Answer:
To determine the amount of heat the gold has absorbed to melt, we simply multiply the mass of the block of ice to the heat of fusion of water which is given above. We calculate as follows:
Heat = 20.0 g (35.4 g)
Heat = 1290 J
When sodium carbonate is dissolved in water, the equation is
.
When carbon dioxide is placed in water, aqueous carbon dioxide is formed: 
<h3>Dissolution of compounds in water</h3>
Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.
Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.
While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.
Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: 
More on the dissolution of substances can be found here: brainly.com/question/28580758
#SPJ1
2. B because it has a lower activation energy.
idk because you have no picture with the lines on it
Answer: Anion. The resulting of losing or gaining electrons would be Anion.
Answer:

Explanation:
Hello,
In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:

Regards.