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:
C. the relative number of atoms of each element, using the lowest whole ratio.
Explanation:
The empirical formula is how we simplify the whole formula to simplify it to its smallest indivisible parts.
It is definitely not the actual number of atoms. If you see an empirical formula, don't think that it's the full thing.
It is also not a representation of a compound to show its atoms' arrangement: this would be a Lewis dot structure, or a ball and stick model, or something similar. We don't use the empirical formula for this purpose.
Answer:
The answer to your question is: C₃H₃O This is my answer.
Explanation:
Data
Sample = 1.3109 g
CxHyOz
CO₂ = 3.2007 g
H₂O = 1.3102 g
Empirical formula = ?
MW CO2 = 44 g
MW H2O = 18 g
For Carbon
44 g -------------------- 12 g
3.2007 g ------------ x
x = (3.2007 x 12) / 44
x = 0.8729 g of Carbon
12 g of C -------------- 1 mol
0.8729 g -------------- x
x = (0.8729 x 1) / 12
x = 0.0727 mol of Carbon
For Hydrogen
18 g ---------------------- 1 g
1.3102 g ------------------- x
x = (1.3102 x 1) / 18
x = 0.0727 g of Hydrogen
1 g ------------------------ 1 mol
0.0727g ---------------- x
x = (0.0727 x 1)/1
x = 0.0727 mol of Hydrogen
For oxygen
g of Oxygen = g of sample - g of Carbon - g of hydrogen
g of Oxygen = 1.3109 - 0.8709 - 0.0727
g of Oxygen = 0.3673
16 g of Oxygen ------------- 1 mol of O
0.3673 g --------------------- x
x = (0.3673 x 1)/ 16
x = 0.0230 mol of Oxygen
Divide by the lowest number of moles
Carbon 0.0727 / 0.023 = 3.1 ≈ 3
Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3
Oxygen 0.0230 / 0.023 = 1
C₃H₃O
Answer:
Answers with detail are given below
Explanation:
1) Given data:
Mass of Rb₃Rn = 76.19 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 478.43 g/mol
Number of moles = 76.19 g/ 478.43 g/mol
Number of moles = 0.16 mol
2) Given data:
Mass of FrBi₂ = 120.02 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 640.96 g/mol
Number of moles = 120.02 g/640.96 g/mol
Number of moles = 0.19 mol
3) Given data:
Mass of Zn₂F₃ = 88.24 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 187.73 g/mol
Number of moles = 88.24 g/ 187.73 g/mol
Number of moles = 0.47 mol
4) Given data:
Number of moles of Sb₄Cl = 1.20 mol
Mass of Sb₄Cl = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 522.49 g/mol
Mass = Number of moles × molar mass
Mass = 1.20 mol × 522.49 g/mol
Mass = 626.99 g
