The important thing to note is the reason why electron react is due to the instability of the electrons. All elements wants to aim the electron configuration of the noble gases. This is the most stable form in which each of the orbitals are sufficiently filled. When it comes to bonding, the order of reactivity is: alkynes > alkenes > alkanes. Alkynes are compounds with triple bonds, alkenes with double bonds and alkanes with single bonds. The single bonds are called saturated hydrocarbons. This is because they have reached stability, so it is quite difficult to react this with reducing or oxidizing agents. Alkynes and alkenes are unsaturated hydrocarbons. They readily react with reducing and oxidizing agents so as to become saturated, as well. The underlying principle for this is that single bonds contain sigma bonds which is the head-on overlapping of electrons. These is the strongest type of covalent bond. Double and triple bonds contain pi bonds which is the side overlapping of electrons orbitals. Hence, these electrons would be easily separated making it more reactive especially during protonation.
The heat released by the substance in the calorimeter is equal to the heat absorbed by water which results to the decrease and increase in temperature, respectively.
We use m Cp ΔT to balance the heat involved
(m Cp ΔT) subs in calorimeter = <span>(m Cp ΔT) water
</span>125 g * Cp * (97.0-23.5 ) C = 250 g *(4.18 J/C g)* (23.5-20)
Cp = 0.398 J/Cg
Answer is B
Because it is a mix of H2O (water) and NaCl (salt)
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>
