When saturated, 400 cm3 of soil has 4% gravitational water (W%), 32% capillary water (W%), and 4% (W%) hygroscopic water. At the permanent-wilting-point (PWP), the sample contains 9.5% water (W%). Calculate the bulk density for the soil (assume a particle density of 2.65 g/cm3). Also, calculate the grams of water, and VOLUMETRIC water content (%), at BOTH saturation AND field capacity. Finally, calculate the mass of the sample at the PWP.​

The rate of disappearance of Br- is 5 times  of the rate of disappearance of BrO₃⁻. Hence, the disappearance rate of Br- in the reaction is 0.201 M/s .

The rate of a reaction is the rate of disappearance of the reactants or the rate of appearance of the products. The rate of a reaction is directly proportional to the molar concentration of the reactants.

For the given reaction, the rate of reaction can be written in the following terms.

-d/dt [BrO₃⁻] = -1/5 d/dt [Br-] =  -1/6d/dt [H+ ] = +1/3d/dt  [Br²] = + 1/3 d/dt[H₂O]

The minus sign indicates the disappearance and + indicates the formation.

Given, the rate of disappearance of BrO₃⁻ = -d/dt [BrO₃⁻]  = 0.041 M/s.

-d/dt [BrO₃⁻]  = 1/5 d/dt [Br-]

then  d/dt [Br-] = 5 -d/dt [BrO₃⁻]  = 5 × 0.041 M/s = 0.201 M/s.

Therefore, the rate of disappearance of Br- in this reaction is 0.201 M/s.

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The term that refers to the passing of a wave through an object is "transmission."

Transmission refers to the process by which a wave passes through an object or medium. In the context of sound, transmission occurs when sound waves travel through different substances, such as air, water, or solids.

When a sound wave encounters an object, it can be transmitted through it, reflected off it, or absorbed by it, depending on the properties of the object and the medium through which the sound is traveling.

For example, when you speak into a microphone, the sound waves produced by your voice travel through the air and are transmitted to the microphone's diaphragm. The diaphragm converts the sound waves into electrical signals, which can then be amplified and reproduced as sound through speakers.

In summary, transmission is the term used to describe the passage of a wave, such as a sound wave, through an object or medium. It is an essential concept in understanding how waves interact with their surroundings and how sound propagates through different materials.

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Answer:

The molar mass is: 18.02 g/mol.

Explanation:

1 mole of Hydrogen= 1.01, so if we have 2 moles of it here, that would be 2.02.

1 mole of Oxygen (that's all we have here)= 16.00

Once you add the two together (2.02+16.00), you will get 18.02.

I hope this made sense! Have a great day!

Answer:

48 moles

Explanation:

Step 1 - Understanding solubility of salts and gases

Solubility is the propertie of a substance to mix itself with another substance. It may happen in a lot of solvents, but we usually talk about solubility in water.

The solubility of a substance will depend on how strongly it interacts with water. In the case of a salt (such as LiCl) we can expect higher solubilities than in the cases of gases. That's because gases do not interact quite strongly with water.

Salts, on the other hand, may dissociate in water to form ions, which interact very strongly with water due to their charge. Not all salts are water-soluble though. It depends on some other features as well. But LiCl, as well as all chlorides (such as NaCl, KCl, and so on) are very soluble in water.

The solubility of salts usually increases with temperature (there are some expceptions to this general rule) whereas the solubility of gases decrease with temperature:

Step 2 - Using solubility to solve the problem

As we just saw, the solubility of a gas decreases with temperature, while the solubility of a salt increases with temperature.

H2 is a gas and LiCl is a solid (all ionic compounds are solid at room temperature), therefore we should expect that by warming the solution up to 75°C, the solubility of H2 will be decreased and of LiCl will be increased.

The result is that H2 gas will scape the solution and we'll see a lot of bubbles, whereas LiCl will remain dissolved in the solution. The correct answer is therefore alternative b) Gaseous H2 bubbles out of the solution.

Answer:

Farenheit

Explanation:

STP is used to measure standard temperature and pressure -- it uses Kelvin for temperature, not Farenheit.

Answer:

58.44277 gram

Explanation:

There are vacant orbitals on the phosphorus atom that allows it to expand its octet.

There is a concept that we would need to consider as we are answering the question that we here and that is the idea of the octet rule. The octet rule states that it is only about eight electrons that can be found on the outermost shell of an atom and as such all of the compounds can be formed in obedience to this rule.

Now we should know that there are no vacant d orbitals that are present on the nitrogen atom and this stems from the fact that it does not have a 3d level as such there are no orbitals that can be able to help the Nitrogen atom so as to be able to expand its octet. This is the reason why its pentafluoride can be easily formed.

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Answer:

Exposure of 2-methyl-2-butene to oxymercuration-demercuration conditions provides 2-methyl-2-butanol.

Explanation:

Please see the attachments below

the answer should be a the the question

Inter-conversion of glucose and fructose occurs with an equilibrium constant of 1.0. Glucose isomerase catalyzes this reaction. The final concentration of fructose at equilibrium from 40mM glucose is a. 40 mM.

Using this formula to find the  final concentration of fructose

Final concentration of fructose  =Equilibrium from glucose/ Equilibrium constant

Where:

Equilibrium constant = 1.0

Equilibrium from glucose = 40 mM

Let plug in the formula

Final concentration of fructose  = 40mM / 1.0

Final concentration of fructose  =  40mM

Therefore we can conclude that the correct option is A.

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The complete question is:

Inter-conversion of glucose and fructose occurs with an equilibrium constant of 1.0. Glucose isomerase catalyzes this reaction. The final concentration of fructose at equilibrium from 40mM glucose is

a. 40 mM

b. 20 mM

c. 10 mM

d. 0 mM​

The Gibbs Free Energy at 298 K, AH = 46 kJ/mol and sº = 0.097 kJ/(K-mol) is 28.906 kJ/mol

It serves as the single master variable that determines whether a given chemical change is thermodynamically possible.

Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously.

Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction.

At 298 K,

AH = 46 kJ/mol

and sº = 0.097 kJ/(K-mol).

ΔG=ΔH−T⋅ΔS

ΔG = 46 - 298 * 0.097

ΔG = 46 - 28.906

ΔG = 28.906 kJ/mol

The Gibbs Free Energy at 298 K, AH = 46 kJ/mol and sº = 0.097 kJ/(K-mol) is 28.906 kJ/mol

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Answer: A

Explanation: just did it

Answer:

b) 14.3%

Explanation:

Hello,

In this case, the experimental error is computed as:

\(error=\frac{exp-theo}{exp} *100\%\)

Whereas exp accounts for the measured value, in this case 1.4 cm, and theo the theoretical value, in this case 1.2 cm. Therefore, the result is:

\(error=\frac{1.4-1.2}{1.4} *100\%\\\\error=14.3\%\)

Thereby, answer should be b) 14.3% (corrected).

Best regards.

Answer:

Mushroom cells

Explanation:

They are mushroom cells because they are together.

The heat change or enthalpy change, ΔH°rxn, for the slow reaction of zinc with water is +417.7 kJ/mol.

The heat change or enthalpy change, ΔH°rxn, for the reaction of zinc (Zn) with water (H₂O) can be calculated using the standard enthalpies of formation for each species involved in the reaction.

Balanced chemical equation for the reaction is;

Zn(s) + 2H₂O(l) → Zn²⁺(aq) + H₂(g)

The standard enthalpy change for the reaction, ΔH°rxn, can be calculated as the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants, each multiplied by their respective stoichiometric coefficients;

ΔH°rxn = Σ(nΔH°f, products) - Σ(mΔH°f, reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively, and ΔH°f is the standard enthalpy of formation.

Assuming standard conditions (25°C and 1 atm), the standard enthalpies of formation for Zn²⁺(aq) and H₂(g) are typically tabulated values. Let's assume their values to be ΔH°f(Zn²⁺(aq)) = -153.9 kJ/mol and ΔH°f(H₂(g)) = 0 kJ/mol, respectively.

The standard enthalpy of formation of water (H₂O) is -285.8 kJ/mol.

Put the values into the equation, we get;

ΔH°rxn = [ΔH°f(Zn²⁺(aq)) + ΔH°f(H₂(g)] - [ΔH°f(Zn(s)) + 2ΔH°f(H₂O(l))]

ΔH°rxn = [-153.9 + 0] - [0 + 2(-285.8)]

ΔH°rxn = -153.9 + 571.6

ΔH°rxn = 417.7 kJ/mol

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Answer:Compounds are chemical substances that contain more than one element. They're created during a chemical reaction where atoms are rearranged into new compound molecules. For example, if carbon atoms react with oxygen atoms they form carbon dioxide molecules.

Explanation:

Answer:

CaCl3 is the limited reagent

Explanation:

Answer:

Option C Removing Na₂CO₃

Option E Heating the container.

Explanation:

The equation for the reaction is given below:

2NaHCO₃ (s) <=> Na₂CO₃ (s) + CO₂ (g) + H₂O (g)

Enthalpy change (ΔH) = 136 KJ

To obtain more CO₂ do the following:

1. Add more NaHCO₃

From chemical equilibrium, adding more NaHCO₃ means the reactant has increase. Thus, it (the reactant) will react to form more product (CO₂)

2. Removing any of the products. Removing any of the products implies that more reactants are in the reaction vessel than the product. Thus, the reactants will react to produce more products (CO₂)

3. Heating the container. Heating the container will lead to an increase in the temperature of the reaction. Since the enthalpy change (ΔH) is positive it means the reaction is endothermic. Thus, heating the container i.e increasing the temperature will favours the forward reaction i.e more products will be obtained.

NOTE: Pressure has no effect in the reaction system since there is no gaseous reactants.

Considering the options given above, C and E gives the correct answer to the question.

Answer:

c. Weak acid

Explanation:

Thus the correct answer is option c. Weak acid.

A strong acid would ionize completely, while a base would give off hydroxide ions (OH⁻).

The standard Gibbs free energy change of the reaction at 25°C is -18,262 J/mol.

Energy is the ability to do work. It exists in many forms, such as kinetic energy (energy of motion), potential energy (stored energy), thermal energy (heat), chemical energy (energy stored in the bonds of molecules), electrical energy (energy carried by electrons), and nuclear energy (energy produced in the nucleus of an atom).

The reaction given is a reversible reaction and the given Kp value is at 25°C. Therefore, we can use the equation ΔG°rxn = -RT ln Kp to calculate the standard Gibbs free energy change of the reaction.
We know that R = 8.314 J/mol K and T = 298 K (25°C)
Therefore, ΔG°rxn = - 8.314 J/mol K x 298 K x ln (2.26 x 10⁴)
ΔG°rxn = -18,262 J/mol
Therefore, the standard Gibbs free energy change of the reaction at 25°C is -18,262 J/mol.

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The grams solutes ini 835 mL of 0.390 M KBr is 38.75 g.

The grams can be calculated as follows:

The first you should calculate the molar

Times the molar of KBr by its valume to get moles

n = Molar x volume

n = 0.390 M x 0.835L

n = 0.32565 moles

thus, you should calculate molar mass so you can calculate the gram

Mass of potasium = 39.10 g/mol

Mass of Bromine = 79.904 g/mol

mass molar = 39.1 g/mol+79.9 g/mol = 119 g/mol

The next step is calculate the gram by time the moles to  its mass molar

gram = n x mass molar

gram = 0.32565 moles x 119 g/mol= 38.75 g.

so, The grams solutes ini 835 mL of 0.390 M KBr is 38.75 g.

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Answer: In addition to building texture, starches are used to contribute stability to food products. This often takes the form of holding water. As previously mentioned, gelatinized starch molecules tend to reassociate with one another.

Answer:

[ICl] = 0.0420 M

[I₂]  = [Cl₂] = 0.0140 M

Explanation:

Step 1: Calculate the initial concentration of ICl

[ICl] = 0.350 mol / 5.00 L = 0.0700 M

Step 2: Make an ICE chart

        2 ICl(g) ⇄ I₂(g) + Cl₂(g)

I        0.0700     0         0

C        -2x          +x        +x

E    0.0700-2x      x          x

The concentration equilibrium constant (Kc) is:

Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²

0.332 = x/0.0700-2x

x = 0.0140

The concentrations at equilbrium are:

[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M

[I₂]  = [Cl₂] = x = 0.0140 M

Answer:

the correct answer is number 7

Answer:

Characteristics...

Explanation:

I hope that's right, if not super sorry. I'm not the smartest. I tried my best if it's wrong

Answer:

* \(\Delta T_B=1.286\°C\)

* \(m=2.5m\)

* \(n=0.05mol\)

* \(M=59.76g/mol\)

Explanation:

Hello,

In this case, considering the boiling point rise problem, we consider its appropriate equation:

\(\Delta T_B=imK_b\)

Whereas i is the van't Hoff factor that for this nonvolatile solute is 1, m is the molality, Kb the boiling point constant of water as it is the solvent and ΔT the temperature difference. In such a way, with the given information we obtain:

- ΔT:

\(\Delta T_B=101.286\°C-100.000\°C\\\\\Delta T_B=1.286\°C\)

- Molality (mol/kg):

\(m=\frac{\Delta T_B}{i*K_b}=\frac{1.286\°C}{1*0.512\°C/m}\\ \\m=2.5m\)

- Moles for 20.02 g (0.02002 kg) of water:

\(n=2.5mol/kg*0.02002kg\\\\n=0.05mol\)

- Molar mass:

\(M=\frac{mass}{moles}=\frac{3.005g}{0.050mol} \\\\M=59.76g/mol\)

Best regards.

Answer:

a. NH3

b.

c. NaCl

d.

e. HCl

f.

The set of reactions that best represents the diagram of a precipitation reaction is given below:

Li₂SO₄ (aq) + AgNO₃ (aq)  ---->

The net ionic equation is given below:

SO₄²⁺ (aq) + 2 Ag⁺ (aq)  ----> Ag₂SO₄ (s) (a precipitate)

Precipitation reactions are reactions in which two soluble solutions when they are mixed result in the formation of an insoluble precipitate.

Precipitation reactions rea usually double replacement reactions.

Double replacement reactions also known as double displacement reactions are reactions in which an exchange of radicals occurs between two metallic cations resulting in the formation of an insoluble compound that forms the precipitate observed.

Considering the given reactions and the products obtained:

NH₄Br (aq) + Pb(C₂H₃O₂)₂ (aq) ---> forms no precipitate

KNO₃  (aq) + CuCl₂(aq) ---->  forms no precipitate

Li₂SO₄ (aq) + 2 AgNO₃ (aq)  ---->  2 LiNO₃ (aq) + Ag₂SO₄ (s) (forms a precipitate)

NaClO₄ (aq) + CaCl₂ (aq)  ---->  forms no precipitate

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