Answer:
1) When 69.9 g heptane is burned it releases 5.6 mol water.
2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>
This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.
- We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>
n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.
<em><u>Using cross multiplication:</u></em>
1.0 mol of heptane releases → 8 moles of water.
0.7 mol of heptane releases → ??? moles of water.
<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>
<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>
<em />
A + B = Products
This is the correct mechanism.
Answer:
The answer to your question is U-234
Explanation:
Data
Thorium-234
beta emission twice
Definition
Beta emission is when a beta particle (electron) is emitted from an atomic nucleus.
First beta emission
²³⁴₉₀Th ⇒ ²³⁴₉₁Pa + e⁻
Second beta emission
²³⁴₉₁ Pa ⇒ ²³⁴₉₂U + e⁻
The atom will be Uranium-324
Answer:
Grinidng chalk into a fine powder
Explanation:
Answer:
Methyl radical
Explanation:
A radical is any specie that contains an odd number of electrons. We must note that the greater the number of alkyl groups which are attached to a carbon atom that bears the odd electrons, the more the degree of delocalization of the odd electrons and consequently the more stable we expect the free radical to be.
Hence the order of free radical stability is; Methyl < Primary < Secondary < Tertiary. Hence, we can easily see that the methyl radical is the least stable free radical.