what happens when sulphur trioxide reacts with concentrated sulphur acid ?

Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.

We have 4 aqueous solutions and we want to determine which has the lowest freezing point.

Freezing-point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.

We can calculate the freezing point depression (ΔT) using the following expression.

ΔT = Kf × b × i

where,

Assuming all the solutions have the same molality, the freezing point depression will be a function of the van't Hoff factor.

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

Thus, NaCl, with the highest Van't Hoff factor, will have the lowest freezing point.

Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.

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We should start with approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution.

To solve this problem, we can use the dilution formula:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

where:

[tex]\( C_1 \)[/tex] is the concentration of the stock solution ([tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).

[tex]\( V_1 \)[/tex] is the volume of the stock solution we need to use (unknown).

[tex]\( C_2 \)[/tex] is the concentration of the final solution ([tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).

[tex]\( V_2 \)[/tex] is the volume of the final solution ([tex]600 ml[/tex] or [tex]0.6 L[/tex]).

We want to find [tex]\( V_1 \)[/tex], so we rearrange the formula:

[tex]\[ V_1 = \frac{C_2V_2}{C_1} \][/tex]

Now, plug in the values:

[tex]\[ V_1 = \frac{(0.1 \text{ M})(0.6 \text{ L})}{18.4 \text{ M}} \][/tex]

[tex]\[ V_1 = \frac{0.06}{18.4} \][/tex]

[tex]\[ V_1 =0.00326 \text{ L} \][/tex]

Since [tex]1 L = 1000 ml[/tex], we convert [tex]\( V_1 \)[/tex] to milliliters

[tex]\[ V_1 = 0.00326 \text{ L} \times 1000 \text{ ml/L} \][/tex]

[tex]\[ V_1 = 3.26 \text{ ml} \][/tex]

Rounding to two decimal places, we find that we need approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex][tex]H\(_2\)SO\(_4\)[/tex] solution.

Volatile organic compounds (VOCs) can evaporate from fracking liquid waste pools, contributing to air pollution. These pools may also contain hazardous air pollutants and heavy metals, requiring careful handling to avoid environmental contamination.

When fracking liquid waste is left in pools on the surface, volatile organic compounds (VOCs) can evaporate into the air and contribute to pollution. These wastewater ponds can contain a variety of pollutants, including hazardous air pollutants such as benzene, toluene, ethylbenzene, and xylene. Moreover, fracking fluid, also known as flowback, can contain chemicals used in the drilling process, heavy metals, and radioactive materials. These substances pose a significant risk to both environmental and human health if they are not properly managed and treated.

The process of hydraulic fracturing or 'fracking' involves injecting high-pressure fluids to fracture shale deposits, which releases trapped gas and oil. The wastewater from this process may return to the water cycle, but the large volume of contaminated water requires careful handling to prevent land and water pollution. As a proactive measure, governments around the world have taken steps, in some cases banning the practice due to the severe risks associated with fracking.

Answer:

The incorrect statement is:

The physical properties of a compound are similar to those of its elements.

Explanation:

The physical properties of a compound depend on its elements but they are not similiar to the physical properties of individual elements.

Take water as an example. Water is made by oxygen and hydrogen (H₂O). At normal , water is liquid and both oxygen and hydrogen are gases.

The solubility product constant (Ksp) of calcium carbonate is used to calculate the solubility of this compound in water. Considering the 1:1:1 molar ratio between CaCO3 and the ions it forms on dissolution, we derive the solubility as 0.0058 g/l.

The subject of your question is the solubility product constant (Ksp) of calcium carbonate (CaCO3). The Ksp is used to calculate the solubility of a compound, such as CaCO3, in a given solvent, which in this case is water. The Ksp is given as 3.36 × 10-9 M² and it is this value that we will use to determine the solubility.

As we know, for the dissolution of calcium carbonate into calcium and carbonate ions, the dissolution reaction can be written as: CaCO3(s) → Ca²+ + CO3²-. Here, it implies that there is a 1:1:1 molar ratio between CaCO3 and the ions it forms on dissolution. Thus, if 's' represents the molarity of CaCO3, then the Ksp expression can be written as Ksp = [Ca²+][CO3²-] = s².

Using the given Ksp value, we can solve the above equation for 's': s = sqrt(Ksp) = sqrt(3.36 × 10-9 M²) = 5.8 × 10-5 M. However, the question asks for the solubility in g/l. To convert molarity (M) to g/l, we multiply by the molecular weight of the compound. For CaCO3, the molecular weight is approximately 100 g/mol. Therefore, the solubility is 5.8 × 10-5 M x 100 g/mol = 0.0058 g/l.

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Answer: B) increasing the surface area of the reactants.

Explanation:

Decreasing the particle size of the reactants in a chemical reaction will increase the rate of reaction by increasing the surface area of the reactants. Hence, option B is correct.

A chemical reaction is a process that leads to the transformation of one set of chemical substances into another. It involves breaking and forming chemical bonds between atoms and molecules to form new substances.

In a chemical reaction, the reacting substances are called reactants, and the newly formed substances are called products.

Decreasing the particle size of the reactants in a chemical reaction will increase the rate of reaction by increasing the surface area of the reactants that are available for the reaction.

When the particle size is reduced, the total surface area of the reactant particles is increased, which makes it easier for the particles to collide and react with one another.

Thus, option B is correct.

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Answer is: pH of solution of sodium cyanide is 11.07.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.068 M.
Ka(HCN) =  4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.068 M - x..
2.04·10⁻⁵ = x² / (0.068 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00116 M.
pOH = -log(0.00116 M) = 2.93.
pH = 14 - 2.93 = 11.07.

The pH of the sodium cyanide solution is 11.56.

Let the cyanide ion be X

We have to set up the ICE table for the problem as follows;

     X^-(aq) + H2O(l)  ⇄ HX(aq) + OH^-(aq)

I    0.068                            x                 x

C   -x                                 + x                +x      

E   0.068 - x                      x                   x

But Kb = Kw/Ka = 1 × 10^-14/4.9 × 10-10

Kb = 2 × 10^-4

So;

Kb = [HX] [OH^-]/[X^-]

2 × 10^-4 = x^2/ 0.068 - x  

2 × 10^-4(0.068 - x) = x^2

1.36 × 10^-5 - 2 × 10^-4x = x^2

x^2 + 2 × 10^-4x - 1.36 × 10^-5 = 0

x= 0.0036 M

Since x = [OH^-] = 0.0036 M

pOH = - log(0.0036 M)

pOH = 2.44

pH = 14 - 2.44 = 11.56

The pH of the sodium cyanide solution is 11.56.

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Explanation:

Plants veins provide energy to the rest of the plant and it also provides structure and support to the plant leaves and also transport water. When plants absorb water and nutrients through their roots, their vascular system come in use to move water and nutrients to the rest part of the plants. There are two types of tissue that make up the plants veins are xylem and phloem. Xylem moves water and minerals from the plants's roots and phloem moves food energy to all parts of plant where the plants need it.    

Answer:

Friction

Explanation:

Friction is a force that slows down moving objects. If you roll a ball across a shaggy rug, you can see that there are lumps and bumps in the rug that make the ball slow down. The rubbing, or friction, between the ball and the rug is what makes the ball stop rolling.

Final answer:

The final pressure of the system is 0.85 atm.

Explanation:

To determine the final pressure of the system, we can use the relationship between volume and pressure known as Boyle's Law. According to Boyle's Law, when the volume of a gas increases, the pressure decreases, and vice versa, at constant temperature and amount of gas.

In this case, the initial volume is 0.75 L and the final volume is 1.1 L. Since the volume increased, we can expect the pressure to decrease. The initial pressure is 1.25 atm. Using Boyle's Law, we can set up the equation:

P1 * V1 = P2 * V2

1.25 atm * 0.75 L = P2 * 1.1 L

Solving for P2, we get: P2 = 1.25 atm * 0.75 L / 1.1 L = 0.85 atm

Therefore, the final pressure of the system is 0.85 atm.

Answer:

0.0135 moles of iodine will be formed.

Explanation:

[tex]2NI_3\rightarrow N_2+3I_2[/tex]

Moles of nitrogen triiodide =[tex]\frac{3.58 g}{394.71 g/mol}=0.0090 mol[/tex]

According to reaction 2 moles of nitrogen triiodide gives 3 moles of iodione gas.

Then 0.0090 mol of nitrogen triiodide will give:

[tex]\frac{3}{2}\times 0.0090 mol=0.0135 mol[/tex]

0.0135 moles of iodine will be formed.

Final answer:

To prepare 0.50 mL of a 0.15 M NaOH solution from a 6.00 M NaOH solution, you would need 12.5 μL of the concentrated solution according to the dilution formula, which is not listed among the provided answer options.

Explanation:

The volume of 6.00 M NaOH solution required to prepare 0.50 mL of 0.15 M NaOH solution can be found using the dilution formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively. Plugging the known values into the equation: (6.00 M)(V1) = (0.15 M)(0.50 mL) ⇒ V1 = (0.15 M)(0.50 mL) / (6.00 M) = 0.0125 mL or 12.5 μL.

Since none of the provided answer choices (a) 12.5 mL, (b) 25 mL, or (c) 30 mL match this result, it seems there might be a typo or a misprint in the question. To prepare 0.50 mL of a 0.15 M solution from a 6.00 M solution, you would need substantially less than 1 mL of the concentrated solution. The likely correct volume needed, based on the question's data, would be 12.5 μL, which is not listed as an option in the question.

Answer : The correct answer is Name = Thallium (Th) , atomic mass = 210 and atomic number = 81 .

Alpha decay :

When atomic nucleus emits an alpha particle that process is known as alpha particle decay . The symbol of alpha particle is [tex] ^4_2He [/tex] . where 4 is atomic mass ( 2 protons + 2 neutrons ) and 2 atomic number .

When alpha particle is released the resultant nuclei (daughter nuclei) have 4 less atomic mass and 2 less atomic number .

Example for nuclear equation of alpha decay can be expressed as:

[tex] {_{90}^{230}Th} \rightarrow {_2^4He} + {_{88}^{226}Ra} [/tex]

Given :

Radioisotope = Bi Atomic mass = 214 Atomic number = 83

When it will release alpha particle , atomic mass will be decreased by 4 and atomic number by 2 .

Atomic mass = 214 - 4 = 210 Atomic number = 83-2 = 81

The atom with atomic number 81 is Thallium (Th) . Hence the daughter nuclei so produced is [tex] ^2^1^0_8_1Th [/tex]

The nuclear reaction can be written as :

[tex] ^{214}_{83}Bi \rightarrow _2^4He + _{81}^{210}Th [/tex]

In a fission reaction, the nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission occurs when a heavy nucleus like uranium absorbs a neutron and splits, emitting additional neutrons, gamma rays, and energy.

The correct description of the changes involved in a fission reaction is: The nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission is a nuclear reaction where the nucleus of an atom, such as uranium-235 or plutonium-239, splits into smaller nuclei after absorbing a neutron.

This process produces additional free neutrons, gamma radiation, kinetic energy of fission fragments, and releases a significant amount of energy. The process is exothermic, and the energy released can be harnessed for electricity generation in nuclear reactors or can be used in nuclear weapons.

Nuclear fission differs from nuclear fusion, which is the combining of two smaller atomic nuclei to form a larger nucleus, also releasing energy. In both processes, large amounts of heat and radiation are emitted. However, for fission, a fissile material like uranium or plutonium is necessary to sustain the reaction, and it involves the breakup of a heavy nucleus into lighter elements.