Answer:
D. Scientific notation is a way of writing large and small numbers
Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Answer:
Humidity is the measure of how much water vapor is in the air.
The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.