Three examples of environmental, industrial and bio-chemistry are listed below:
- Environmental chemistry: Contamination, Atmospheric Deposition, and Soil Pollution.
- industrial chemistry: industrial inorganic chemicals, industrial organic chemicals, and agricultural chemicals
- bio-chemistry: genetic, immunology, and enzymology
<h3>Meaning of Chemistry</h3>
Chemistry can be defined as a branch of science which is concerned with the substances matter is composed of, their properties and reactions,
Chemistry also deals with the use of such reactions to form new substances.
In conclusion, Three examples of environmental, industrial and bio-chemistry are listed anove
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Answer:
B) Electrons are located in the cloud-like areas around the nucleus.
Explanation:
The quantum mechanical model of the atom does not consider the path through which an electron travels. It rather estimates the probability of where electrons can be found at each energy level.
The region of maximum probability of where an electron is located is sometimes called an electron cloud or orbital. Each orbital of an atom and the electrons accomodated are described completely by a set of four quantum numbers.
Hydrogen sulfide gas is reacted with dioxygen gas to produce gaseous sulfur dioxide and water. It is false.
When hydrogen sulfide gas reacts with dioxygen gas, it produces water and solid Sulphur and not gaseous sulfur. The chemical equation of the given substances is :
2H2S + O2 → 2S + 2H20
It is a redox reaction as both Oxidation and reduction happens in it. Oxidation of hydrogen sulphur to sulphur happens and reductions of oxygen to water takes place.
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Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point