What is formed from 2 pieces of different metals stuck together lengthwise?

Answer:

Answer is B

Explanation:

During a redox reaction, the number of electrons gained is equal to the number of electrons lost, demonstrating the law of conservation of charge.

In an oxidation-reduction reaction, also known as a redox reaction, the number of electrons gained is equal to the number of electrons lost. This principle is known as the law of conservation of charge which states that the total charge before a reaction must equal the total charge after the reaction.

So, when one substance loses electrons (oxidation), another must gain them (reduction). Therefore, the correct answer is A) equal to the number of electrons lost.

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The equilibrium constant is a number that shows the extent to which reactants are converted into products. The equilibrium constant of the reaction is 0.76.

The equilibrium constant is a number that shows the extent to which reactants are converted into products.

We know that to obtain the equilibrium constant; K = k1/k-1 where;

k-1 = rate constant of reverse reaction

k1 = rate constant of forward reaction.

Hence;

K = 297 l·mol–1·min–1/393 l·mol–1·min–1

K = 0.76

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The molar solubility of CaSO4 in a solution containing 0.100 M Na2SO4 is 2.4 x 10−4 M, calculated using the Ksp value and the concentration of sulfate ions due to the common ion effect.

The molar solubility of CaSO4 in a solution containing 0.100 M Na2SO4 can be calculated using the solubility product constant (Ksp) and the concept of common ion effect. The solubility product expression for CaSO4 can be written as Ksp = [Ca²+ ][SO4²−]. Since the sodium sulfate solution already contains sulfate ions, we will consider the concentration of [SO4²−] as 0.100 M given in the question. The equation simplifies to Ksp = [Ca²+]*0.100. Therefore, the molar solubility is [Ca²+]= Ksp/0.100 which is 2.4 x 10−5/0.100 = 2.4 x 10−4 M. This suggests that in a 0.100 M Na2SO4 solution, additional CaSO4 would start precipitating when the [Ca²+] reaches 2.4 x 10−4 M.

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The molar solubility of CaSO₄ in a solution containing 0.100 M Na₂SO₄, with a Ksp of 2.4 × 10⁻⁵ is 2.4 × 10⁻⁴ M, which is lower than in pure water due to the common ion effect.

Calculating the molar solubility of calcium sulfate (CaSO₄) in a solution containing 0.100 M sodium sulfate Na₂SO₄ given the solubility product constant (Ksp) for CaSO₄ is 2.4 × 10⁻⁵.

Because Na₂SO₄ dissociates into 2 Na⁺ and 1 SO₄²⁻ in solution, the concentration of SO₄²⁻ due to Na₂SO₄ is 0.100 M. Since the solution is already saturated with sulfate ions, the molar solubility of CaSO₄ will be different from its solubility in pure water. The common ion effect will reduce the solubility of CaSO₄ in the solution.

To find the molar solubility of CaSO₄, let's set up the equation based on the Ksp expression:

Ksp = [Ca²⁺] [SO₄²⁻]

Since the concentration of SO₄²⁻ from Na₂SO₄ is already 0.100 M, and the Ksp for CaSO₄ is 2.4 × 10⁻⁵, the molar solubility of CaSO₄ in this solution is calculated as:

Ksp = (s) × (0.100 M)

2.4 × 10⁻⁵ = (s) × (0.100 M)

s = 2.4 × 10⁻⁴ M

Therefore, the molar solubility of CaSO₄ in a 0.100 M Na₂SO₄ solution is 2.4 × 10⁻⁴M.

Answer :The correct answer for (C₄H₁₀ )₃ N is C) TERTIARY AMINE .

The substitution level of nitrogen in amine or in simple words , description of carbon atoms attached to a given nitrogen in any amine is of 4 types :

1 ) Primary amine :

When the given nitrogen is attached to only one carbon then substitution level is known as PRIMARY amine.

Example : CH₃NH₂ : The nitrogen is attached to only one carbon (CH₃) . Hence it is primary amine

2) Secondary amine :

When the given nitrogen is attached to two carbons .

Example : (CH₃)₂NH : The nitrogen is attached to two carbon ( CH₃)₂ , hence it is secondary amine .

3) Tertiary amine :

When the given nitrogen is attached to three carbon atom .

Example : (CH₃)₃NH : The nitrogen is attached to 3 carbons atoms (CH₃)₃ . Hence it is a tertiary amine

4) Quaternary amine : when the nitrogen attached to 4 carbon atoms and amine posses a positive charge on it , its is known as quaternary amine .

Example : (CH₃)₄N⁺ : The nitrogen is attached to 4 carbons and nitrogen . hence it is quaternary amine .

(Image attached )

The given compound is (C₄H₁₀ )₃ N has 3 groups of C₄H₁₀ attached to N , means 3 carbons attached to N , ( image attached ) . Hence it can be said that (C₄H₁₀ )₃ N is TERTIARY AMINE .

Hence correct option is C) Tertiary amine .

For light 3 to function, switch D does not have to be on. This diagram shows a parallel circuit that provides more than one way for the current to return to the power source.

Answer:

Potassium dicarbonatedifluoroplatinate (II)

Explanation:

Hello,

Based on the IUPAC rules, the given compound is a complex called potassium dicarbonatodifluoroplatinate (II) as long as the carbonate ion is present twice as a monodentate ligand therefore it is preceded by a "di" prefix. In addition, two fluorines are also present with the "di" prefix as well as a platinum which complete the anionic section including +2 as the platinum oxidation state.

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Complete the following table.

Acid Molarity Moles of H⁺ released per liter

HCl 1

H2SO4 1

H3PO4 1

H2SO4 0.5

H3PO4 3

HNO3 2

[Answer]

1

2

3

1

9

2

Answer:

The answer is smooth and Involuntary.

Explanation:

Smooth muscles are involuntary which means they automatically do a function without the brain telling it to.

The colors of the compounds in the beakers could potentially be: [CoF6]3– (green), [Co(NH3)6]3+ (yellow-orange), and [Co(CN)6]3– (red), although colors can vary based on different conditions.

The question asks us to identify the compounds present in three beakers based on the color of the solution. While this is generally impossible to answer with absolute certainty without more information or additional tests, we can make a fair guess based on some known color-characteristics of these compounds.

[CoF6]3– is generally green due to the color of the Cobalt(III) ion.

[Co(NH3)6]3+ is usually yellow to orange in color.

FInally, [Co(CN)6]3– commonly has a strong red color.

Note that in reality, actual colors can vary depending on concentration, temperature, and other factors.

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The wavelength of a hockey of mass 0.17kg traveling at a speed of37 m/s can be calculated by using De Broglie's equation and it comes out to be  1.05m×10⁻³⁴m.

The De Broglie's equation is an equation which shows that a particle has a wave length and hence particle can show wave nature. Any matter can have two nature one is particle nature and other is wave nature.

The wave nature in a particle was given by De Broglie Particle nature of a matter can be described by the phenomenon of black body radiation. Photoelectric effect two shows particle nature nature of light.

Mathematically,

[tex]{\lambda}=\frac{h}{mv}[/tex]

where,

[tex]{\lambda}[/tex]=wavelength=?

m=mass of hockey=0.17kg

h=Plank's constant

v=velocity of hockey=37 m/s

Substituting all given values in the equation

[tex]\lambda[/tex]= (6.626×10⁻³⁴)÷(0.17kg ×37 m/s )=1.05m×10⁻³⁴m

Thus the wavelength of motorcycle is 1.05m×10⁻³⁴m.

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Answer:

The correct answer is 1. Moseley started his investigations by reviewing the work of those who contributed to the previous versions of the table.

Explanation:

Henry Gwyn Jeffreys Moseley was a British physicist. His most important scientific contribution was the proof of the correctness of the concept of atomic numbers in chemistry.

In 1913, using X-ray spectroscopy, he found a systematic relationship between the wavelength and the atomic number. Previously, it was assumed that the atomic number was an arbitrary number based on the order of the atomic masses, but had to be changed to bring an element into the right place in the periodic table. Moseley's discovery showed that atomic numbers had an experimentally measurable basis. In addition, he showed that there were gaps in numbers 43, 61 and 75 (known today as the radioactive elements technetium and promethium, as well as the stable but rare rhenium). His work was further proof of the then controversial atomic theory.

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

GPE is determined by the height of the object and its mass. When you cut the height in half, the GPE is reduced by half because the potential energy is directly proportional to the height.

For example, if a book has a certain GPE at a certain height, cutting the height in half will result in the book having half the GPE it had before.

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Molten barium chloride is separetes:
BaCl₂(l) → Ba(l) + Cl₂(g), 
but first ionic bonds in this salt are separeted because of heat: 
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

The electrolysis of molten barium chloride involves reduction of barium ions and oxidation of chloride ions, requiring two electrons to move through the circuit for one unit of the reaction.

During the electrolysis of molten barium chloride (BaCl2), barium ions (Ba2+) are reduced at the cathode, and chloride ions (Cl-) are oxidized at the anode. The half-reactions, showing the movement of electrons through the circuit, are as follows:

The overall cell reaction is obtained by combining the half-reactions and balancing the electrons:

Ba2+(l) + 2Cl-(l) → Ba(l) + Cl2(g)

For each mole of barium ions reduced, two electrons are involved in the transfer. Similarly, for each mole of chlorine gas produced, two electrons are given up by chloride ions. As such, there would be two electrons that moved through the circuit for one unit of the reaction to occur.

The percent yield of CO2 is calculated by dividing the actual yield (28.16 g of CO2) by the theoretical yield determined from the stoichiometry of the balanced chemical equation and the limiting reactant, and then multiplying by 100 to get the percentage.

The question asks for the percent yield of CO2 produced from the reaction of C8H18 with O2. First, we need to consider the balanced chemical equation:

Therefore, the percent yield can be calculated by dividing the actual yield of CO2 (28.16 g) by the theoretical yield (calculated in step 4), and then multiplying by 100 to get a percentage.

The answer is:

the mass of the excess reactant  leftover after the reaction has reached completion is 90.135 grams

The explanation:

According to the reaction equation:

 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

when m is the mass of C₂H₆ m(C₂H₆) = 152 g

So we need to get the number of moles of C₂H₆

n(C₂H₆) = mass C₂H₆ / molar mass of C₂H₆(M)

and when the molar mass of C₂H₆ = 30 g/mol

so, by substitution:

n(C₂H₆) = m(C₂H₆) / M(C₂H₆).

n(C₂H₆) = 152 g / 30 g/mol.

n(C₂H₆) = 5.067 mol.

Then

when the mass of O₂ m(O₂) = 231 g

so we need to get the number of moles of O₂

when nO₂ = mass O₂/ molar mass of O₂

when molar mass of O₂ = 32 g /mol

So, by substitution:

n(O₂) = 231 g / 32 g/mol.

n(O₂) = 7.218 mol

So O₂ is the limiting reactant

according to the chemical reaction we can get the molar ratio between the O₂and C₂H₆:

n O₂ : n C₂H₆    →  7.218 : n C₂H₆

      7 : 2                   7      :   2

∴ n(C₂H₆) = 2 * 7.218 mol / 7

∴ n(C₂H₆) = 2.0625 mol

The number of moles remaining n(C₂H₆) = 5.067 mol - 2.0625 mol

∴ n (C₂H₆) = 3.0045 mol

So the mass remains = moles remains * molar mass of C₂H₆

∴ m (C₂H₆) = 3.0045 mol * 30 g/mol = 90.135 g