Occurs at a specific temperature
Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in

<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =



<u>= 34.72 g</u>