The oxidation state of the elements in the compounds are:
CoH₂:
FeBr₃:
<h3>What is the oxidation states of the elements in the given compounds?</h3>
The oxidation states of the elements in each of the given compounds is determined as follows:
Cobalt dihydride, CoH₂
Co = +2
H = -1
Iron (iii) bromide, FeBr₃
Fe = +3
Br = -1
In conclusion, the oxidation state of the elements are charges they have in the compound.
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Answer:
Mass = 357.7 g
Explanation:
Given data:
Mass of Fe = 250 g
Mass of oxygen = 120 g
Mass of iron(III) oxide produced = ?
Solution:
Chemical equation:
4Fe + 3O₂ → 2Fe₂O₃
Number of moles of Fe:
Number of moles = mass/molar mass
Number of moles = 250 g/ 55.8 g/mol
Number of moles = 4.48 mol
Number of moles of O₂ :
Number of moles = mass/molar mass
Number of moles = 120 g/ 32 g/mol
Number of moles = 3.75 mol
Now we will compare the moles of reactants with product.
Fe : Fe₂O₃
4 : 2
4.48 : 2/4×4.48 = 2.24
O₂ : Fe₂O₃
3 : 2
3.75 : 2/3×3.75= 2.5
Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.
Mass of Fe₂O₃:
Mass = number of moles × molar mass
Mass = 2.24 mol × 159.69 g/mol
Mass = 357.7 g
Answer:
Number of protons
Explanation:
There are three sub atomic particles. These are;
- Protons
- Electrons
- Neutrons
Among these three particles, only one determines the identity of the element. This is the Protons. The number of protons which is also called the atomic number determines the identity of an element. For instance, atom with one proton is Hydrogen and n other element can have atomic number of one.
Answer:
Celsius is currently a derived unit for temperature in the SI system, kelvin being the base unit. ... The two main reference points of the Celsius scale were the freezing point of water (or melting point of ice) being defined as 0 °C and the boiling point of water being 100 °C.
Explanation:
Hope it helps
Ok then! So mitosis is when a cell splits and doesn't lose/gain any chromosomes. In meiosis the chromosomes join and split evenly at the cell's "poles". Chromosomes will be lost evenly through this process.
