Calculate the voltage generated by a hydrogen - oxygen fuel cell at 73.5°C
when the partial pressures of hydrogen and oxygen are 19.8 atm.

An atom with too many neutrons relative to protons is said to be unstable and can undergo radioactive decay to become more stable. There are several types of radioactive decay, including alpha decay, beta decay, and gamma decay.

In alpha decay, the unstable atom emits an alpha particle, which is a helium nucleus consisting of two protons and two neutrons. This results in a new nucleus with two fewer neutrons and two fewer protons.

In beta decay, the unstable atom emits a beta particle, which is either an electron or a positron. When an atom emits an electron, one of its neutrons is converted into a proton, and the atomic number of the atom increases by one. When an atom emits a positron, one of its protons is converted into a neutron, and the atomic number of the atom decreases by one.

In gamma decay, the unstable atom emits a gamma ray, which is a high-energy photon. Gamma decay does not change the number of protons or neutrons in the nucleus but instead releases excess energy.

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D. Al of the following is a reactant in the chemical equation

In a chemical equation, the substance or substances to the left of the arrow are referred to as reactants. A material that is present when a chemical reaction first begins is known as a reactant. Products refer to the material or substances to the right of the arrow. A material that is present following a chemical reaction is known as a product.

Methane (CH4) and oxygen (O2) are the reactants and carbon dioxide (CO2) and water are the products in this chemical process. (H2O). This illustration demonstrates that chemical bonds may form and break during a chemical process. The forces that keep the atoms of a molecule together are known as chemical bonds.

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Starting with 8.00 g of Phosphorus-32 (32P) with a half-life of 14.0 days, after 98.0 days, 0.125 g of  32P will remain.

The half-life of a radioactive isotope is the time required for half of the original sample to decay. In this case, the half-life of 32P is 14.0 days, which means that after 14.0 days, half of the 32P will decay, leaving 4.00 g.

To find out how much 32P remains after 98.0 days, we need to determine the number of half-lives that have passed. Dividing 98.0 days by 14.0 days gives us 7.

Therefore, after 7 half-lives, the amount of 32P that remains can be calculated as:

Amount remaining = (1/2)⁷ x 8.00 g = 0.125 g

Therefore, after 98.0 days, 0.125 g of 32P will remain.

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The reaction that takes place when 2 teaspoons of supernatant liquid is added to the purple camote peel extract soaked with rubbing alcohol overnight is the formation of a purple pigment.

The purple pigment is created when the alcohol and steel wool vinegar react with the camote peel extract to break down the cell walls and release the pigment. This reaction is further enhanced by the addition of the supernatant liquid, which helps to dissolve the pigment and make it more easily visible.

The reaction that takes place when 2 teaspoons of supernatant liquid is added to the flower alcoholic extract of bougainvillea petal soaked with rubbing alcohol overnight is the formation of a pinkish-red pigment.

The pinkish-red pigment is created when the alcohol and steel wool vinegar react with the petal extract to break down the cell walls and release the pigment. This reaction is further enhanced by the addition of the supernatant liquid, which helps to dissolve the pigment and make it more easily visible.

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The wave property that should be tested in this experiment is absorption, which refers to the extent to which a material can absorb sound waves. The correct answer is option a.

By testing how different materials interact with sound waves from the straw horn, the student can gain insight into the properties of those materials and their ability to absorb sound.

A. Wave property: absorption

Method: playing the straw horn in a room with hard surfaces and a room with soft surfaces

To test this property, the student should play the straw horn in a room with hard surfaces, such as walls and floors made of concrete or tile, and a room with soft surfaces, such as walls and floors made of carpet or drapes.

By comparing the sound produced in each room, the student can observe how the sound waves interact with different materials and how effectively each material absorbs the sound.

This method allows the student to investigate how different materials absorb sound waves and how this affects the sound produced by the straw horn. This information can be valuable in understanding how sound travels in different environments and how to optimize sound quality in different settings.

The correct answer is option a.

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The net ionic equation when the ammonium nitrate is dissolved in the water :

NH₄NO₃(s) + H₂O(l) ⇄ NH₄⁺(aq)  + NO₃⁻(aq)

The component that will ionizes in the aqueous solution that is the ammonium ion. The nitrate ion is that does not ionize in the aqueous solution.

The acid-base hydrolysis in equilibrium that is the established when the ammonium nitrate is dissolved in the water, the net ionic equation is as :

NH₄NO₃(s) + H₂O(l) ⇄ NH₄⁺(aq)  + NO₃⁻(aq)

The ions has the equal and the oppisite charges. They both can combine in the electrically neutral ratio of the 1:1.  The net ionic equation can be depicts by the molecules and the ions.

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The result is negative, this means the reaction is not spontaneous under standard conditions. In other words, a lithium-sulfur battery cannot be constructed under standard conditions.

To calculate the potential for the reaction of Li(s) with S(s), we need to use the reduction potentials and the Nernst equation:

Ecell = Ereduction(cathode) - Ereduction(anode)

where Ereduction is the reduction potential, cathode is the reduction half-reaction occurring at the cathode (where reduction occurs) and anode is the oxidation half-reaction occurring at the anode (where oxidation occurs).

In this case, Li(s) is the anode and S(s) is the cathode. So, we need to flip the sign of the reduction potential for the anode:

Ecell = E(S2-/S) - (-E(Li+/Li))

Ecell = 0.48 V - 3.05 V

Ecell = -2.57 V

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The answer is (d) cannot solve using 5% approximation rule.

The balanced equation for the reaction is:

C(g) + e(g) ⇌ 2W(g)

The equilibrium constant expression is given by:

Kc = [W]^2 / [C][e]

At equilibrium, let's assume that x moles of C react with x moles of e to produce 2x moles of W. Therefore, the equilibrium concentrations are:

[C] = (3.5 - x) mol/L

[e] = (x) mol/L

[W] = (2x) mol/L

Substituting these values :

Kc = [(2x)^2] / [(3.5 - x)(x)]

Simplifying this expression:

4x^2 + 2.34x - 8.19 = 0

Solving this quadratic equation :

x = (-2.34 ± sqrt(2.34^2 - 4(4)(-8.19))) / (2(4))

x = (-2.34 ± 3.64) / 8

We can ignore the negative root as it does not make physical sense. Therefore:

x = 0.4575 mol/L

Thus, the concentration of e at equilibrium is:

[e] = 0.4575 mol/L

Therefore, the answer is (d) cannot solve using 5% approximation rule.

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In the equation 2h2 + o2 + 2h2o, the two hydrogen molecules (H2) react with one oxygen molecule (O2) to form two molecules of water (H2O). This reaction is known as combustion and it requires a certain ratio of hydrogen to oxygen in order for the reaction to take place.

Here correct answer is D)

In this equation, the ratio of hydrogen to oxygen is 2:1. This means that for every one liter of hydrogen, two liters of oxygen are needed in order for the reaction to take place.

In answer to the questions, a) one liter of hydrogen would react with two liters of oxygen, b) one liter of hydrogen would react with 22.4 liters of oxygen, c) 22.4 liters of hydrogen would react with one liter of oxygen, and d) two liters of hydrogen would react with one liter of oxygen.

This equation is a great example of the law of conservation of mass, as the total number of atoms on each side of the equation remain the same

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D. Solution A will take longer to dissolve and might have some undissolved salt remaining at the bottom of the solution due to the low temperature.

In Solution A, the salt is in large clumps, the solvent is cold water, and the agitation is slow stirring. These factors contribute to a slower dissolution process. Large clumps have less surface area exposed to the solvent, cold water has less energy to break the ionic bonds between salt ions, and slow stirring provides less agitation to promote dissolution.

Consequently, it will take longer for the salt to dissolve in Solution A, and there might be undissolved salt remaining at the bottom.

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Metamorphic rock that breaks into sheets when broken is called "foliated metamorphic rock." Foliated metamorphic rocks are characterized by their layering or alignment of minerals, which occurs during the metamorphic process. This distinct feature makes it possible for the rocks to break along these layers, forming sheets when fractured.

The process of metamorphism, which involves the transformation of pre-existing rocks due to changes in temperature, pressure, or mineral composition, causes the minerals within the rock to reorient themselves. This reorientation leads to the alignment of minerals, creating a parallel orientation of platy or elongated minerals within the rock.

Examples of foliated metamorphic rocks include slate, phyllite, schist, and gneiss. Slate, for instance, is derived from shale and is characterized by its fine-grained texture and well-defined cleavage, allowing it to break easily into thin sheets.

Phyllite, formed from slate, exhibits a slightly coarser texture and a glossy sheen due to the re-crystallization of its constituent minerals. Schist, a medium to coarse-grained rock, is characterized by its platy minerals like micas, which also contribute to its sheet-like breaking pattern. Lastly, gneiss, formed at even higher metamorphic grades, displays distinct banding due to the segregation of its minerals but may also break into sheets depending on the mineral alignment.

In summary, a metamorphic rock that breaks into sheets when broken is known as a foliated metamorphic rock. This unique property is a result of the alignment of minerals during the metamorphic process, creating distinct layers that allow the rock to fracture along these planes. Examples of foliated metamorphic rocks include slate, phyllite, schist, and gneiss, each exhibiting varying degrees of foliation and textural features due to different metamorphic conditions.

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The product is HI

There are six product molecule

Hydrogen is the limiting reactant

There is one iodine molecule in excess

To determine the limiting reactant in a chemical reaction, you need to compare the number of moles of each reactant present to the stoichiometry of the balanced chemical equation.

The limiting reactant is the reactant that is completely consumed in a chemical reaction, and which therefore limits the amount of product that can be formed. The other reactant, which is not completely consumed, is called the excess reactant.

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Answer:

In Valence Bond Theory (VBT), the water molecule is formed by overlapping of two hydrogen 1s orbitals with two hybridized oxygen orbitals. The oxygen atom in the water molecule has two unpaired electrons in two 2p orbitals and two paired electrons in two 2s orbitals. It hybridizes the 2s and 2p orbitals to form four hybridized sp3 orbitals. These four sp3 hybridized orbitals point towards the corners of a tetrahedron.

The two hybridized orbitals of oxygen containing unpaired electrons overlap with the 1s orbitals of two hydrogen atoms. This overlapping results in the formation of two O-H sigma (σ) bonds. The two remaining hybridized orbitals containing the paired electrons do not participate in bond formation.

The bond angle in the water molecule is 104.5°, which is less than the tetrahedral angle (109.5°) because the two lone pairs of electrons on the oxygen atom exert greater repulsion than the two bonding pairs. This causes the bonding pairs to be pushed closer together, resulting in a smaller bond angle.

The number of moles of krypton in a cylinder containing 17 L of krypton at 22.8 atm and 112 degrees Celsius is 6.47 moles.

There seems to be a typo in the question as it states that the cylinder contains Argon (Ar) but then asks for the number of moles of Krypton (Kr). Assuming the gas in the cylinder is Krypton, we can use the ideal gas law to calculate the number of moles:

PV = nRT

where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the gas constant (0.082 L·atm/mol·K), and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin:

T = 112°C + 273.15 = 385.15 K

Now we can plug in the values and solve for n:

n = PV/RT

n = (22.8 atm)(17 L)/(0.082 L·atm/mol·K)(385.15 K)

n ≈ 20.3 moles

Therefore, there are approximately 20.3 moles of Krypton in the cylinder.

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The volume of the balloon after the dry ice sublimes will be 3.40 L at STP.

The balanced chemical equation for the sublimation of solid CO₂ is:

CO₂(s) → CO₂(g)

At STP (standard temperature and pressure), which is 0°C (273.15 K) and 1 atm (101.325 kPa), one mole of any ideal gas occupies 22.4 L of volume. We can use this information to calculate the volume of CO₂ gas produced by the sublimation of 7.00 g of dry ice.

First, we need to convert the mass of dry ice to moles of CO₂ using the molar mass of CO₂, which is 44.01 g/mol:

7.00 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 0.159 moles CO₂

Next, we can use the ideal gas law to calculate the volume of CO₂ gas produced:

PV = nRT

where P is the pressure (1 atm), V is the volume we want to find, n is the number of moles of CO₂ (0.159 moles), R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature (273.15 K):

V = nRT/P = (0.159 mol)(0.08206 L·atm/mol·K)(273.15 K)/(1 atm) = 3.40 L

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a.It requires 1066 mL of 0.20 M KOH  to reach the equivalence point.

b.The equivalence point, the concentration of [Cl-], [K+], and [[tex]NH_{3}[/tex]] in the solution, is 0.175 M.

What is the Equivalence point?

The chemical equivalent between the added titrant and the sample analyte is called the equivalence point in a titration.

a. We need to know how many moles of [tex]NH_{4}Cl[/tex] are in the solution to calculate the volume of 0.20 M KOH needed to achieve the equivalence point.

First, we can determine how many moles  [tex]NH_{4}Cl[/tex] are present in the solution:

moles [tex]NH_{4}Cl[/tex] = mass / molar mass

moles [tex]NH_{4}Cl[/tex] = 11.4 g / 53.49 g/mol (molar mass of [tex]NH_{4}Cl[/tex])

moles [tex]NH_{4}Cl[/tex] = 0.2132 mol

At the equivalence point, all the [tex]NH_{4}Cl[/tex] has interacted with the KOH, resulting in an equal amount of moles of [tex]NH_{3}[/tex] and [tex]H_{2} O[/tex]. This suggests that 0.2132 moles of KOH are also needed to react with [tex]NH_{4}Cl[/tex] The volume of 0.20 M KOH required to react with 0.2132 mol can be determined using the equation for the reaction between [tex]NH_{4}Cl[/tex] and KOH:

[tex]NH_{4}Cl[/tex] + KOH → [tex]NH_{3}[/tex] + [tex]H_{2}O[/tex] + KCl

moles KOH = moles [tex]NH_{4}Cl[/tex]

                   = 0.2132 mol

volume of KOH = moles KOH / concentration of KOH

                          = 0.2132 mol / 0.20 mol/L

                           = 1.066 L or 1066 mL

Therefore, 1066 mL of 0.20 M KOH is required to reach the equivalence point.

b. At the equivalence point, an equal amount of moles of KOH and [tex]NH_{4}Cl[/tex] interacted to create [tex]NH_{3}[/tex], [tex]H_{2}O[/tex], and KCl.

We may determine the concentration of [Cl-] and [K+] in the solution following the reaction at the equivalence point by assuming volumes are additive:

moles KCl = moles [tex]NH_{4}Cl[/tex]

                 = 0.2132 mol

volume of solution = 150 mL + volume of KOH added

                                = 150 mL + 1066 mL

                                = 1216 mL

                                 = 1.216 L

[Cl-] = moles KCl / volume of solution

[Cl-] = 0.2132 mol / 1.216 L

[Cl-] = 0.175 M

[K+] = moles KCl / volume of solution

[K+] = 0.2132 mol / 1.216 L

[K+] = 0.175 M

The fact that the reaction between [tex]NH_{4}Cl[/tex]and KOH is a one-to-one reaction can be used to compute the concentration of [[tex]NH_{3}[/tex]]. As a result, 0.2132 mol of NH3 is likewise created at the equivalence point. Using the overall volume of the solution, we can get the [[tex]NH_{3}[/tex]] concentration:

[[tex]NH_{3}[/tex]] = moles [tex]NH_{3}[/tex]/ total volume of solution

[[tex]NH_{3}[/tex]] = 0.2132 mol / 1.216 L

[[tex]NH_{3}[/tex]] = 0.175 M

Therefore, at the equivalence point, the concentration of [Cl-], [K+], and [[tex]NH_3}[/tex]] in the solution is 0.175 M.

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1. Moles of CO₂ is 0.8047.

2. Moles of C is 1.6094.

3. C:H ratio is 1:3.

4. The Emperial formula is c6H6.

5. Emperical formula mass is 78g/mol.

1. Moles of CO₂ = 24.41 g / 44.01 g/mol = 0.5548 mol; Moles of H₂O = 14.49 g / 18.02 g/mol = 0.8047 mol

2. Moles of C = 0.5548 mol (1 C atom in CO₂); Moles of H = 0.8047 mol * 2 (2 H atoms in H₂O) = 1.6094 mol

3. C:H ratio = 0.5548:1.6094 ≈ 1:3 (divide by smallest value), but 1:2.89 is closer, which gives a ratio of 6:6 (multiply by 3 to get whole numbers)

4. Empirical formula: C₆H₆

5. Empirical formula mass: (6 * 12.01) + (6 * 1.01) = 78 g/mol. Molecular formula: (246 g/mol) / (78 g/mol) = 3; C₆H₆ * 3 = C₁₂H₁₂ (molecular formula)

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The volume of a 2.00 M solution that contains 3.75 moles of solute can be calculated using the formula V = n/C, where V is the volume of the solution, n is the number of moles of solute, and C is the concentration of the solution in units of M (moles per liter).

Plugging in the given values, we get V = 3.75 moles / 2.00 M = 1.88 L. Therefore, the volume of the solution is 1.88 liters. In 100 words, the volume of a solution can be determined by knowing the amount of solute present and the concentration of the solution.

This is because the concentration of a solution is defined as the amount of solute dissolved in a given volume of solution. Using the formula for concentration, we can determine the amount of solute dissolved in a given volume of solution.

Then, by rearranging the formula to solve for volume, we can determine the volume of the solution required to dissolve a specific amount of solute.

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The difference in the rate of sugar and acid reaction to the reaction between KI(aq) and Pb(NO₃)₂(aq) can be explained by the type of chemical bonding present in each case. In the case of sugar and acid, the reaction is a covalent bond breaking and forming process that occurs gradually and can take time to complete.

Covalent bonds are relatively strong and require more energy to break, which can result in slower reaction rates.

On the other hand, the reaction between KI(aq) and Pb(NO₃)₂(aq) involves the formation and breaking of ionic bonds. Ionic bonds are relatively weaker than covalent bonds and require less energy to break, resulting in faster reaction rates.

Additionally, the presence of water in the reaction between KI(aq) and Pb(NO₃)₂(aq) can also speed up the reaction by facilitating the movement of ions and increasing their collision frequency.

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The percentage composition by mass of Y in the hypothetical compound X2Y5Z3 is 34.53%.

To calculate the percentage composition by mass of Y in the hypothetical compound X2Y5Z3, we first need to calculate the total molar mass of the compound:

Total molar mass = (2 moles of X x 13.25 amu/mole) + (5 moles of Y x 69.23 amu/mole) + (3 moles of Z x 109.34 amu/mole)
Total molar mass = 26.50 amu + 346.15 amu + 328.02 amu
Total molar mass = 700.67 amu

Next, we can calculate the percentage composition by mass of Y:

percentage composition by mass of Y = (mass of Y / total molar mass) x 100%
percentage composition by mass of Y = (5 moles of Y x 69.23 amu/mole / 700.67 amu) x 100%
percentage composition by mass of Y = 34.53%

Therefore, the percentage composition by mass of Y in the hypothetical compound X2Y5Z3 is 34.53%.

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The mass of 6.9 moles of nitrous acid (HNO₂) is 324.3 grams.

To calculate the mass of 6.9 moles of nitrous acid (HNO₂), follow these steps:

1. Determine the molar mass of HNO₂.
2. Multiply the molar mass by the given moles (6.9 moles) to find the mass.

Step 1: Determine the molar mass of HNO₂.
HNO₂ consists of 1 hydrogen atom, 1 nitrogen atom, and 2 oxygen atoms.
- The atomic mass of hydrogen (H) is 1 g/mol.
- The atomic mass of nitrogen (N) is 14 g/mol.
- The atomic mass of oxygen (O) is 16 g/mol.

Molar mass of HNO₂ = (1 x 1) + (1 x 14) + (2 x 16) = 1 + 14 + 32 = 47 g/mol.

Step 2: Multiply the molar mass by the given moles (6.9 moles).
Mass of HNO₂ = 6.9 moles × 47 g/mol = 324.3 g.

So, the mass of 6.9 moles of nitrous acid (HNO₂) is 324.3 grams.

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The solubility of the gas increases to 0.780 g/L when the pressure is increased to 131 kPa.

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Thus, we can use the following equation to calculate the new solubility:

S₂ ÷ S₁ = P₂ ÷ P₁

where S₁ is the initial solubility, S₂ is the new solubility, P₁ is the initial pressure, and P₂ is the new pressure.

Plugging in the given values, we have:

S₂ ÷ 0.650 g/L = 131 kPa ÷ 109 kPa

Solving for S₂, we get:

S₂ = (0.650 g/L) × (131 kPa ÷ 109 kPa)

S₂ = 0.780 g/L

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The chemical equation for KOH dissociating in water to form ions is:

[tex]\rm KOH (aq) \rightarrow K^+(aq) + OH^-(aq)[/tex], which shows that KOH is a base.

A chemical equation is an illustration of a chemical reaction's reactants and products.

Equation for the dissociation of KOH:

[tex]\rm KOH (aq) \rightarrow K^+(aq) + OH^-(aq)[/tex]

In the above mentioned reaction, potassium ions ([tex]\rm K^+[/tex]) and hydroxide ions ([tex]\rm OH^-[/tex]) are generated by the dissociation of KOH.

Therefore, KOH is a base because it produces hydroxide ions ([tex]\rm OH^-[/tex]) when it dissociates in water.

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In constructive interference, the waves add together.

Interference is the phenomenon where two or more waves interfere with each other to form a resultant wave of greater, lower or same amplitude.

In constructive interference, the waves combine in such a way that their amplitudes are reinforced, resulting in a wave with a larger amplitude than the individual waves.

So, the correct answer is: a. they add

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In fire, the energy released is slower as compared to the explosion in which the energy released is faster and more damaging.

Fires and Explosions are phenomena that releases a high amount of heat and light into their surrounding. Both of them causes the surroundings to burn down if they are not performed or caused in a controlled environment.

However, the main difference between the two is the rate at which the energy is released. In a fire, the energy which is released be it heat energy or light energy, the energy is released slowly through combustion as compared to explosions. Fires basically involve a sustained combustion process.

In an explosion the energy that is released at an extreme rate, it creates shockwaves that can cause damage significantly to its surrounding. Explosions are a one-time event.

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If an electron is released during radioactive decay, the type of decay that has taken place is beta decay.

In beta decay, a neutron within the nucleus is converted into a proton, releasing an electron (also called a beta particle) in the process.

In alpha decay an alpha particle is emitted from the atomic nucleus and a new atomic nucleus is formed. So, no release of electron is there.

In gamma decay the unstable nuclei release excess energy by continuous electromagnetic process. This does not involve release of electron.

The electromagnetic decay also do not involve the release of an electron.

Thus option b is the correct answer.

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The addition of ammonia increase the solubility of slightly soluble salt AgCl as : ammonia forms very soluble complex ion by coordinating to Ag and removing it from solution. This shifts the solubility equilibrium to right.

When ammonia (NH₃) is added to a solution containing AgCl, it can coordinate with silver ions (Ag+) to form a complex ion called [Ag(NH₃)₂]+, which is highly soluble in water. This complex ion removes the Ag+ ions from the solution, thereby decreasing the concentration of Ag+ in the solution. According to Le Chatelier's principle, this will shift the equilibrium of AgCl dissolution reaction to the right, resulting in increase in the solubility of AgCl.

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While balancing equation write the physical state of reactants and products as well as any reaction conditions.

Reactants are the substances that are present at the start of a chemical reaction. They are typically the substances that are used up during the reaction and are converted into different products. Reactants are usually written on the left side of a chemical equation, while the products are written on the right side. Reactants are essential components of any chemical reaction and are essential in order for the reaction to take place. Reactants are also known as substrates or starting materials.

Balancing the Following Equations:

1) CuSO4(s) + 2KI(aq) → Cu2I2(s) + K2SO4(aq) + I2(g)

2) 2NH3(g) + O2(g) → 2NO(g) + 2H2O(g)

3) 3Fe2O3(s) + 4CO(g) → 6Fe(s) + 3CO2(g)

4) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

5) 2Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq)

6) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

7) 2MnO2(s) + 4HCl(aq) → 2MnCl2(aq) + 2H2O(l) + Cl2(g)

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Answer:

please provide more información or a photo

Explanation:

Of you want me to hwlp you please have more infor like a picture