Answer:
6.3×10⁻³ moles of Cu
Explanation:
The relation is, that 1 mol of particles are contained by 6.02×10²³ particles.
This is Avogadro's number that explains that: one mole determines the number of fundamental units that are contained in a constant number but they do not depend on the type of material or the type of particle, and this quantity is 6.02 × 10²³
We can make a rule of three:
6.02×10²³ atoms are contained 1 in mol of Cu
Therefore, 3.8×10²¹ atoms will be contained in (3.8×10²¹ . 1) / NA = 6.3×10⁻³ moles
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 />
Valence electrons in NH2Cl:
=1(5)*2(1)*1(7)
=14
So there are 14 valence electrons in NH2Cl
As each stick represents an valence electron pair.
And there are 14 valence electrons, so there can be 7 valence electron pairs.
And for 7 valence electron pairs, 7 sticks needed.
The answer is 67.82 g/mol
