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3 years ago

Ions are atoms that have gained or lost which subatomic particles?

2 answers:

7 0

Protons cannot be lost without the atoms becoming an entirely different element. Elements can infact have different numbers of neutrons within the same element, but neutrons are not related to the electrical charge. The answer is c. electrons.

melisa1 [442]3 years ago

7 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

There are 3 subatomic particles present in an atom. They are: protons, electrons and neutrons.

Protons: They carry positive charger and are found in nucleus.
Electrons: They carry negative charge and are found around the nucleus.
Neutrons: They dos not carry any charge and are found in nucleus.

An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.
When an atom gains electrons, it will form a negative ion known as anion.

Hence, the correct answer is Option C.

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When writing equations the mass on the left of the arrow must

sp2606 [1]

Answer:

When writing equation the mass on left side of equation must be equal to the mass on right side. True

Explanation:

The chemical reactions always follow the law of conservation of mass.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

For example:

In photosynthesis reaction:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass because total mass is equal on both side of equation.

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3 years ago

Mercury is a liquid meteal but it is not hard? how can we say it is a metal

NNADVOKAT [17]

What's the problem ? Hardness is not the definition of a metal. You need to expand your thinking. EVERY element is solid, liquid, and gas, over different ranges of temperature ... including all of the metals. There are only TWO elements that are liquid AT ROOM TEMPERATURE, and mercury is one of them. But on a mild day at the south pole, mercury is solid too.

7 0

3 years ago

The antibiotic prontosil, developed by bayer pharmaceuticals in 1932, was used in saving the life of president roosevelt's son f

Alja [10]

Prontosil is a compound produced by the coupled reaction of an aryldiazonium ion and an aromatic compound.

<h3>What are diazonium compounds?</h3>

These are organic compounds in which there are ionic interactions between the azo group (-N₂⁺) and an anion X⁻.

The general structure is RN₂⁺X⁻.

R is the lateral chain that might be an aromatic ring, among other options.

N₂⁺ is the azo group

X⁻ is the anion

The azo group characterizes as being unstable and reactive. This property is because one of the N atoms has a positive charge.

-N⁺≡ N

<h3>What is the coupling reaction of aryldiazonium compounds?</h3>

Aryldiazonium salt reactions can occur in two ways,

Substitution reactions
Coupling reactions

Coupling reactions are the aromatic electrophilic substitution, where the aryldiazonium ion acts as an electrophile for an activated aromatic compound to attack it.

The coupling reaction occurs at the azo group level.

In the exposed example,

the benzene ring with sulfur bonded to oxygen atoms is the coupling component
the benzene ring with NH₂ and the azo group is the diazonium ion

In the attached files you will find the drawings.

You can learn more about diazonium compounds at

brainly.com/question/9061271

brainly.com/question/9389852

brainly.com/question/13147442

5 0

2 years ago

One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa

Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = - nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K; T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0

3 years ago

Do all electrodes make good canthodes and anodes? why or why not

Sedbober [7]

An electrode is an electrical conductor used to make contact with a nonmetallic part of a circuit The word was coined by William Wheel at the request of the scientist Michael Faraday from the Greek words electron, meaning amber and hods, a way.

4 0

3 years ago