The correct answer is 53.
It would be neutral if they were both at 53, however, since there's 54 electron, then it becomes negatively electrified and becomes -1. You just delete one number from it and you'll see that it's 53.
Given data:
Diameter of the gold wire (d) = 0.175 cm
Length of the gold wire (l) = 1.00 * 
cm
Volume of the given gold wire = volume occupied by a cylindrical object\
= π * 
*l
here r = radius of the cross section
where : r = d/2
Hence the volume V = π * 
*l
= 
Density of gold = 19.3 g/cm3
Weight of the gold wire = Density * volume
= 
* 
= 
I believe Calcareous is not a hydrogenous sediment. Hydrogenous sediments are example of marine sediments that are formed directly from chemical processes in sea water. They include, manganese nodules, phophorites, metal sulfides, evaporites and carrbonates. In shallower areas, such as on continental shelves and near islands, rock salt, calcium salts and sulfates may settle on the ocean floor.
Answer:
NH3(g) + H2O(1) → NH4+(aq) + OH (aq)
HF(aq) + H2O(1) → H3O+(aq) + F (aq)
Explanation:
Acid-base reactions are chemical reactions involving acids and bases. Acids tend to ionize/dissociate in water, a property which determines their strength. Ionization of an acid refers to the acid losing its hydrogen ion (H+) in water solution. An acid ionizes or dissociates to form a conjugate base.
A strong acid is so because it ionizes completely in water i.e. loses all its hydrogen ion (H+) while a weak acid partially ionizes in water.
In the chemical reactions;
1) NH3(g) + H2O(1) → NH4+(aq) + OH (aq)
H20 loses its hydrogen ion (H+) in this reaction to form an anion (OH-). Hence, water (H20) is an acid in this case which ionizes to form a conjugate base (OH-). This is an example of ionization of acid.
2) HF(aq) + H2O(1) → H3O+(aq) + F (aq)
Hydrogen fluoride (HF) loses its hydrogen ion (H+) in the presence of water to form anion (F-). The HF is the acid while F- is it's conjugate base. Thus, an example of ionization of acid
