In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.
O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13
CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044
Answers: O2 = 0.13
CO2 = 0.044
1) A Car engine.............. Hope it helps, Have a nice day :)
Answer:
homogeneous and heterogeneous mixtures
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
pure substances can be divided into two groups; elements and compounds
