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Answer:
we need 6.0 moles of zinc (Zn)
Explanation:
Step 1: Data given
Number of moles ZnO produced = 6.0 moles
Step 2: The balanced equation
2 Zn + O2 → 2 ZnO
For 2 moles Zinc we need 1 mol Oxygen to produce 2 moles Zinc oxide
Step 3: Calculate moles zinc
For 2 moles Zn we need 1 mol O2 to produce 2 moles ZnO
For 6.0 moles 2nO produced, we need 6.0 moles of zinc (Zn) and 3.0 moles of O2 to react.
This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.
- The alkanes are non- polar solvents.
- The alkanes are immiscible in water but freely miscible in other non-polar solvent .
- The alkanes are consisting of weak dipole dipole bonds can not breaks the strong hydrogen bond.
- The alkanes having only carbon (C) and hydrogen (H) atom which is bonded by a single bonds only.
- The alkanes posses weak force of attraction that is weak van der waals force of attraction.
The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.
learn about butane
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Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ = ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 2 x 3.64 x 10⁻³ M/s
⇒ = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s
Two. O and E are both vowels
