The ionic formula of sodium oxide would be Na20
Answer: 0.0069L
Explanation:
2H2O(l) ---->O2(g) + 4H+(aq) + 4e-
no of moles= it/eF
NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)
Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)
= 0.0002798 moles= 2.798x 10 ^-4moles
Using ideal gas equation,
P V = n R T
Where, P is the pressure,
V is the volume,
n is the number of moles,
R is the gas constant, and T is the temperature
We have, 1 bar = 0.986923 atm
Substituting the values,
V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L
Volume of O2 produced = 0.0069L
Answer:
changing shape under heat and pressuremelting completely, then cooling
Explanation:
Answer:
See explaination
Explanation:
Reaction between p- toluidine and o- vaniline to form imine.This reaction does occur in the presence of a acidic source.
Here the -OH group in the o- vaniline is quite acidic so it acts a internal acidic catalyst. Hence the reaction occurred in the absence of any external acidic source.The reaction mechanism is given at the attached file.
please kindly see attachment.
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>