The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal.
According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.
With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.
The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.
To learn more about Markovnikov's rule
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Answer:
The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K
Explanation:
Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.
The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase
The answer is <span>ReGallium or Indium</span>
