What is the percent of nitrogen (N) by mass
in a 55.0 g sample of Fe(NO3)3? The molar
mass of Fe(NO3)3 is 241.86 g/mol.

The weight of fluorine released in to the air over 6 month will be 0.365 kg

Freon is a non-combustible gas that is used as a refrigerant in air conditioning applications.

This freon undergoes an evaporation process over and over again to help produce cool air that can be circulated throughout your AC system

Total number of weeks = 6 x 4 = 24

Mass leak rate of freon = 31.68 g/week

Mass leak rate of flourine = fluorine mass in freon / molecular mass of freon x leak rate

                                          = 19 x 3 / 118.5 x  31.68

                                          =  15.23 gm/week

Total fluorine leak in 6 month = 15.23 x 24 = 365.72 gm = 0.365 kg

​

Hence, The weight of fluorine released in to the air over 6 month will be 0.365 kg

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Explanation:

For those who like to step beyond the challenge, the mass of the materials could be measured to check the methodology. Prior to separating the mixture, measure the mass of the total mixture. After separating and drying all the materials, measure the mass of each material. Add the three masses together. The total should equal the initial mass if all materials were recaptured during the separation process.

Iron particles will be drawn towards a magnet when one is used. In this manner, aluminum powder and iron powder can be separated. Due to the fact that iron is magnetic but the other three are not.

Magnets are used in magnetic separation, a waste management technique, to remove metal from the trash. As the materials are gathered together and segregated before processing, this is most frequently seen in single and mixed streams of recycling.

This method can be used to eliminate unwanted metals from products. All materials are kept pure by it. Recycling facilities frequently use magnetic separation to isolate metals, purify ores, and separate components from recycling.

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Answer:

Sodium

Calcium

Magnesium

Aluminium

Zinc

Iron

Copper

Explanation:

In quantum mechanics, the angular momentum quantum number "l" defines the shape of the atomic orbital. The l value is an integer ranging from 0 to (n-1) where n is the principal quantum number.

Therefore, for an electron in the L(n=2) shell of an atom, the possible values of the angular momentum quantum number l would range from 0 to 1, since n=2.

This is because the L shell is the second shell, which has n=2. Therefore, it can have subshells with l=0 and l=1, also known as the s and p subshells respectively.

The angular momentum quantum number also has an effect on the energy of the electron, with higher l values having higher energy.

Thus, the possible values of the angular momentum quantum number l for an electron in the L(n=2) shell of an atom are l=0 and l=1.

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Answer:

The K type thermocouple is used in the test to measure the temperature, especially in the burning chamber..

Two thermometers need to be used during the smoking process. One thermometer measures the air temperature of the smoker. The other thermometer is used to monitor the food product temperature..

Explanation:

Hope it helps you..

J-just correct me if I'm wrong, okay?..(;ŏ﹏ŏ)

B-but..

Y-your welcome in advance!!..

(;ŏ﹏ŏ)(ㆁωㆁ)

Answer:       electron on the shell  of the atom

Explanation:  that what I trying to remember

Cytochrome c is a small protein molecule located in the mitochondrial membrane and serves as an electron carrier in the electron transport chain.

Cytochrome c is a protein that plays an important role in the electron transport chain. It transfers electrons between Complex III and Complex IV in the electron transport chain, helping to create a proton gradient across the inner mitochondrial membrane.

This proton gradient is used to produce ATP, the energy currency of the cell. Without cytochrome c, the electron transport chain would not be able to function efficiently and produce the energy needed for cellular processes.

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The events in the order of top to bottom for a digital recording to become sound waves through a speaker include:

This is defined as a pattern of disturbance caused by movement of energy along a medium.

The order of events which produces the sound heard by audience can be seen above.

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The amount of current in this wire is equal to 1.813 C.

Several electrons =  3.4 × \(10^{19}\) electrons.

Time = 3 seconds.

Charge of electron = 1.6 × \(10^{-19}\) C

First, we would calculate the entire quantity of charge using this formula to get the flow of charge in this wire;

Q = Nc

Q = 3.4 × \(10^{19}\) × 1.6 × \(10^{-19}\)

Q = 5.44 C

In mathematics, the following formula yields the charge quantity:

Q = It

The amount of the fee is Q.

The current is I.

The time, t, is expressed in seconds.

Making I the formula's subject, we have:

I = Q ÷ t

By changing the parameters in the formula, we obtain;

I = 5.44 ÷ 3

I = 1.813 C

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The of for  the decay of uranium-238 to lead-206 is 4.5 × yr. The rock was formed 2.00 x 10⁹ yrs .

Option B is correct.

At = A₀ x (1/2)^(t / half life)

where

At = amount remaining after time = t has elapsed

A₀ = initial amount

(Ut / U₀) = (1/2)^(t / half life)

log(Ut / U₀) = log[(1/2)^(t / half life)]

log(Ut / U₀) = (t / half life) x log(1/2)

t = half life x log(Ut / U₀) / log(1/2)

Let  Pb₀ = 0

Ut = 1mg.. ie.. we have 1mg of U₂₃₈ today

Pbt = 0.313mg

Ut + (U converted to lead) = U₀

0.313mg Pb x (1 m mole Pb / 207.2 mg Pb) x (1 m mole U / 1 m mole Pb) x (238.03mg U / 1mmole U) = 0.3596mg U

Ut = 1mg

U₀ = 1 mg + 0.3596mg = 1.3596mg

t = 4.5x10⁹ yr x log (1mg / 1.3596mg) / log(1/2) = 2.00x10⁹ yrs

Radium-226 is formed when uranium-238 breaks down in a nuclear fission reaction, resulting in radon gas (radon-222). Radon rots to frame a progression of girl nuclides, the majority of which are alpha-molecule delivering isotopes, for example, polonium-210.

Uranium-238, the most common isotope in uranium metal, has a half-existence of around 4.5 billion years; that is, a portion of the particles in any example will rot in that measure of time.

A chain reaction, also known as an atomic explosion, is the result of this. Three new neutrons and some binding energy are released when a uranium-235 atom takes in a neutron and splits into two new atoms. A uranium-238 atom either loses or absorbs two neutrons, stopping the reaction.

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Answer:

bismuth

Explanation:

i think if not i am so sorry

The new boiling point is found essentially the same way as the freezing point, except replacing the freezing point quantities with boiling point quantities:

Where,

DeltaTb, is the change in boiling point temperature

Kb, is the molal boiling point

m, is the molality

We have to calculate first which is the molality of the solution.

Now, we replace the know data into the first equation:

The boiling point of a solution will be the boiling temperature of water alone plus the temperature delta found:

Tb=100°C + 0.46°C=100.46°C

The boiling point of the solution will be 100.46°C

C/ If Fe is greater than Fg, the negatively charged oil droplet will move freely toward the negatively charged plate.

In the Millikan oil droplet experiment, the negatively charged oil droplets are subjected to an electric field created by the charged plates. The electric force (Fe) acts on the oil droplet in a direction opposite to the gravitational force (Fg). When Fe is greater than Fg, the electric force overcomes the gravitational force, causing the negatively charged oil droplet to experience an upward force. As a result, the oil droplet moves freely upward toward the negatively charged plate.

Option B is incorrect because if Fe is equal to Fg, the forces balance each other, resulting in a stationary droplet. However, the question states that Fe is increased until it is greater than Fg, implying that the droplet is no longer stationary but moves in response to the electric force.

Therefore, option C is the correct answer, as it describes the effect of an electric field on the motion of a negatively charged oil droplet in the Millikan oil droplet experiment.

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Answer:

3.Test for unsaturation is carried out by either Baeyer's test or bromine water test.

(a) What is Baeyer's reagent?

Ans:

Baeyer's reagent is an alkaline solution of Potassium Permanganate \(\bold{KMnO_4}\) .

(b) How does it detect presence of double bond in organic compound? Write the reaction involved.

Ans:

When a dilute alkaline solution of \(\bold{KMnO_4}\) is added to the given unsaturated organic compound, if the purple color of permanganate solution disappears, the presence of double bond is indicated.

For example:

2\(\bold{KMnO_4}\) + 2KOH \(\longrightarrow\)\(\bold{2K_2MnO_4 + H_2O+[O]}\)

\(\bold{CH_2=CH_2(ethene) +H_2O+[O]\longrightarrow CH_2OH-CH_2OH (ethylene \:glycol :Colourless)}\)

(c) Write the product when ethyne is treated with Baeyer's reagent.

Ans:

Product will not produce as ethyne doesn't contain double bond.

(d) Benzene (an aromatic hydrocarbon) does not give Baeyer's test though it has double bonds. Why?

Ans:

As Benzene is an unsaturated compound. The structure of benzene contains 3 alternate single double bonds.

(e) Write the reaction for bromine water test for ethyne.

Ans:

Reaction:

\(\bold{CH_2CH_2(ethene)+Br_2( red \: color) \longrightarrow CH_2Br-CH_2Br}\)  

(1,2-dibromoethane (colorless}

Kolbe electrolytic method can be used to prepare alkane, alkene and alkyne.

(a) At which electrode, alkane, alkene or alkyne is obtained?

Ans:

alkane, alkene or alkyne is obtained at anode.

(b) Which electrolyte should be chosen for alkane, alkene or alkyne for the method? Suggest an example of electrolyte for each.

Ans:

alkane: Potassium or sodium salt of Carboxylic acid

alkene: Potassium or sodium salt of dicarboxylic acid

alkyne: Potassium or sodium maleate

(c) If potassium salt of dicarboxylic acid is taken for the method, which hydrocarbon is prepared?

Ans: If potassium salt of dicarboxylic acid is taken for the method, Alkene hydrocarbon is prepared.

(d) Write the detail process for the electrolysis of aqueous potassium maleate.

Ans: Attachment

(e) Suggest a demerit of the method.

Ans:

Demerits:

The reaction between aqueous sodium phosphate and calcium nitrate can be written as shown below;

Na3PO4(aq) + 3Ca(NO3)2(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

To get the net ionic equation, we will first write the balanced ionic equation, and then cancel out the spectator ions. The balanced ionic equation is given as shown below;

3Na⁺(aq) + PO₄³⁻(aq) + 3Ca²⁺(aq) + 6NO₃⁻(aq) → Ca₃(PO₄)₂(s) + 6Na⁺(aq) + 6NO₃⁻(aq)

We then cancel out the spectator ions to obtain the net ionic equation as shown below;

3Na⁺(aq) + PO₄³⁻(aq) + 3Ca²⁺(aq) → Ca₃(PO₄)₂(s)

The balanced molecular equation is given as follows;

Na3PO4(aq) + 3Ca(NO3)2(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

The complete ionic equation can be obtained as shown below;

3Na⁺(aq) + PO₄³⁻(aq) + 3Ca²⁺(aq) + 6NO₃⁻(aq) → Ca₃(PO₄)₂(s) + 6Na⁺(aq) + 6NO₃⁻(aq)

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Answer:

2.03 × 10⁻⁴ moles

Explanation:

To convert atoms to moles, use Avogadro's number (6.022 × 10²³).

(1.22 × 10²⁰ atoms)/(6.022 × 10²³ atoms/mol) = 2.03 × 10⁻⁴ mol

Answer:

7.35×1023 atoms

Explanation:

Answer:

Copper Sulphate

Explanation:

Cu (Copper)

SO4 (Sulphate)

Given :

Formula applied :

\(\boxed {Q = mc \triangle T}\)

Let's solve for c !

⇒ 237,000 = 20 × c × 15

⇒ c = 237,000 ÷ 300

⇒ \(\boxed {c = 790 J kg^{-1} K^{-1}}\)

∴ The specific heat capacity of granite is 790 J kg⁻¹ K⁻¹.

Answer:

C) Lighting a matchstick

Explanation:

Mixing an acid and base will cause a neutralization reaction, a type of double displacement.

Photosynthesis is just a series of chemical reactions, but don't deal with the use of O2, but rather it's creation.

Reacting sodium and chlorine would make NaCl in a synthesis reaction.

Answer:

C) Lighting a matchstick

Explanation:

The equilibrium constant for a base ionization reaction is called the base ionization constant. This corresponds to option b.

The base ionization constant, also known as the acid dissociation constant (Ka) for bases, is a quantitative measure of the extent to which a base dissociates or ionizes in water.

It represents the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium for the ionization reaction of a base.

The base ionization constant is denoted as Kb, and it is specific to the particular base being considered. It helps determine the strength of a base and provides valuable information about its behavior in aqueous solutions. By comparing the values of Kb for different bases, their relative strengths and reactivity can be assessed.

Options a, c, and d are incorrect because they do not accurately represent the term commonly used for the equilibrium constant of a base ionization reaction. Therefore, the correct option is B.

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The standard emf of the  voltaic cell based on Ag+(aq)/Ag(s) and Fe3+(aq)/Fe2+(aq) half-cells is 0.03 V.

To determine the reactions at the cathode and anode, we need to know the standard reduction potentials of the given half-cells. The standard reduction potentials are as follows:

Ag+(aq) + e- → Ag(s) E° = +0.80 V
Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V
Since the higher reduction potential will occur at the cathode and the lower one at the anode, we have:
Cathode reaction: Ag+(aq) + e- → Ag(s)
Anode reaction: Fe3+(aq) + e- → Fe2+(aq)
The cathode reaction involves the reduction of silver ions to silver metal, and the anode reaction involves the reduction of ferric ions to ferrous ions. And the standard emf of the cell is:

E°cell = E°cathode - E°anode

E°cell = +0.80 V - (+0.77 V)

E°cell = 0.03 V

In this voltaic cell, the cathode reaction is Ag+(aq) + e- → Ag(s), and the anode reaction is Fe3+(aq) + e- → Fe2+(aq). The standard emf of the cell is 0.03V.

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A voltaic cell is based on Ag+(aq)/Ag(s) and Fe3+(aq)/Fe2+(aq) half-cells. The cathode reaction is \(Ag^+(aq) + e^{-}\ - > Ag(s)\) and the anode reaction is \(Fe^{3+}(aq)\ - > Fe^{2+}(aq) + e^{-}\).

Determine the cathode and anode reactions

To determine the cathode and anode reactions in the voltaic cell based on the \(Ag^+(aq)/Ag(s)\ and\ Fe^{3+}(aq)/Fe^{2+}(aq)\) half-cells, we need to compare the reduction potentials of each half-reaction. The reaction with the higher reduction potential will occur at the cathode (where reduction takes place), while the reaction with the lower reduction potential will occur at the anode (where oxidation takes place).

The standard reduction potentials for the half-reactions are as follows:

\(Ag^+(aq) + e^{-}\ - > Ag(s) \ \ \ \ \ \ \ \ E\° = +0.80 V\)

\(Fe^{3+}(aq) + e^{-} \ - > Fe^{2+}(aq) \ \ \ \ \ \ \ \ E\° = +0.77 V\)

Since the standard reduction potential for \(Ag^+(aq)/Ag(s)\) is higher than that of \(Fe^{3+}(aq)/Fe^{2+}(aq)\), the \(Ag^+(aq)/Ag(s)\) half-reaction will occur at the cathode (reduction) and the \(Fe^{3+}(aq)/Fe^{2+}(aq)\) half-reaction will occur at the anode (oxidation).

A.) Cathode reaction:

\(Ag^+(aq) + e^{-}\ - > Ag(s)\)

B.) Anode reaction:

\(Fe^{3+}(aq)\ - > Fe^{2+}(aq) + e^{-}\)

It's important to note that the oxidation half-reaction is typically written in the reverse direction (as an oxidation) to emphasize the loss of electrons.

Therefore, the cathode reaction is \(Ag^+(aq) + e^{-}\ - > Ag(s)\) and the anode reaction is \(Fe^{3+}(aq)\ - > Fe^{2+}(aq) + e^{-}\).

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Answer:

15°

Explanation:

At the lowest temperature the gas will have the lowest rate of diffusion because diffusion deals with the motion of particles.

The gas with the lowest temperature will be the slowest to diffuse.

Answer:

15º

Explanation:

The rate of diffusion is lower when the temperature is lower. Since 15º is the lowest temperature of the answers, it has the lowest rate of diffusion.

Answer:

D

Explanation:

I think it is D

Answer:

To test if a lipid is saturated or unsaturated iodine is added. If the iodine changes from brown to clear the lipid is unsaturated. If the iodine does not change colors the lipid is saturated. To test for the degree of lipid saturation iodine is added to the unsaturated lipid.

Explanation:

Emerging scientific ideas are new theories or ideas that are gaining attention in the scientific community, but have not yet been fully accepted or confirmed.

Emerging ideas refer to the new and innovative ideas or theories that have yet to gain full scientific acceptance. While a scientific consensus is a view or theory that has been universally accepted and confirmed by multiple experiments or research, an emerging scientific idea is a new and unproven theory or idea that is gaining attention in the scientific community. These emerging ideas may also be referred to as scientific hypotheses. In contrast to scientific consensus, emerging scientific ideas have not yet been subjected to rigorous testing and confirmation.

They are generally proposed to explain new observations or experimental results, which have not yet been fully understood or explained by established scientific theories. Emerging scientific ideas can have the potential to challenge the current scientific consensus. If an emerging scientific idea is found to be valid, it can ultimately lead to the establishment of a new scientific consensus. For example, the emerging scientific idea of the Higgs boson particle led to the discovery of a new field in particle physics, which is now an established scientific consensus.

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The formula for ammonium sulfide is (NH4)2S. There are 20.00 hydrogen atoms in 2.50 mol of ammonium sulfide.

To determine the number of hydrogen (H) atoms in 2.50 mol of ammonium sulfide, we need to consider the stoichiometry of the compound. From the formula (NH4)2S, we can see that each ammonium ion (NH4+) contains four hydrogen atoms.

So, in 1 mole of ammonium sulfide, we have 2 moles of ammonium ions, which means there are 2 * 4 = 8 hydrogen atoms.

To find the number of hydrogen atoms in 2.50 mol of ammonium sulfide, we can use the following conversion:

Number of hydrogen atoms = (2.50 mol) * (8 hydrogen atoms/1 mol)

Number of hydrogen atoms = 20.00

Therefore, there are 20.00 hydrogen atoms in 2.50 mol of ammonium sulfide.

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Answer:

The Lewis dot diagram is supposed to have dots on each side. What's incorrect is that there isn't a dot on the bottom, only the left and right side and the top. What's correct about this is that there are 5 outer valence electrons, and they correctly put 5 dots, even though they're in the wrong place.

Explanation:

The given Lewis dot diagram of nitrogen is correct for the number of valence electrons for nitrogen. However, they are not correctly placed because two of them are lone pairs and only these two are shown as paired. Remaining three electrons have to be represented as single electrons.

Lewis dot structure of an element represent its valence electrons shown  as dots around the atom. The lone pair of electrons are shown as paired dots and other electrons as single dots.

For compounds, a Lewis diagram shows the bonded and non bonded electrons between bonded atoms. Nitrogen is 7th element and it contains 7 electrons in which 5 are valence electrons.

Nitrogen contains one lone pair of electrons. They have to be shown as paired dots in the Lewis structure. In the given diagram there are two pairs of electrons which is incorrect. Three electrons have to be shown as single dots.

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Answer:

-1300. kJ  

Explanation:

We have two equations:

1. P₄(s)       +3O₂(g) ⟶ P₄O₆(s);                 ΔH₁  = -1640.1 kJ

2. P₄O₁₀(s)               ⟶ P₄(s)      + 5O₂(g); ΔH₂ =  2940.1 kJ

From these, we must devise the target equation:

3. P₄O₆(s)  + 2O₂(g) ⟶ P₄O₁₀(s);               ΔH = ?

The target equation has P₄O₆(s) on the left, so you reverse Equation 1.

When you reverse an equation, you reverse the sign of its ΔH.

4. P₄O₆(s)                 ⟶ P₄(s)       +3O₂(g); ΔH₁  =  1640.1 kJ

Equation 4 has P₄ on the right. That is not in the target equation.

You need an equation with P₄ on the left, so you reverse Equation 2.  

5. P₄(s)      + 5O₂(g)  ⟶ P₄O₁₀(s);               ΔH₂ = -2940.1 kJ

Now, you add equations 4 and 5, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 3:

4. P₄O₆(s)                      ⟶ P₄(s)        + 3O₂(g);      ΔH₁ =   1640.1 kJ

5. P₄(s)       + 2(5)O₂(g) ⟶ P₄O₁₀(s);                       ΔH₂ = -2940.1 kJ

3. P₄O₆(s)   + 2O₂(g)     ⟶ P₄O₁₀(s);                       ΔH   = -1300.  kJ

ΔH for the reaction is -1300. kJ

assuming all of the xenon has reacted, The formula of the product is XFe2

In this situation, xenon and fluorine are reacting to form a binary compound. The reaction can be viewed by the below balanced chemical equation:

Xe + 2F2 -> XeF2

The reactants, xenon and fluorine, are initially present at partial pressures of 1.24 atm and 10.10 atm, respectively.Since the reactants are initially present at a total partial pressure of 1.24 atm + 10.10 atm = 11.34 atm, and the final partial pressure of fluorine is 7.62 atm, the partial pressure of the xenon fluoride compound must be 11.34 atm - 7.62 atm = 3.72 atm.

we can set up the following equation:1.24 atm + 10.10 atm = 3.72 atm + 7.62 atm

Solving for the number of moles of fluorine gives:

10.10 atm - 3.72 atm = 7.62 atm

= 6.38 atm

Since the pressure of a gas is directly proportional to the number of moles of the gas present,

n = P / (R * T

= 6.38 atm / (0.08206 Latm/molK * 298 K / 100 mL)

= 0.5 molThe number of moles of xenon present can be calculated in a similar way:

n = P / (R * T

= 1.24 atm / (0.08206 Latm/molK * 298 K / 100 mL)

= 0.05 mol

Since the ratio of moles of xenon to moles of fluorine in the product is 1:2, the formula of the product must be XeF2.

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