Answer:
(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane
(d) Heptane is more volatile than octane
Explanation:
We can use Raoult's Law to solve this problem.
It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,
(a) Vapour pressure of each component
Let heptane be Component 1 and octane be Component 2.
(i) Moles of each component
(ii) Total moles
(iiii) Mole fractions of each component
(iv) Partial vapour pressures of each component
(b) Total pressure
(c) Mass percent of each component in vapour
The ratio of the mole fractions is the same as the ratio of the moles.
If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane
(d) Enrichment of vapour
The vapour is enriched in heptane because heptane is more volatile than octane.
Answer:
if were talking about berylium then the answer should be 2 because it wants to move 2 spaces back (losing electrons) to become a nobel gas so the answer is 2+
Explanation:
it has 2 valence electrons YW :)
A composition reaction produces a single substance from multiple reactants. A decomposition reaction produces multiple products from a single reactant.
Hope This Helped
Answer:
look at peaks:
sharp peak at ~ 1700 = carbonyl
broad peak at ~ 3300 = alcohol
both = mixture
Explanation:
In the first spectrum, the broad peak occurring in the range of 3300-3700 cm-1 corresponds to OH bond stretch frequency. Carbonyl C=O bond stretch has a frequency range of 1670-1820 cm-1. The absence of any peaks in this region in the given spectrum rules out the presence of carbonyl compound in the reaction mixture. So, the reaction mixture contains only pure alcohol.
The second spectrum shows a peak around 1700 cm-1 corresponding to the carbonyl compound along with the broad peak around 3300 cm-1 corresponding to alcohol. So, the reaction mixture contains both alcohol and carbonyl compounds.
A table containing different functional groups and their corresponding Infra Red frequencies is given below for further references
Answer:
5SiO2 + 2CaC2 --> 5Si + 2CaO + 4CO2
4NH3 + 5O2 --> 4NO + 6H2O