The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
OCl- + I- → OI-1 +Cl-.
This rapid reaction gives the following rate data:

[OCl-](M) [I]- (M) Rate (M/s)
1.5×10^3 1.5×10^3 1.36×10^4
3.0×10^3 1.5×10^3 2.72×10^4
1.5×10^3 3.0×10^3 2.72×10^4

Write the rate law for this reaction.
Calculate the rate constant with proper units.
Calculate the rate when [OCl-]= 1.8×10^3 M and [I-]= 6.0×10^4 M .

Electromagnetic radiation can be arranged by wavelength, frequency, or energy. In increasing order, by wavelength, it is: X-ray, Ultraviolet, Visible, Infrared, Microwave, Radio waves. In increasing order, by frequency and energy, it is: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray.

The types of electromagnetic radiation listed can be arranged by wavelength, frequency, and energy. The relationship among these three properties is that as wavelength increases, frequency and energy decrease, and vice versa.

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Answer:

Room B contains the greater mass of air

Explanation:

According to the given data, the volume of both rooms is the same, so let's consider both volumes as V. An open door connects both the room, which infer that pressure for both the rooms is also the same, let it be P.

Considering the above conditions, the general gas equation for the air in both rooms can be given as:

[tex]PV=n_{A}RT_{A}\\ \\ PV=n_{B}RT_{B}[/tex]

Here, n represents the moles of air, R is the gas constant, and T is the temperature. Taking the ratio of both the above equations we get

[tex]\frac{PV}{PV}= \frac{n_{A}RT_{A}}{n_{B}RT_{B}}\\[/tex]

moles of the gas are mass per molecular mass, as the molecular mass is the same in both the cases so n can be replaced by m, as follows

[tex]\frac{PV}{PV}= \frac{m_{A}RT_{A}}{m_{B}RT_{B}}\\[/tex]

By simplifying we get,

[tex]\frac{m_{A}}{m_{B}}=\frac{T_{B}}{T_{A}}[/tex]

According to the given conditions, TA is greater than TB. So from the derived relation, it can be seen that the mass of air in room B is greater than the mass of air in room A.

The ranking of the bonds from strongest to weakest is: bond between atoms in a metal, bond between sodium and chloride in NaCl, bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms. The bond energies for each bond are different, with the bond between atoms in a metal having the highest bond energy, followed by the bond between sodium and chloride in NaCl, the bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms having the lowest bond energy.

The ranking of the bonds from strongest to weakest is as follows:

The bond energy, also known as bond dissociation energy, is the energy required to break a bond. Here are the bond energies for each bond:

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The strongest bond is the bond between atoms in a metal, followed by the bond between sodium and chloride in NaCl. The bond between hydrogen and oxygen in water is weaker than the previous two, and the van der Waals bond between adjacent hydrogen atoms is the weakest. The bond energy is high for the bonds in a metal and NaCl, moderate for the bond in water, and low for the van der Waals bond between hydrogen atoms.

The bonds can be ranked from strongest to weakest as follows:

The bond energy for each bond is:

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Answer:

Blood agent

Explanation:

chemical warfare agents are of four major types:

nerve, blister, choking and blood agents

Nerve agents - it attacks the central nervous system and can enter the body through inhalation

Blister agents - it attacks skins and it is rapidly absorbed by the skin cells

chocking agents -  as name denoted, it is related to the respiratory system.

Blood agent-  it attacks the circulatory system of the body. it prevents the circulation of blood into the tissue and organs

Answer: the rate of the reaction will increase.

Explanation: when the concentration of a reaction is increased, this will increase the effective collisions of reactants which in turn increases the rate of the reaction

Doubling the concentration of both reactants A and B in the rate equation rate = k[A][B], which is first order with respect to each reactant, will quadruple the reaction rate. This is a characteristic of second-order reactions.

The student is asking about the effect on the reaction rate when concentrations of reactants in a given rate equation are altered. In the scenario where A reacts with B to give C and D, described by the rate equation rate = k[A][B], if the concentration of both A and B is doubled, the reaction rate will change. This is because the given reaction is first order with respect to both A and B, meaning it is a second-order reaction overall.

If you double the concentration of A to 2[A], the rate equation becomes k(2[A])[B], thereby doubling the reaction rate. Similarly, if B's concentration is also doubled to 2[B], the rate equation now reads k(2[A])(2[B]), which means you multiply the initial rate by four. Thus, doubling the concentrations of both A and B will quadruple the reaction rate due to the multiplicative effect in the rate law.

Remember, rate laws are determined experimentally, and in experiments comparing different trials, doubling the concentration often leads to a reaction rate that is proportional to that change. This reflects a direct relationship between reactant concentrations and reaction rates, as observed in first-order reactions for each reactant.

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Answer:

Well in comparison of Ni (II) and Ni (IV), Ni (IV) is a stronger vation with the +4 charge so it will attract the more oxygen ions in the aqueous solution. That is why Ni (IV) will be more strongly hydrated.

Explanation:

Final answer:

In an aqueous solution, the Ni (IV) cation is more strongly hydrated than the Ni (II) cation due to its higher charge density, which attracts more water molecules into its hydration sphere.

Explanation:

In an aqueous solution containing Ni (II) and Ni (IV) salts, the Ni (IV) cation is expected to be more strongly hydrated. This outcome is attributed primarily to the charge density. The Ni (IV) has a higher charge (+4) compared to Ni (II) which has a +2 charge. In terms of hydration, water molecules, which act as dipoles, are more strongly attracted to ions with a higher charge density. This means that the Ni (IV) ion, with its higher charge, attracts and binds water molecules more strongly than the Ni (II) ion.

This strong attraction results in a greater degree of hydration for the Ni (IV) ion as it pulls more water molecules into its hydration sphere. This process is crucial in understanding the properties of solutions, especially in predicting the behavior of ions in biological and chemical systems.

Final answer:

To find the volume of a platinum sample with a mass of 11.2 g, divide the mass by the density of platinum, which is 21.45 g/cm3, resulting in a volume of approximately 0.5221 [tex]cm^3[/tex].

Explanation:

To determine the volume of a platinum sample with a mass of 11.2 g, given the density of platinum is 21.45 g/cm3, we can use the formula for density which is:

Density = Mass / Volume.

By rearranging the formula to solve for volume, we get:

Volume = Mass / Density.

Substituting the given values:

Volume = 11.2 g / 21.45 g/cm3 = 0.5221  [tex]cm^3[/tex].

Therefore, the volume of the platinum sample is approximately 0.5221  [tex]cm^3[/tex].

Answer:

a relief valve is inserted in the downstream line from the pressure regulator.

Explanation:

Whenever dry nitrogen from a portable cylinder is used in service and installation practice the item of most importance in consideration of safety is a relief valve is inserted in the downstream line from the pressure regulator.

The most important item to consider for safety when using dry nitrogen from a portable cylinder in service and installation practice is a pressure regulator, which controls the flow and release of the gas. It is also crucial to inspect the system for leaks regularly.

When using dry nitrogen from a portable cylinder in service and installation practice, the most important item to consider for safety is a pressure regulator. A pressure regulator is used to control the flow and release of nitrogen gas from the cylinder. It ensures that the pressure does not exceed safe limits and reduces the risk of an explosion or other accidents.

Additionally, it is crucial to always check for leaks in the system before using dry nitrogen. Leaks can result in the accumulation of nitrogen gas, which can displace oxygen in the air and lead to asphyxiation. Therefore, regularly inspecting the system and using appropriate leak detection methods, such as soapy water or a leak detection solution, is essential for safety.

In summary, when working with dry nitrogen from a portable cylinder, the two most important safety considerations are the use of a pressure regulator and regular inspection for leaks.

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Answer: The new water level when silver metal is submerged is 18.24 mL

Explanation:

We are given:

Volume of graduated cylinder = 15.0 mL

To calculate volume of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of silver metal = 10.5 g/mL

Mass of silver metal = 34.0 g

Putting values in above equation, we get:

[tex]10.5g/mL=\frac{34.0g}{\text{Volume of silver metal}}\\\\\text{Volume of silver metal}=\frac{34.0g}{10.5g/mL}=3.24mL[/tex]

New water level = Volume of silver metal + Volume of graduated cylinder

New water level = (3.24 + 15.0) mL = 18.24 mL

Hence, the new water level when silver metal is submerged is 18.24 mL

Answer:

Pressure and volume

Explanation:

The ideal gas equation is given below:

PV = nRT

P = nRT /V

If nRT is constant,

P will be inversely proportional to V(P & 1/V)

Therefore, pressure and volume are inversely proportional if all factors remains constant

In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant. This is known as Boyle's Law.

In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant.

This is known as Boyle's Law, which states that when the volume of a given amount of gas is decreased, the pressure increases, and vice versa.

For example, if the volume of a gas is reduced by half, the pressure will double, and if the volume is doubled, the pressure will be halved.

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Answer:

The balanced reaction is this:

2 C₄H₁₀ (g)  +  13 O₂(g) →  8 CO₂(g) +  10 H₂O (g)

Explanation:

Combustion is a chemical reaction whose reagents are oxygen, usually in excess and a hydrocarbon to generate carbon dioxide and water in the form of steam, as products.

Butane is considered as a reactant and it is a sort of alkane, in this case with 4 C (prefix but).

O₂ is needed for the complete combustion of butane.

Answer:

The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.

Explanation:

a) The molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]

b) The molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]

c) The molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]

(a) 1.22 M sugar (C12H22O11) solution:

Given:

- Molarity (\(M\)) of sugar solution = 1.22 M

- Density of solution = 1.12 g/ml

We'll assume that the density given is for the solution and not just for water.

1. Calculate the mass of 1 L (1000 ml) of solution:

[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]

[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]

[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]

2. Convert the mass of the solution to kilograms:

[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]

3. Calculate the moles of sugar (C12H22O11):

[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]

[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]

4. Calculate the molality ([tex]\(m\)[/tex]):

[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]

Let's perform the calculation.

For the 1.22 M sugar solution:

1. Calculate the mass of 1 L (1000 ml) of solution:

[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]

[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]

[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]

2. Convert the mass of the solution to kilograms:

[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]

3. Calculate the moles of sugar (C12H22O11):

[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]

[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]

4. Calculate the molality ([tex]\(m\)[/tex]):

[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]

[tex]\[ m = \frac{{1.22 \, \text{mol}}}{{1.12 \, \text{kg}}} \][/tex]

[tex]\[ m \approx 1.089 \, \text{mol/kg} \][/tex]

So, the molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]

(b) 0.87 M NaOH solution:

Given:

- Molarity ([tex]\(M\)[/tex]) of NaOH solution = 0.87 M

- Density of solution = 1.04 g/ml

We'll follow the same steps as above to calculate the molality.

Let's calculate for the NaOH solution.

For the 0.87 M NaOH solution:

1. Calculate the mass of 1 L (1000 ml) of solution:

[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]

[tex]\[ \text{mass of solution} = 1.04 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]

[tex]\[ \text{mass of solution} = 1040 \, \text{g} \][/tex]

2. Convert the mass of the solution to kilograms:

[tex]\[ \text{mass of solution in kg} = \frac{{1040 \, \text{g}}}{{1000}} = 1.04 \, \text{kg} \][/tex]

3. Calculate the moles of NaOH:

[tex]\[ \text{moles of NaOH} = \text{Molarity} \times \text{volume in L} \][/tex]

[tex]\[ \text{moles of NaOH} = 0.87 \, \text{mol/L} \times 1 \, \text{L} = 0.87 \, \text{mol} \][/tex]

4. Calculate the molality (\(m\)):

[tex]\[ m = \frac{{\text{moles of NaOH}}}{{\text{mass of solvent in kg}}} \][/tex]

[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]

[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]

[tex]\[ m \approx 0.8365 \, \text{mol/kg} \][/tex]

So, the molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]

(c) 5.24 M NaHCO3 solution:

Given:

- Molarity (\(M\)) of NaHCO3 solution = 5.24 M

- Density of solution = 1.19 g/ml

Let's follow the same steps as above to calculate the molality for the NaHCO3 solution.

For the 5.24 M NaHCO3 solution:

1. Calculate the mass of 1 L (1000 ml) of solution:

[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]

[tex]\[ \text{mass of solution} = 1.19 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]

[tex]\[ \text{mass of solution} = 1190 \, \text{g} \][/tex]

2. Convert the mass of the solution to kilograms:

[tex]\[ \text{mass of solution in kg} = \frac{{1190 \, \text{g}}}{{1000}} = 1.19 \, \text{kg} \][/tex]

3. Calculate the moles of NaHCO3:

[tex]\[ \text{moles of NaHCO3} = \text{Molarity} \times \text{volume in L} \][/tex]

[tex]\[ \text{moles of NaHCO3} = 5.24 \, \text{mol/L} \times 1 \, \text{L} = 5.24 \, \text{mol} \][/tex]

4. Calculate the molality (\(m\)):

[tex]\[ m = \frac{{\text{moles of NaHCO3}}}{{\text{mass of solvent in kg}}} \][/tex]

[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]

[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]

[tex]\[ m \approx 4.4034 \, \text{mol/kg} \][/tex]

So, the molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]

Answer: 6.7*10^-7 m

Explanation:

The full explanation is shown in the image attached. The energy of the photon is obtained by dividing the bond energy by the Avogadro's number. Using the Plank's equation, we cash obtain the frequency or wavelength of radiation required by substituting into the given equation appropriately.

Final answer:

To calculate the shortest wavelength of light capable of dissociating the Br-I bond, the bond energy given (179 kJ/mol) is first converted to Joules per photon. Then, applying the equation E = hc / λ with Planck's constant and the speed of light gives the shortest wavelength needed to break the bond.

Explanation:

The question asks to calculate the shortest wavelength of light capable of dissociating the Br–I bond in iodine monobromide, given that the bond energy is 179 kJ/mol. To solve this, we apply the equation relating the energy of a photon (E) to its wavelength (λ), using Planck's constant (h) and the speed of light (c).

The energy of the photon needed can be calculated using the equation E = hc / λ, where h = 6.626 x 10-34 J·s (Planck's constant) and c = 3.00 x 108 m/s (speed of light). First, convert the bond dissociation energy from kJ/mol to Joules (J) by multiplying by 1000 and dividing by Avogadro's number (6.022 x 1023 mol-1), giving the energy required per molecule.

Finally, rearrange the equation to solve for λ: λ = hc / E. Plugging in the values, we find the shortest wavelength of light capable of breaking the Br–I bond. Remember, since the question gives the bond dissociation energy in kJ/mol, conversion to Joules is necessary for the calculation to proceed correctly.

Answer:

it contains 11 electrons

Explanation:

Answer:

25.0 g of sodium carbonate are present in 220 ml of the solution.

Explanation:

Hi there!

I have found the complete question on the web:

Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.

First, let's find how many moles of sodium carbonate have a mass of 25.0 g.

The molar mass of Na₂CO₃ is 106 g/mol.

So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:

25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃

The solution contains 1.07 mol sodium carbonate in 1 liter.

So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:

0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).

25.0 g of sodium carbonate are present in 220 ml of the solution.

Final answer:

To calculate the volume in milliliters of a sodium carbonate solution, you need to know the concentration of the solution.

Explanation:

To calculate the volume in milliliters of a sodium carbonate solution, we need to know the concentration of the solution. Without that information, we are unable to provide an accurate calculation. Please provide the concentration of the sodium carbonate solution in the question.

Answer: Option A. A candy bar

Option B. stretched rubber band

Option D. A roller coaster at the top of a hill

Explanation:

Answer:

The correct answers are options A. "A candy bar", B. "A stretched rubber band" and D. "A roller coaster at the top of a hill".

Explanation:

Potential energy is defined as the force that any body for its location at a certain position. Three examples of objects having potential energy are a candy bar, a stretched rubber band and a roller coaster at the top of a hill. These three objects are not in motion and its energy is only "potential", rising from the arrangement of its particles, the gravitational force, electric charges, among other factors.

Answer:

Molality for this solution is 0.31 m

Explanation:

Molality → Mol of solute / kg of solvent

Mass of solute = 1.85 g (CO(NH₂)₂

% by mass, means mass of solute in 100 g of solution

Mass of solution = Mass of solute + Mass of solvent

As 1.85 g is the mass of solute, the mass of solvent would be 98.15 g to complete the 100g of solution.

Let's convert the mass of solvent (g to kg)

1 g = 1×10⁻³ kg

98.2 g .  1×10⁻³ kg / 1g = 0.0982 kg

Let's convert the mass of solute to moles (mass / molar mass)

1.85 g / 60 g/mol = 0.0308 mol

Molality = 0.0308 mol / 0.0982 kg → 0.31 m

To find the molality of the solution, we calculate the number of moles of urea (0.031 mol) and the mass of water (0.10315 kg). The molality is then computed as moles of urea divided by the mass of water, which comes out to be approximately 0.30 m.

To calculate the molality of the solution, we need to first know the number of moles of the solute (urea) and the mass of the solvent (water) in kilograms.

The mass % of the solution is given as 1.85%, meaning that we have 1.85g of urea in 100g of the solution. Considering that the molar mass of urea (CO(NH2)2) is about 60.06 g/mol, the number of moles of urea can be calculated as follows: Moles of urea = mass / molar mass = 1.85g / 60.06 g/mol ≈ 0.031 mol.

Since the density of the solution is given as 1.05 g/mL, the mass of the solution for 100 mL (or 100g since density = mass/volume) would be 100 mL * 1.05 g/mL = 105g. In this 105g of the solution, 1.85g is urea, so the mass of water (solvent) would be 105g - 1.85g = 103.15g or 0.10315 kg (since 1g = 0.001 kg).

Molality is the number of solute's moles divided by the mass of solvent (in kg). Thus, the molality of the solution would be: Molality = moles of urea / mass of water in kg = 0.031 mol / 0.10315 kg ≈ 0.30 mol/kg or 0.30 m.

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Answer:

Molarity for the solution is 1.48 M

Explanation:

Molarity involves the moles of solute that are contained in 1L of solution. It is a sort of concentration. The most usual.

7.73 by mass involves the mass of solute that are contained in 100 g of solution, so this one of ethylene glycol contains 7.73 g of it.

Let's determine the moles of solute

7.73 g / molar mass of ethylene glycol = moles

7.73 g / 62 g/mol = 0.124 mol

If the mass of solution is 100 g, we can determine the volume with density.

Density = Mass / Volume

Volume = Mass / Density

Volume = 100 g / 1.19 g/mL = 84.03 mL

In conclusion, 0.124 moles are contained in 84.03 mL

Molarity is mol/L, so let's convert the volume in L

84.03 mL . 1L / 1000 mL = 0.08403 L

0.124 mol / 0.08403 L = 1.48 M

Answer: a:6 electrons, b:10electrons, c: 2electrons

Explanation:

1.Principle quantum number(n)

n is the number of shell, it is a positive integer. The maximum number of electrons in a shell is 2(n2)

2.Azimuthal quantum number(l)

l is the number of subshell in the principal shell

The name of subshell are s, p, d, f

The number of nodes in each subshell is

s is l=0, p is l=1, d is l=2, f is l=3.

Each shell can have 2 x l + 1 sublevels, and each sublevel can accommodate maximum of 2 electrons and have two electrons.

A. n=2, l=1 then 2 x 1 + 1= 3 subshell, 3*2= 6 electrons in the p subshell

b. 3d, d is l = 2 then 2 x 2 + 1 = 5 sublevels, 5*2=10 electrons in the d subshell

c. 4s, s is l=0 then 2 x 0 + 1 = 1 sublevel, 1*2=2 electrons in the s subshell.

Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Decomposition of silver chloride is represented as:

[tex]2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)[/tex]

Thus the coefficient in front of AgCl when the equation is properly balanced is 2.

Answer:

the coefficient in front of AgCl(s) when the equation is properly balanced is 2.

Explanation:

The decomposition of solid silver chloride (AgCl) into its elements when exposed to sunlight can be represented by the following balanced chemical equation:

2AgCl(s) → 2Ag(s) + Cl2(g)

In this balanced equation:

The coefficient in front of AgCl(s) is 2.

The coefficient in front of Ag(s) is 2.

The coefficient in front of Cl2(g) is also 2.

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Answer:

(a) F⁻(g) ⇒ F⁺(g)         ΔE=2009 KJ/mol

(b) Na⁺(g) ⇒ Na⁻(g)   ΔE=-548.6 KJ/mol

Explanation:

Hess's Law

"Energy changes for a reaction is independent of steps or route involved."

Now the conversion F⁻ to F⁺ is a two step reaction. In first step F⁻ loses an electron e⁻ to become neutral atom.

F⁻ ⇒ F + e⁻   ΔH=328 KJ/mol (i)

Kindly note this energy is derived from electron affinity, which is the energy changes while adding an electron to neutral atom and its same while removing an electron from ion.

In second step,

Another electron is removed from F.

F ⇒ F⁺ + e⁻    ΔH= 1681 KJ/mol (ii)

This energy is 1st ionization energy for F, which is the energy required to remove first electron from outer most shell of neutral atom.

Now the total energy change can be found by applying Hess's law and adding equations (i) and (ii);

F⁻ ⇒ F + e⁻   ΔH=328 KJ/mol

F ⇒ F⁺ + e⁻    ΔH= 1681 KJ/mol

F⁻ ⇒ F⁺ + 2e⁻  ΔH= 2009 KJ/mol

This reaction involves two steps in first Na⁺(g) gains electron and become neutral atom while in second step Na(g) accepts another electron to become Na⁻(g).

Na⁺(g) + e⁻⇒ Na(g)    ΔH=-495.8 KJ/mol   (i)

Na(g) + e⁻ ⇒ Na⁻(g)   ΔH=-52.8 KJ/mol     (ii)   (By Applying Hess's Law)

Na⁺(g) + 2e⁻ ⇒ Na⁻(g)  ΔH=-548.6 KJ/mol

Kindly note as there is no molar change in above mentioned reactions therefore enthalpy changes(ΔH) and internal energy changes(ΔE) are same.

Answer:

8.6 mL

Explanation:

Glacial acetic acid (or anhydrous acetic acid) can be thought of as pure acetic acid, and has a density of 1.05 g/cm³ (or g/mL).

To answer this problem first we convert moles of acetic acid (CH₃CO₂H) to grams, using its molar mass (60 g/mol):

Now we convert grams of acetic acid to mL, using its density:

Explanation:

A.

The first student will be on the lower bunk on the first floor because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available.

B.

7 students are in the TOP bunks because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available. Therefore, all the rooms up till the third floor (Remember, third floor has 3 suites), so the first floor is filled - 1 person on the top bunk, 2 floor is filled- 4 persons and the third floor; the first suite is filled - 1 person and the second suite is a little partially filled- 1 person.

C.

Following the criteria 1, 2 and 3, the 21st student occupies the third suite on the third floor because all the floors (1 and 2) are occupied so the third suite on the third floor is still vacant.

D.

From the criteria there are therefore 10 persons at the TOP bunk. All the rooms up till the third floor are filled, so the first floor is filled - 1 person on the top bunk, second floor is filled (2 suites) - 4 persons and the third floor; the first suite and second suite is filled - 4 persons; the thirs suite has 6 persons present so 1 person is at the top bunk.

Answer: It is because tyrosine kinases and BTK have similar solubilities

Explanation:

In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.

In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)

Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.

Answer:

Same within the groups...

Different within the periods...

Explanation:

The outer configuration is same with a group because of their arrangement in the periodic table. If we see group 1a of alkalies, all of them have 1 electron in their outer shell and this order goes on increasing  up-to noble gases that have 8 electrons in their outer shell each.

This arrangement is called periodicity of the periodic table and elements are arranged according to the periodic law.

Along a period, these properties of outer configuration goes on increasing than within the group.

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol[/tex]

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

[tex]0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol[/tex]

The chemical equation for the reaction of NaOH and sulfuric acid follows:

[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = [tex](4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol[/tex]

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sodium sulfate

Moles of sodium sulfate = [tex]1.462\times 10^{-3}moles[/tex]

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

[tex]\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M[/tex]

Moles of excess sulfuric acid = [tex]3.498\times 10^{-3}mol[/tex]

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

[tex]\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M[/tex]

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

[tex]\text{Molarity of NaOH}=\frac{0}{0.050}=0M[/tex]

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

After the reaction between solutions of H2SO4 and NaOH is complete, the remaining concentrations are 0.0317 M H2SO4, 0 M NaOH, and 0.0836 M Na2SO4 are formed.

The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O. This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate (Na2SO4) and two moles of water.

Using this stoichiometry, the moles of H2SO4 and NaOH in the solutions can be calculated with moles = concentration x volume (in L). That gives us 0.020 L * 0.248 mol/L = 0.00496 mol for H2SO4 and 0.015 L * 0.195 mol/L = 0.002925 mol for NaOH.

Since one mole of H2SO4 reacts with two moles of NaOH, all of the NaOH will react, and you will have 0.00496 mol - 2* 0.002925 mol = 0.00111 mol of H2SO4 left. The concentration of H2SO4 then is 0.00111 mol / 0.035 L (sum of volumes) = 0.0317 M.

NaOH is completely reacted, so its concentration is 0 M. From the reaction stoichiometry, 0.002925 mol of Na2SO4 is formed, so its concentration is 0.002925 mol / 0.035 L = 0.0836 M.

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Answer:

Option b. 0.048 M

Explanation:

We have the molecular weight and the mass, from sulcralfate.

Let's convert the mass in g, to moles

1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.

Molarity is mol /L

Let's convert the volume of solution in L

10 mL . 1L/1000 mL = 0.01 L

4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L

Answer:

The wavelength in m = 3.208 m

The wavelength in nm =[tex]3.208\times 10^9 nm[/tex]

The wavelength in Å = [tex]3.208\times 10^{10} \AA[/tex]

Explanation:

To calculate the wavelength of light, we use the equation:

[tex]\lambda=\frac{c}{\nu}[/tex]

where,

[tex]\lambda[/tex] = wavelength of the radiation

c = speed of light = [tex]3.0\times 10^8m/s[/tex]

[tex]\nu[/tex] = frequency of wave

We have :

Frequency of the wave = [tex]\nu =93.5MHz=93.5\times 10^6 Hz[/tex]

[tex]1 Hz = 1s^{-1}[/tex]

Wavelength of the wave = [tex]\lambda [/tex]

[tex]\lambda =\frac{c}{\nu}[/tex]

[tex]\lambda=\frac{3\times 10^8 m/s}{93.5\times 10^6 s^{-1}}=3.208 m[/tex]

[tex]1 m = 10^9 nm[/tex]

[tex]3.208 m= 3.208\times 10^9 nm[/tex]

[tex]1 m = 10^{10} \AA[/tex]

[tex]3.2086 m= 3.208\times 10^{10} \AA[/tex]

The wavelength in m = 3.208 m

The wavelength in nm =[tex]3.208\times 10^9 nm[/tex]

The wavelength in Å = [tex]3.208\times 10^{10} \AA[/tex]

The wavelength in m = 3.208 m

The wavelength in nm = [tex]3.208*10^9 nm[/tex]

The wavelength in Å = [tex]3.208*10^{10}[/tex] Å

It is equal to the speed (v) of a wave train in a medium divided by its frequency (f). It is given by:

[tex]\lambda=\frac{c}{f}[/tex]

where,

[tex]\lambda[/tex] = wavelength of the radiation

c = speed of light = [tex]3.0*10^8m/s[/tex]

f = frequency of wave = 93.5 MHz = [tex]93.5*10^6Hz[/tex]

Calculation of wavelength:

[tex]\lambda=\frac{c}{f}\\\\\lambda=\frac{3.0*10^8m/s}{93.5*10^6s^{-1}} \\\\\lambda=3.208m=3.208*10^9nm=3.208*10^{10}[/tex]Å.

Thus, The value for wavelength in different units are:

The wavelength in m = 3.208 m

The wavelength in nm = [tex]3.208*10^9 nm[/tex]

The wavelength in Å = [tex]3.208*10^{10}[/tex] Å

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