The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
Answer:
3.47 mol MnSO4
Explanation:
54.94 + 32.07 + (16 × 4) = 151.01g/mol MnSO4
542.32g / 151.01g/mol = 3.47 mol MnSO4
It has a pb of 69c from google
The best answer is C
Since identical atoms would have the same electronegativity, thus making its electronegativity difference close to zero, if not equal to zero (O), making the bond between the two to be covalent. Thus, a chemical bond formed between two identical atoms is a covalent bond.
