Order of increasing atomic mass!
hope this helps :)
Answer:
Answer is given below:
Explanation:
Given Data:
heat = 3.4kJ
work done is = 1.9 kJ
To Find:
ΔE=?
Formula:
ΔE = q + w
Solution:
ΔE = q + w
ΔE = 3.4 kJ + 1.9kJ
ΔE = 5.3 kJ
b. Alkyne series
c. Saturated hydrocarbons
d. Unsaturated hydrocarbons
e. Aromatic hydrocarbons
B. Pandemic
C. Epidemiology
D. Toxicology
The answer is A
The bacteria that cause disease are known as pathogens. The term pathogen is used to refer to all types of disease causing microorganisms. These include bacteria, fungi, protozoa and fungi.
Bacteria. Microscopic organisms that come in many shapes and sizes. Some bacteria that cause disease in man are Salmonella typhi which causes typhoid and Streptococcus pyogens which causes sore throat.
Virus. A virus is a microscopic entity much smaller than even bacteria and can only exist inside a host such as a cell. It cannot live on its own. Some viruses that cause disease are HIV which causes AIDS and Rhino virus which causes colds.
Fungi. These are a group of unicellular or multicellular microscopic organisms that live by feeding on organic matter. A type of fungus that causes disease is Trichophyton mentagrophyte which is responsible for athlete's foot.
Protozoa. A group of one celled organisms which live in water. Entamoeba histolytica is a protozoa, an amoeba which causes amoebic dysentery in man.
Ca²⁺
Ionization energy is defined as the minimum amount of energy required to knock out the electron from valence shell of an atom in its gaseous state. While, second Ionization energy is defined as the amount of energy required to knock out the second electron from an ion containing +1 charge in gaseous state.
Among given options Ca²⁺ is the correct choice because the calcium has lost two electrons i.e. first electron was removed by providing first ionization energy i.e.
Ca + 1st IE → Ca¹⁺ + 1 e⁻
and second electron is was removed by providing second ionization energy i.e.
Ca ¹⁺ + 2nd IE → Ca²⁺ + 1 e⁻
is formed by providing the second ionization energy to remove an electron.
Further Explanation:
The energy that is needed to remove the most loosely bound valenceelectrons from the isolated neutral gaseous atom is known as the ionization energy. It is denoted by IE. The value of IE is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.
Ionization energy is further represented as first ionization, second ionization and so on. When the first electron is removed from a neutral, isolated gaseous atom, the energy needed for the purpose is known as the first ionization energy, written as . Similarly, when the second electron is removed from the positively charged species (cation), the ionization energy is called the second ionization energy and so on.
The neutral atom corresponding to is calcium. If second ionization energy is supplied to calcium atom, it results in the removal of two electrons and thus is formed. So can be formed by providing second ionization energy. to the neutral atom.
The neutral atom corresponding to is nitrogen. If second ionization energy is supplied to nitrogen, it results in the formation of, not . So cannot be formed by providing the second ionization energy to the neutral atom.
The neutral atom corresponding to is iron. If second ionization energy is supplied to the iron atom, it results in the formation of , not . So cannot be formed by providing the second ionization energy to the neutral atom.
The neutral atom corresponding to is sulfur. If second ionization energy is supplied to the sulfur atom, it results in the formation of , not . So cannot be formed by providing the second ionization energy to the neutral atom.
Therefore, the only ion that can be formed by supplying the second ionization energy is .
Learn more:
1. Which is the oxidation-reduction reaction:
2. What is the mass of 1 mole of viruses: brainly.com/question/8353774
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Periodic classification of elements
Keywords: second ionization energy, Ca2+, N3-, Fe3+, S2-, IE1, IE2, first electron, second electron, neutral atom., nitrogen, calcium, iron, sulfur.
Answer:
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are
Answer:
Answer:
\boxed{\text{(a) 0.00 V; (b) 0.424 V}}
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}
Explanation:
its right trust