Answer:
Cracking.
Explanation:
A chemical reaction can be defined as a reaction in which two or more atoms of a chemical element react to form a chemical compound. An example of a chemical reaction involving hydrocarbons is cracking.
Hydrocarbon can be defined as an organic compound that comprises of hydrogen and carbon only.
In Organic chemistry, cracking refers to the process of lowering the boiling point of a heavy, complex or long-chain hydrocarbon such as kerogens by exposing it to different environmental treatments such as hydrogen enriched catalysts, pressure or high temperatures, in order to produce smaller, lighter and more useful molecules (alkanes and alkenes) such as gasoline, diesel fuel, etc.
Glucose is a simple sugar with the molecular formula C6H12O6. It is a carbohydrate.
Answer:
The over all charge on atom will be +2.
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
For example
X is the element having 12 protons 10 electrons and 14 neutrons.
The number of protons and electrons are not equal which means two electrons are lose by the given atom and it is present in the form of cation.
The over all charge on atom will be +2.
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
