Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
Answer:
The length of an edge of this unit cell is 407.294 pm
Explanation:
Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.
To determine the edge length, a relationship between the radius of the atom and edge length is used.
X = R√8
Where;
X is the length of an edge of this unit cell
R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m
X = 144 X 10⁻¹²√8
X = 407.294 X 10⁻¹² m
X = 407.294 pm
Therefore, the length of an edge of this unit cell is 407.294 pm
Answer:
there are 9 Chlorine atoms
Explanation:
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Initial concentration
= ?
Final concentration
= 12.5 grams
Applying in the above equation, we get that:-
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
D there is one kind of cell of which all living things are made