S8 solid sulfur (i.e 8 sulfur atoms covalently bound together), and O2 is a molecule of oxygen gas ( two oxygen atoms bounds together)
No, P4 S8 and O2 are not polyatomic compounds. P4 and S8 refer to elements forming molecules, while O2 refers to a diatomic molecule. Polyatomic compounds consist of two or more different elements bonded together in a molecule.
Sulfur is generally S8. S8 + 8 O2 → 8 SO2
The answer to the chemistry question would be false. 21.4g S8 is not equal to 171.2g S. Grams is a measurement of weight, and in order to be exact, the weight needs to be the same.
1/8 S8 + O2 --> SO2 , delta H degree f = -296.9 kJ
S8 is the symbol of an allotrope of sulfur, a chemical element - not a compound.
Type of reaction for S8 + O2
No, P4 S8 and O2 are not polyatomic compounds. P4 and S8 refer to elements forming molecules, while O2 refers to a diatomic molecule. Polyatomic compounds consist of two or more different elements bonded together in a molecule.
First, balance the chemical equation: S8 + 8 O2 -> 8 SO2. Calculate the moles of each reactant using their molar masses. The limiting reactant is the one that produces the least amount of SO2, which is S8 in this case. Therefore, use the stoichiometry of the balanced equation to calculate the mass of SO2 produced from 31.5g of S8.
Sulfur is generally S8. S8 + 8 O2 → 8 SO2
The answer to the chemistry question would be false. 21.4g S8 is not equal to 171.2g S. Grams is a measurement of weight, and in order to be exact, the weight needs to be the same.
H2o + co2 ----> o2 + carbohydrate
S8 refers to the molecular formula of octasulfur, which consists of eight sulfur atoms bonded together in a cyclic structure. It is the most stable and common allotrope of sulfur, forming a crown-shaped molecule. In this form, the sulfur atoms are arranged in a ring, making S8 a distinctive and well-studied compound in chemistry.
S8 + 8 O2 → 8 SO2 burning sulfur in the presence of oxygen
1/8 S8 + O2 --> SO2 , delta H degree f = -296.9 kJ
S8 + 12O2 = 8SO3.Since there are 8 sulfur on the left, you must have 8 on the right, so 8SO3. Now you have 24 oxygen on the right, so you need 24 on the left. Multiplying O2 by 12 balances it.
s8
The oxidation number of S in S8 is ZERO!