The pair of elements that forms a bond with the least ionic character is

A. Na and Cl
B. K and Cl
C. O and Cl
D. Mg and Cl

J.J. Thomson's experiment did not disprove the theory that atoms are indivisible. Instead, his experiment demonstrated the presence of subatomic particles, specifically electrons, within atoms.

The atom's indivisibleness was not refuted by J. J. Thomson's experiment. Instead, his research demonstrated that electrons, a type of subatomic particle, are present in atoms. The "plum pudding" model of the atom was created as a result of Thomson's research from the late 19th and early 20th centuries, which advanced knowledge of atomic structure.

Thomson studied the behavior of electrically charged particles (electrons) inside a vacuum tube in his well-known cathode ray tube experiment. A stream of negatively charged particles (electrons) traveled from the cathode (negative electrode) to the anode (positive electrode) when a voltage was applied across the tube. Thomson came to the conclusion that these particles were the basic building blocks of atoms.

The old belief that atoms were indestructible was refuted by Thomson's discovery of electrons within them, which also supported the notion of subatomic particles inside the atom. His "plum pudding" concept proposed that atoms had a neutral overall charge because electrons were contained within a positively charged "pudding" or matrix.

Later investigations, like those by Ernest Rutherford, helped to clarify our understanding of atomic structure and revealed the existence of a positively charged nucleus that contains protons as well as neutral particles known as neutrons. The existence of indivisible atoms was therefore not refuted by Thomson's experiment, but it did introduce the idea of subatomic particles within atoms, particularly electrons.

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The Arrhenius theory limits acids and bases to those that produce hydrogen or hydroxide ions in aqueous solutions, making ammonia a base only when it reacts with water and not recognizing it as an inherent base due to its structure. The Brønsted-Lowry theory overcomes this limitation by including substances like ammonia that accept hydrogen ions, further expanded by the Lewis model where ammonia can donate an electron pair.

The Arrhenius theory is limited in its definition of acids and bases because it requires the presence of hydroxide ions in the formula of bases and hydrogen ions in the formula of acids. This aspect of the theory does not account for substances like ammonia (NH3) which do not release hydroxide ions directly into solution.

Ammonia becomes a base according to the Arrhenius model only by its reaction with water where it forms ammonium (NH4+) and hydroxide (OH-) ions. Water donates a hydrogen ion to the ammonia, making water the Brønsted-Lowry acid and ammonia the Brønsted-Lowry base. Thus, ammonia is a base in the Arrhenius sense only when dissolved in water, and reveals the limitations because it does not contain hydroxide in its structure.

In the broader Brønsted-Lowry theory, ammonia is clearly a base since it accepts a hydrogen ion from a water molecule. Similarly, the Lewis model recognizes ammonia as a base since it can donate a pair of electrons to a hydrogen ion.

Answer:

Au3+

Explanation:

The standard cell notation for an electrolytic cell with aluminum and gold electrodes  Au(s)| Au³⁺  ||   Al³⁺| Al (s)

A device in which electrical energy is converted into chemical energy or chemical energy is converted into electric energy is called an Electrolytic cell

The electrolytic cell consists of two metallic electrodes and an electrolyte.

In this cell nomenclature, the electrode to the left of the salt bridge is always assumed to be the anode, and the accompanying half-equation is always stated as an oxidation , while right side is cathode and the half equation is Reduction.

Reaction at Anode: Au(s) → Au³⁺(aq) + 3e⁻  ..........(oxidation)

Reaction at Cathode: Au³⁺(aq) + 3e⁻ →Au(s) ..............(reduction)

The standard cell notation for an electrolytic cell with aluminum and gold electrodes

                         Au(s)| Au³⁺  ||   Al³⁺| Al (s)

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Answer:

No, potassium chloride will not dissolve completely. Only 28.15 g of potassium chloride will get dissolved.

Explanation:

Amount od potassium chloride added to 50 g of boiling water = 250.0 g

The solubility of potassium chloride in boiling water is 56.3 g/100 g water.

Amount of potassium chloride soluble in 100 g of boiling water = 56.3 g

Amount of potassium chloride soluble in 1 g of boiling water = [tex]\frac{56.3 g}{100}[/tex]

Amount of potassium chloride soluble in 50.0 g of boiling water :

[tex]\frac{56.3 g}{100}\times 50.0 g=28.15 g[/tex]

Amount of potassium chloride left undissolved = 250.0 g - 28.15 g = 221.85 g

28.15 g of potassium chloride will dissolve and remaining 221.85 g og potassium chloride will not.

The molarity of the solution is 0.279 M.

To find the molarity of the solution, we need to calculate the number of moles of glucose and the volume of the solution in liters. The molarity is defined as moles of solute per liter of solution.

First, we calculate the number of moles of glucose:

Moles of glucose = mass of glucose / molar mass of glucose

Moles of glucose = 5.10 g / 180 g/mol = 0.028 moles

Next, we calculate the volume of the solution in liters:

Volume of solution = 100.5 ml / 1000 = 0.1005 L

Finally, we plug in the values into the molarity formula:

Molarity = moles of glucose / volume of solution

Molarity = 0.028 moles / 0.1005 L = 0.279 M

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In the formation of one molecule of aluminum oxide (Al2O3), a total of 6 electrons are transferred from two aluminum atoms to three oxygen atoms to maintain electrical neutrality.

The formation of one molecule of aluminum oxide (Al2O3) involves the transfer of electrons between aluminum atoms and oxygen atoms. Each aluminum atom loses three electrons, becoming Al3+ ions, while each oxygen atom gains two electrons to become O2- ions. To maintain electrical neutrality, the compound must have equal numbers of positive and negative charges.

Therefore, to form one molecule of Al2O3, two aluminum atoms (2 Al) will lose a total of 6 electrons (2 Al × 3 e-), and three oxygen atoms will gain a total of 6 electrons (3 O × 2 e-) during the electron transfer process. The complete transfer involves 12 electrons changing places but, in effect, the number of electrons transferred is the total on one side of the reaction which is 6.

Answer:

8

Explanation:

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There are 8 atoms of oxygen in the chemical formal 2Ca(CIO[tex]_2[/tex])[tex]_2[/tex]. Atoms are incredibly tiny, measuring typically 100 picometers across.

A particle called an atom has a nucleus made up of neutrons and protons that is encircled by an electron cloud. The fundamental unit of the chemical components is the atom, and the protons in an atom serve as a means of differentiating one chemical element from another.

Every atom with 11 protons, for instance, is sodium, while any atom with 29 protons becomes copper. The element's isotope is determined by the amount of neutrons in it.

Atoms are incredibly tiny, measuring typically 100 picometers across. Over a thousand carbon atoms make up an average human hair. Because it's smaller than the visible light spectrum's smallest wavelength, people cannot view atoms using standard microscopes. There are 8 atoms of oxygen in the chemical formal 2Ca(CIO[tex]_2[/tex])[tex]_2[/tex].

Therefore, there are 8 atoms of oxygen in the chemical formal 2Ca(CIO[tex]_2[/tex])[tex]_2[/tex].

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Answer:

Electrical energy is produced from oxidation reactions.

Explanation:

In fuel cells elecric energy is produced from the oxidation of reactants, that fuel is often Hydrogen, that is storaged in the fuel cell and then it just grabs oxygen from the air and converts that chemical reaction of oxidation in electric current.

Final answer:

The combustion of 1.00 mol of CH4 produces 44.01 grams of CO2, as one mole of methane combusts to create one mole of carbon dioxide, and the molar mass of CO2 is 44.01 g/mol.

Explanation:

The mass of CO2 produced by the combustion of 1.00 mol of CH4 can be calculated by looking at the balanced chemical equation:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

From the equation, it is clear that one mole of methane (CH4) produces one mole of carbon dioxide (CO2) upon combustion. To convert moles of CO2 to grams, we use the molar mass of CO2, which is 44.01 g/mol.

Calculation:

1.00 mol CH4 × (44.01 g CO2 / 1 mol CO2) = 44.01 g CO2

Therefore, the combustion of 1.00 mol of CH4 will produce 44.01 grams of CO2.

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

[tex] ln \frac{N}{N_0} = - k*t [/tex] ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

[tex] T _\frac{1}{2} = \frac{ln2}{k} [/tex] ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of [tex] T_\frac{1}{2} [/tex] in above formula (equation 2) :

[tex] 28.8 yrs = \frac{ln 2}{ k } [/tex]

Multiply both side by k

[tex] 28.8 yrs * k = \frac{ln 2 }{k} * k [/tex]

Dividing both side by 28.8 yrs

[tex] \frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs} [/tex]

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

[tex] ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t [/tex]

[tex] ln (0.0971 ) = -0.0241 yrs ^-^1 * t [/tex]

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

[tex] \frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1} [/tex]

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

96.9 years

Given:

Question:

How many years would pass before the strontium-90 concentration would drop to 1.0 ppm?

The Process:

In the calculations of half-lives, the expressions are the following:

[tex]\boxed{ \ N = \frac{N_o}{2^n} \ }[/tex]

where [tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ }[/tex] are used. In these expressions;

Step-1: find out the number of half-lives (n)

[tex]\boxed{ \ N = \frac{N_o}{2^n} \ } \rightarrow \boxed{ \ 2^n = \frac{N_o}{N} \ }[/tex]

[tex]\boxed{ \ 2^n = \frac{10.3}{1.0} \ }[/tex]

[tex]\boxed{ \ 2^n = 10.3 \ }[/tex]

[tex]\boxed{ \ n \cdot \ln{2} = \ln{10.3} \ } \rightarrow \boxed{ \ n = \frac{\ln{10.3}}{\ln{2}} \ }[/tex]

We get n = 3.364572

Step-2: find out in how many years would pass before the strontium-90 concentration would drop to 1.0 ppm

[tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ } \rightarrow \boxed{ \ t = n \times t_{1/2} \ }[/tex]

t = 3.364572 x 28.8

t = 96.9

Thus in 96.9 years will pass before the concentration of strontium-90 will drop to 1.0 ppm.

- - - - - - - - - -

Notes:

The half-life of radioactive decay is the period of time required for half of the initial amount of the substance to disintegrate. The shorter the half-life of radioactive decay, the higher the rate of radioactive decay and the more radioactivity. The half-life is the characteristic property of each element.

Answer: acid + base → salt + water

Explanation: Neutralization is a type of double displacement reaction in which exchange of ions take place. Neutralization is a chemical reaction between an acid and a base where ions exchange and lead to formation of salt and water.

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O[/tex]

HCl is an acid which donates [tex]H^+[/tex] ions when dissolved in water. [tex]NaOH[/tex] is base which donates [tex]OH^-[/tex] ions when dissolved in water. They combine to form [tex]NaCl[/tex] which is salt and [tex]H^+[/tex] and [tex]OH^-[/tex] ions combine to form water [tex]H_2O[/tex].

Answer:

false

Explanation:

Final answer:

Thermal expansion is the increase or decrease in size of a body due to a change in temperature. Solid metals undergo greater thermal expansion than liquids. This is because the closely packed atoms or molecules in solid metals are pushed farther apart by the increase in temperature, resulting in a larger size for the whole body.

Explanation:

Thermal expansion: Thermal expansion is the increase, or decrease, of the size (length, area, or volume) of a body due to a change in temperature. It occurs in all dimensions - length, area, volume - and is not limited to solid metals. However, solid metals undergo greater thermal expansion than liquids do. The increased thermal expansion in solid metals can be attributed to their closely packed atoms or molecules, which are pushed farther apart by the increase in temperature, resulting in a larger size for the whole body.

Answer: After 20 seconds, 16 grams of unobtanium will be present, after 40 seconds, 8 grams of unobtanium will be present after 60 seconds, 4 grams of unobtanium will be present, after 80 seconds, 2 grams of unobtanium will be present and after 100 seconds, 1 grams of unobtanium will be present.

Explanation: Half life is the time in which half of the reaction is completed. Thus the half of the substance will be decomposed and half of it will remain.

Amount of the substance left after n half lives will be=[tex]\frac{A}{2^n}[/tex]

where A= initial amount of substance

n=no of half lives=[tex]\frac{\text{given time}}{\text{half life}}[/tex]

a) t= 20 seconds

no of half lives=[tex]\frac{20}{20}=1[/tex]

amount of the substance left after 1 half life will be=[tex]\frac{32}{2^1}[/tex]=16 g.

b)  t= 40 seconds

no of half lives=[tex]\frac{40}{20}=2[/tex]

amount of the substance left after 2 half lives will be=[tex]\frac{32}{2^2}[/tex]=8g.

c) t= 60 seconds

no of half lives=[tex]\frac{60}{20}=3[/tex]

amount of the substance left after 3 half lives will be=[tex]\frac{32}{2^3}=4g[/tex]

d) t= 80 seconds

no of half lives=[tex]\frac{80}{20}=4[/tex]

amount of the substance left after 4 half lives will be=[tex]\frac{32}{2^4}=2g[/tex]

e) t= 100 seconds

no of half lives=[tex]\frac{100}{20}=5[/tex]

amount of the substance left after 5 half lives will be=[tex]\frac{32}{2^5}=1g.[/tex]

The half-life of a radioactive element is defined as the time required by the specific isotope to decrease by half of its original value. The unobtanium after given time will be present as:

Half-life is the time required by the unobtanium is the half of the reaction is completed. Half of the substance will be decomposed, such that:

Amount of substance left n half-lives = [tex]\dfrac{\text A}{{2}^{\text n}}[/tex]

where A= initial amount of substance

Now,

n = number of half-lives = given time /half-life

Given,

1. Time = t = 20 seconds

2.Time = t = 40 seconds

3.Time = t = 60 seconds

4.Time = t = 80 seconds

5.Time = t = 100 seconds

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