Answer:
Mw = 179.845 g/mol
Explanation:
∴ w = 26.2 g
∴ 1 mol = 6.02 E23 molecules.......Avogadro's number
⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol
⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol
<h3>
Answer:</h3>
1.93 g
<h3>
Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
It is a combustion reaction because when methane burns with oxyzen it produces carbon dioxide,water and heat and light.
After 3 half-lives, 125 grams of the parent isotope will remain.
Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7B185%2C000%7D%7B2.303%20%5Ctimes%208.314%7D%20%5B%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%200.7289%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%2010%5E%7B0.7289%7D%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%205.356)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356
