Write a balanced equation sodium chloride and silver nitrate for the precipitation reaction 2 Which product in the above reaction is the precipitate?

Answer:

1. single replacement

Explanation:

Single replacement can be represented by AB+C ⇄B+AC, which matches the picture

Double replacement and double displacement are the same kind of reaction and can be represented by AB+CD ⇄BD+AC

Decomposition reaction can be represented by AB ⇄A+B

Answer:

A. inverse

Explanation:

Boyle's Law is a relationship between pressure and volume. In this relationship, pressure and volume have an inverse relationship when temperature is held constant. If there is a decrease in the volume there is less space for molecules to move and therefore they collide more often, increasing the pressure.

After considering the given data we conclude that the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K is [tex]4.8*10^{-17}[/tex].

To evaluate the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K, we can apply the following steps:
Write the balanced chemical equation for the reaction.
[tex]N_2(g) + 2NO_2(g) - - - > 2NO_2(g)[/tex]
Write the expression for Kp for each of the two reactions given.
[tex]Kp_1 = [NO]^2/[N_2][O_2][/tex]
[tex]Kp_2 = [NO_2]^2/[NO]^2[O_2][/tex]
Here,[tex][NO], [N_2], [O_2], and [NO_2][/tex]are the partial pressures of the respective species at equilibrium.
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex] into the expression for Kp for the desired reaction.
[tex]Kp = Kp_2/Kp_1 = ([NO_2]^2/[NO]^2[O_2])/([NO]^2/[N_2][O_2]) = ([NO_2]^2[N_2])/[NO]^4[/tex]
Restructure the expression for Kp to solve for the unknown partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex], and the given partial pressures of N2 and O2, into the expression for [NO].
[tex][NO] = \sqrt(Kp_1[N_2][O_2])[/tex]
Stage the calculated value of [NO] into the expression for[tex][NO_2][/tex] to obtain the equilibrium partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
[tex][NO_2]^2 = (2.4*10^{12} )(4.0*10^{-31} )([N_2][O_2])^2/[N_2][/tex]
[tex][NO_2]^2 = 9.6*10^{-19} [N_2][O_2]^2[/tex]
[tex][NO_2] = \sqrt(9.6*10^{-19} [N_2][O_2]^2)[/tex]
[tex][NO_2] = 3.1*10^{-10} atm[/tex]
Stage the evaluated values of [NO] and[tex][NO_2][/tex] into the expression for Kp to obtain the equilibrium constant.
[tex]Kp = ([NO_2]^2[N_2])/[NO]^4[/tex]
[tex]Kp = (9.6*10^{-19} )(3.1*10^{-10} )^2/[/\sqrt(4.0*10^{-31*1} atm)]^4[/tex]
[tex]Kp = 4.8*10^{-17}[/tex]
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The order of the increasing pH of the solution is Hydrochloric acid, Acetic acid, Ammonia, and Sodium hydroxide.  

Molecular formula of compounds:  Ethane: C2H6, 1-Butyne: C4H6, Propene: C3H6 and Cyclooctane: C8H16

Rank of solutions in order of increasing pH:

Hydrochloric acid (HCl) is a strong acid that dissociates completely in water. Its pH is very low (less than 1) and it is strongly acidic. It is an acidic solution.  It's a 1.0 M solution of hydrochloric acid.

Ammonia (NH3) is a weak base that does not completely dissociate in water. It is a weak base with a pH greater than 7 but less than 10. It is a basic solution.  It's a 1.0 M solution of ammonia.

Acetic acid (CH3COOH) is a weak acid that does not completely dissociate in water. It has a pH of around 3.5-4.0, making it a weakly acidic solution. It is an acidic solution.  It's a 1.0 M solution of acetic acid.

Sodium hydroxide (NaOH) is a strong base that dissociates completely in water. It has a very high pH (greater than 14), making it a strongly basic solution. It is a basic solution.  It's a 1.0 M solution of sodium hydroxide.

So, from the above explanation, the order of the increasing pH of the solution is

Hydrochloric acid, Acetic acid, Ammonia, and Sodium hydroxide.  

Step 2: Molecular formula of compounds:

Ethane: C2H6, contains 2 carbon atoms

1-Butyne: C4H6, contains 4 carbon atoms

Propene: C3H6, contains 3 carbon atoms

Cyclooctane: C8H16, contains 8 carbon atoms

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Answer: B. nebula

Explanation:

Answer:

B. nebula

Explanation:

Birth takes place inside hydrogen-based dust clouds called nebulae.

-TheUnknownScientist

Answer:

The killers who choose poison for killing the victim are the one who does not want to confront the victim. When a knife or a bullet is used to kill a person he may struggle and can cause harm to killer also.  

Explanation:

Poison is the most easiest way to kill a person without any struggle. The poison can be given to a person in a juice or through an injection. The poison entered in the body of victim will cause his heart to cease gradually and he will not have energy to struggle with the killer to save his life.

The Mesopotamians discovered copper (Cu) in 9000 BC.

The half-reaction Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) is a reduction half-reaction, while the half-reaction Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex] is an oxidation half-reaction.

The given half-reactions are:

Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq)

Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex]

The oxidation and reduction reactions are defined as follows:

Oxidation reaction: A half-reaction that includes the loss of electrons is referred to as an oxidation reaction. The oxidation number of the species involved in the reaction is increased in this process.

Reduction reaction: A half-reaction that involves gaining electrons is referred to as a reduction reaction. The oxidation number of the species involved in the reaction is decreased in this process.

Now let us identify which half-reaction is oxidation and which is reduction:

Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) (reduction reaction)

Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex](oxidation reaction)

Thus, the half-reaction Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) is an example of a reduction half-reaction, while the half-reaction Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex] is an example of an oxidation half-reaction.

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Answer:

The answer to this question is HCl(aq) + H2O(l) = H3O+(aq) + Cl-(aq). HCl is considered a strong acid meaning it is 100% ionized in solution. The equilibrium constant is therefore very large.

Answer:

the lungs and heart give oxygen to the cells and the sensory system gives messages to the brain

Explanation:

The enthalpy change of the reaction is -456.8 kJ.

The balanced chemical equation of the reaction is as follows:

Ca(s) + 1/2O2(g) + CO2(g) → CaCO3(s)

Given reactions:

Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.1 kJ/mol

CaCO3(s) → CaO(s) + CO2(g) ΔH = 178.3 kJ/mol

The target reaction is a combination of the two given reactions, hence the enthalpy change of the target reaction is the sum of the enthalpies of the two given reactions. It means that:

ΔHrxn = ΔH1 + ΔH2

We have:ΔH1 = -635.1 kJ/mol

ΔH2 = 178.3 kJ/mol

Therefore,ΔHrxn = ΔH1 + ΔH2= (-635.1 kJ/mol) + (178.3 kJ/mol)= -456.8 kJ/mol

Therefore, the enthalpy change of the reaction is -456.8 kJ.

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Answer:

Radiometric dating

Explanation:

Radiometric dating would be the most effective way since it is a technique that can establish the age of objects older than a few thousand years.

Barium nitrate and sodium chloride do not react with each other because both of them are soluble in water. The chemical equation for the reaction between barium nitrate and sodium chloride is given below.

Ba(NO₃)₂ + 2NaCl → BaCl₂ + 2NaNO₃

The reaction between barium nitrate and sodium chloride is a double displacement reaction, where barium cation is exchanged with sodium cation and nitrate anion is exchanged with chloride anion. But the reaction does not occur due to the solubility of barium nitrate and sodium chloride in water.The solution of barium nitrate and sodium chloride will remain clear and colorless with no precipitation forming. In fact, it is a method of testing the presence of sulfate ions in a solution. A small amount of barium nitrate is added to the solution to form barium sulfate. Since barium sulfate is insoluble, it forms a white precipitate.

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Answer:

valence electrons

Explanation:

i just took the test

To calculate the binding energy per nucleon for each isotope of silver, we need to determine the total binding energy of each isotope and divide it by the number of nucleons (protons and neutrons) in each isotope.

The binding energy per nucleon represents the average energy required to remove a nucleon from the nucleus. It is a measure of the stability of the nucleus. To calculate the binding energy per nucleon, we need to determine the total binding energy and the number of nucleons for each isotope. First, we calculate the total binding energy for each isotope using the mass-energy equivalence equation E = mc², where E is the energy, m is the mass, and c is the speed of light.

For the first isotope, 10747Ag, with an atomic mass of 106.905095 g/mol, we multiply the atomic mass by the Avogadro's number to obtain the mass of one mole of atoms. Then we divide the mass by the number of atoms in one mole to get the mass of one atom. The difference between the mass of one atom and the mass of the individual nucleons (protons and neutrons) gives the mass defect. Multiplying the mass defect by c² gives the binding energy of one atom. Finally, dividing the binding energy by the number of nucleons in the isotope gives the binding energy per nucleon.

We repeat the same calculations for the second isotope, 10947Ag, with an atomic mass of 108.904754 g/mol. The resulting values represent the binding energy per nucleon for each isotope of silver.

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The bromine atom in BrF3 is sp3d hybridized, in BrO2- it is sp3 hybridized, and in BrF5 it is sp3d2 hybridized.

The hybridization of bromine in each of the following compounds is as follows:

a) BrF3:

The central atom, bromine (Br), in BrF3 undergoes sp3d hybridization. It forms three sigma bonds with three fluorine atoms and has two lone pairs of electrons. The three sigma bonds are formed by overlapping of the sp3d hybrid orbitals of bromine with the 2p orbitals of the fluorine atoms.

b) BrO2-:

The central atom, bromine (Br), in BrO2- undergoes sp3 hybridization. It forms two sigma bonds with two oxygen atoms and has two lone pairs of electrons. The two sigma bonds are formed by overlapping of the sp3 hybrid orbitals of bromine with the 2p orbitals of the oxygen atoms.

c) BrF5:

The central atom, bromine (Br), in BrF5 undergoes sp3d2 hybridization. It forms five sigma bonds with five fluorine atoms and has one lone pair of electrons. The five sigma bonds are formed by overlapping of the sp3d2 hybrid orbitals of bromine with the 2p orbitals of the fluorine atoms.

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals in a molecule. It helps to explain the molecular geometry and bonding in the molecule. The hybridization of an atom depends on the number of sigma bonds it forms and the number of lone pairs of electrons it possesses.

In summary, the bromine atom in BrF3 is sp3d hybridized, in BrO2- it is sp3 hybridized, and in BrF5 it is sp3d2 hybridized. The hybridization of bromine determines the shape and geometry of the molecules and provides insights into the nature of their chemical bonding.

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Answer:

2 NO2 -> 2 O2 + N2

is a decomposition reaction

The most common hazard from exposure to non-ionizing radiation is tissue heating.

The primary risk associated with exposure to non-ionizing radiation lies in the potential for tissue heating. Non-ionizing radiation lacks the necessary energy to dislodge electrons from atoms or molecules, yet it can still transfer energy to matter by inducing molecular vibration.

This vibration, in turn, can elevate the temperature of the exposed tissue, consequently giving rise to a spectrum of health concerns, including:

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Answer:

B. One element changes into another.

Explanation:

We can see from the equation that uranium is changed to thorium and helium. Transmutation is the process by which one element is changed to another through radioactive decay, nuclear bombardment, or other similar processes.

Answer:b

Explanation:

C-14 is an isotope of the element carbon and it differs from the carbon atom in the fact that it has two more neutrons. The correct option is B) C-14 has two more neutrons.An isotope is referred to as the element of the same atomic number, and isotopes possess a different number of neutrons in their nucleus.

The correct option is B

For example, the carbon-14 isotope of carbon has 6 protons and 8 neutrons, whereas the carbon-12 isotope of carbon has 6 protons and 6 neutrons .This column of the periodic table represents the halogen family. This is a family of reactive elements. Members of the halogen family differ in that they have different atomic numbers and mass numbers; they have the same number of valence electrons. Hence, the correct option is C) They have different atomic numbers and mass numbers; they have the same number of valence electrons.

The halogens are a family of highly reactive non-metallic elements from Group 17 (the seventh column) of the periodic table. The halogens are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).Young stars, just beginning their life in the galaxy, would contain mostly hydrogen and helium. Hence, the correct option is D) hydrogen and helium. During the star formation process, these two gases are pulled together by gravity to create nuclear fusion reactions. As a result of this reaction, the hydrogen is converted into helium.

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Answer:

first one is right.

there are 117 elements, so 2nd option is wrong.

elements are made up of one type of atoms, so 3rd is also wrong.

most elements are metal. so 3rd is wrong

Answer:

i think its 62.50 :) hope this helped you out

The new concentration of the solution is 0.30 M.

To calculate the new concentration of the solution, we need to consider the dilution process.

Given information:

Initial volume of the HCl solution = 20 mL

Initial concentration of the HCl solution = 0.60 M

Final volume of the diluted solution = 40 mL

The number of moles of HCl in the initial solution can be calculated using the formula:

Moles = Concentration × Volume (in liters)

Initial moles of HCl = 0.60 M × (20 mL / 1000 mL/L) = 0.012 moles

During dilution, the moles of solute remain constant. Therefore, the moles of HCl in the diluted solution will also be 0.012 moles.

The final concentration of the solution can be calculated using the formula:

Final Concentration = Moles / Volume (in liters)

Final concentration = 0.012 moles / (40 mL / 1000 mL/L) = 0.30 M

Therefore, the new concentration of the solution after dilution is 0.30 M.

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A  layer represents the radiative zone which is at the centre of the cross-sectional diagram.

The layer of a star that lies just outside the core, to which radiant energy is transferred from the core in the form of photons.

In the Sun, the region between the solar core at 0.2 of the Sun's radius and the outer convection zone at 0.71 of the Sun's radius is referred to as the radiation zone, although the core is also a radiative region.

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B

A is inner core B is outer core and there is a few other layers but from what I know looking at this the second layer has always been the radioactive zone.

If you get this right I would love brainiest

thats kinda hot not even gon'

Answer:

B

Explanation:

Can't really explain it sorry buy heating the solvent tends to increase the solubility

The step that would increase the solubility of the gas in the liquid solvent is to increase the temperature of the gas and liquid.

Solubility is the mixing of any substance into another substance and the fully mixing of the ions of the substance to another substance's ions.

Solubility depends upon temperature, increasing temperature increase the solubility.

In the mixing of gas and liquid, the liquid is solvent and gas is solute, always.

Thus, the correct option is B.

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The reaction of (CH₃)₃CO- with iodomethane or 1-chlorobutane will increase the reaction rate due to an increased frequency of collisions and an increased concentration of electrophiles.

The reaction between (CH₃)₃CO- (tert-butoxide ion) and iodomethane or 1-chlorobutane is an example of an SN₂ nucleophilic substitution reaction. In SN₂ reactions, the rate-determining step involves the simultaneous attack of the nucleophile (tert-butoxide ion) on the electrophile (iodomethane or 1-chlorobutane) with the inversion of configuration.

When the concentration of iodomethane is increased in the reaction mixture, it will lead to an increase in the reaction rate. This increase occurs due to the collision theory and the effect of concentration on the reaction kinetics.

According to the collision theory, for a reaction to occur, the reactant molecules must collide with sufficient energy and proper orientation. By increasing the concentration of iodomethane, there are more iodomethane molecules available in the reaction mixture. Consequently, there will be an increased frequency of collisions between the nucleophile (tert-butoxide ion) and the electrophile (iodomethane), increasing the likelihood of successful collisions leading to a reaction.

Additionally, the rate of an SN₂ reaction is dependent on the concentration of both the nucleophile and the electrophile. In this case, as the concentration of iodomethane increases, the concentration of the electrophile increases. This increase in electrophile concentration will result in a higher reaction rate, as more electrophiles are available for reaction with the nucleophile.

Therefore, increasing the concentration of iodomethane or 1-chlorobutane in the reaction mixture will increase the reaction rate due to an increased frequency of collisions and an increased concentration of electrophiles.

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Answer:

what type of question?

Explanation:

whats your favorite color???