Force is a disturbance that results in a wave.
I think the number of protons in the nucleus the number of valence electrons atomic mass...
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>
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Strong electron withdrawing groups prevents Friedel Crafts reaction because when a withdrawing group attracts the electrons decrease the availability of the electronic pair and the enough electronic density of it in order to make possible the aromatic electrophilic substitution.
Strong electron donating groups like NH2 doesn’t undergo Friedel Crafts reaction because NH2 is a Lewis base that means an electron donor. Due to Friedel Crafts reaction is an acid base reaction, the AlCl3 will be attacked by the lone pair available from NH2 producing a salt. The same occurs with phenol because the lone pair of electrons available in the OH group so Friedels Crafts doesn’t undergo with benzene attached to the strong electron donating groups NH2 or OH.
True, H2O is a compound with 2 hydrogens and one oxygen
