Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Mass / volume = density
30.943g / 35ml = 0.88408571g/ml
If you drop a bath bomb into water, then it will fizz because a chemical reaction is taking place.
I think it is 1620 (lxwxh) x 10 to get to millimeters
According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.
With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.
The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.
To learn more about Markovnikov's rule
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