Answer:
Decomposition
Explanation:
A decomposition reaction is a type of reaction in which a compound is broken down into its constituent elements sometimes under the influence of heat.
When iron (III) hydroxide is heated,new products are formed according to the equation; 2Fe(OH)3 -----------> Fe2O3 + 3H2O.
This is a thermal decomposition reaction.
Answer:
You can view more details on each measurement unit: molecular weight of Copper(I) Oxide or grams The molecular formula for Copper(I) Oxide is Cu2O. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Copper(I) Oxide, or 143.0914 grams.
Explanation:
Answer:
Precipitation
Explanation:
In the water cycle, water experiences different phase changes from one state to another in nature.
The cycling affords water to in solid, liquid and the vapor form.
From the cycle show, W represents precipitation.
- During precipitation, water in the atmosphere begins to fall.
- U is evaporation
- X is transpiration.
- V is the condensation.
Answer:
If you are meaning O2^2- ion, well, don’t forget that ions have a charge, that has to be specified when referring to them. It is a polyatomic ion just because it consists of more than one atom, irrespective of the fact that these atoms are of the same element or not. It was given that name because at the early times chemistry was founded as a science, it was found that with respect to other oxygenated substances, peroxides contained more oxygen than expected: Latin prefix per- gives the terms it is attached to the meaning of being increased, enhanced, and similar.
Explanation:
Answer:
The concentration of chloride ion is
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the content 100 ppm suggests the presence of 100 mg of in 1 L of solution.
The molar mass of is equal to the molar mass of Cl atom as the mass of the excess electron in is negligible as compared to the mass of Cl atom.
So, the molar mass of is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of present in 100 mg (0.100 g) of is calculated as shown below:
So, there is of present in 1 L of solution.