nitrogen gas is 51 I.

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Answer:

answer in picture

Explanation:

Answer:

pH = 0.984

Explanation:

Molarity LiOH = 2.1x10⁻²M

HCl will react with LiOH as follows:

HCl + LiOH → H₂O + LiCl

1 moles of HCl reacts per mole of LiOH

Moles of each reactant in solution are:

HCl = 0.477L ₓ (0.17mol / L) = 0.08109 moles HCl

LiOH = 0.253L ₓ (2.1x10⁻² mol / L) = 5.313x10⁻³ moles of LiOH.

That means LiOH is the limiting reactant and excess moles of HCl that will remain in solution are:

0.08109 mol - 0.005313mol = 0.0758 moles HCl

As HCl dissociates in water as H⁺ and Cl⁻ ions, you will have in solution 0.0758 moles of H⁺

pH = -log [H⁺] and [H⁺] = moles H⁺ / L of solution.

Volume of the mixture in liters is: 0.477L + 0.253L = 0.730L.

That means [H⁺] is 0.0758 moles of H⁺ / 0.730L = 0.1038M

Replacing:

pH = -log [H⁺]

pH = -log [0.1038]

Explanation:

just take the given moles and multiply it with the ratio of molecules to moles to get your answer

The number of nitrogen molecules in 6.0 moles of N2 is = 3.6 × 10²⁴

Avogadro's number( 6.02 × 10²³) is equal to the number of molecules in 1 mole of substance.

That is,

1 mole of N2 = 6.02 × 10²³

6 moles of N2 = 6 × 6.02 × 10²³

= 36.12 × 10²³

= 3.6 × 10²⁴

Therefore, the number of nitrogen molecules in 6.0 moles of N2 is = 3.6 × 10²⁴

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3.88 moles of TiF4 will result from the reaction as F2 will act as a limiting reagent.

The balanced chemical equation for the reaction between titanium (Ti) and fluorine (F2) to form titanium tetrafluoride (TiF4) is:

Ti + 2 F2 → TiF4

The stoichiometric ratio of Ti to F2 in this reaction is 1:2, which means that one mole of Ti reacts with two moles of F2 to produce one mole of TiF4.

Given that 7.76 moles of F2 are reacted with 14.38 moles of Ti, we need to determine which reactant is limiting and calculate the amount of TiF4 that can be produced based on the limiting reactant.

The amount of Ti required to react with 7.76 moles of F2 is:

7.76 moles F2 x (1 mole Ti / 2 moles F2) = 3.88 moles Ti

Since there are 14.38 moles of Ti available, Ti is in excess and F2 is the limiting reactant.

The amount of TiF4 produced from the reaction of 7.76 moles of F2 is:

7.76 moles F2 x (1 mole TiF4 / 2 moles F2) = 3.88 moles TiF4.

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There are approximately \(3.011 * 10^{23}\) atoms of N, \(4.817 * 10^{24}\) atoms of H, \(1.505 * 10^{23}\) atoms of C, and \(4.517 * 10^{23}\) atoms of O in 0.250 moles of \((NH_4)_2CO_3\).

The molecular formula of ammonium carbonate (\((NH_4)_2CO_3\)) contains 2 nitrogen atoms (N), 8 hydrogen atoms (H), 1 carbon atom (C), and 3 oxygen atoms (O).

To determine the number of atoms in 0.250 moles of \((NH_4)_2CO_3\), we can use Avogadro's number, which is \(6.022 * 10^{23}\) atoms per mole.

Number of N atoms = 2 x Avogadro's number x 0.250 = \(3.011 * 10^{23}\)atoms of N

Number of H atoms = 8 x Avogadro's number x 0.250 = \(4.817 * 10^{24}\) atoms of H

Number of C atoms = 1 x Avogadro's number x 0.250 = \(1.505 * 10^{23}\) atoms of C

Number of O atoms = 3 x Avogadro's number x 0.250 = \(4.517 * 10^{23}\) atoms of O

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Answer:

Answer: Carbon dioxide and Water

Explanation:

» Butane is an alkane. When alkanes undergo combustion [ oxidization ], carbondioxide and water are formed.

Steel ball bearings and marbles can be separated through manual sorting due to their difference in appearance.

Before answering the question, let us define the given choices.

The given separation techniques are not suitable for the separation of steel balls bearings, and marbles. Instead, the sorting separation technique must be used. This is used for manually sorting coarse materials based on their physical characteristic, which in this case is their difference in appearance.

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The correct balanced reaction for the given cell diagram is option B:

2 Al(s) + 3 Cd²⁺(aq) → 2 Al³⁺(aq) + 3 Cd(s)

The balanced reaction for the given cell diagram can be determined by looking at the half-reactions occurring at each electrode. The cell diagram is written in the form:

Al(s) | Al³⁺(aq) || Cd²⁺(aq) | Cd(s)

The half-reaction at the anode (left side) is the oxidation half-reaction, and the half-reaction at the cathode (right side) is the reduction half-reaction.

From the cell diagram, we can identify the half-reactions as follows:

Anode (oxidation): Al(s) → Al³⁺(aq) + 3e⁻

Cathode (reduction): Cd²⁺(aq) + 2e⁻ → Cd(s)

To balance the number of electrons transferred in both half-reactions, we need to multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:

2 Al(s) → 2 Al³⁺(aq) + 6e⁻

3 Cd²⁺(aq) + 6e⁻ → 3 Cd(s)

Now, we can combine the balanced half-reactions to obtain the balanced overall reaction:

2 Al(s) + 3 Cd²⁺(aq) → 2 Al³⁺(aq) + 3 Cd(s)

Therefore, the correct balanced reaction for the given cell diagram is option B:

2 Al(s) + 3 Cd²⁺(aq) → 2 Al³⁺(aq) + 3 Cd(s)

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