Answer:
The cis double bond present in unsaturated fatty acids acids results in lower melting point when compared to saturated fatty acids of the same chain length.
Explanation:
Melting point of a fatty acids are affected by the length and degree of unsaturation of the hydrocarbon chain.
At room temperature, saturated fatty acids with hydrocarbon chain lengths between 12-24 are waxy solids whereas unsaturated atty acids of the same chain length are liquids. This is due to the nature of the packing of the fatty acid molecules in the saturated and unsaturated compounds.
In the saturated compounds, the molecules are tightly packed side by side with minimal steric hindrance and maximal van der Waals forces of attraction between molecules. However, in unsaturated fatty acids, the cis double bond introduces a bend or kink in the molecules which then interferes with the tight packing of the molecules and reducing interaction between molecules. Therefore, less energy is required to cause a disorder in the arrangement of unsaturated fatty acids, leading to a lowering of melting point.
The correct name for the hydrocarbon would be option 2. 2 - methyl - 2 - pentene.
Some policies they might do is to put limits on water usage, like making sure that people don't use too much water in baths and when they are tending to their gardens.
False They can function as both. An example is Aluminium Oxide. These kind of substances are called "Amphoteric", they can behave as both acids and bases.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
