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 />
Answer:
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
Explanation:
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Rate = ![\frac{1}{k[A_{0}]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%7D)
Half life =![K\frac{1}{[A_{0}]^{(n-1)} }](https://tex.z-dn.net/?f=K%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%20%7D)
![\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }](https://tex.z-dn.net/?f=%5Cfrac%7B%28halflife_%7B1%7D%29%7D%7B%28halflife_%7B2%7D%29%7D%3D%5Cfrac%7B%5BA_%7B2%7D%5D%5E%7B%28n-1%29%7D%7D%7B%5BA_%7B1%7D%5D%5E%7B%28n-1%29%7D%20%7D)
where n = order of reaction
Putting values
![\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B109%7D%7B231%7D%3D%5Cfrac%7B%5B0.132%5D%5E%7B%28n-1%29%7D%7D%7B%5B0.280%5D%5E%7B%28n-1%29%7D%7D)
Hence n = 2
![halflife=\frac{1}{k[A_{0}]}](https://tex.z-dn.net/?f=halflife%3D%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%7D)
Putting values
K = 0.0328
Answer:
d = 3.29 g/cm³
Explanation:
Given data:
Density of material = ?
Mass of material = 56.6 g
Volume of material = 17.2 cm³
Solution:
Formula;
d = m/v
d = density
m = mass
v = volume
by putting values,
d = 56.6 g/ 17.2 cm³
d = 3.29 g/cm³
