Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.
Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.
1 Cal ---------- 4.184 J
? Cal ---------- 130.0 J
130.0 x 1 / 4.184 => 31.07 Cal
hope this helps!
<span>0.48 grams.
Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced.
To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So
0.0075 mol * 63.546 g/mol = 0.476595 g.
Round the results to 2 significant figures, giving 0.48 grams.</span>
The temperature of a reaction causes its rate of reaction to increase because the heat inputted into the solution excites the electrons that make up the solution, therefore making them move faster, colliding more often with other molecules of the solution. This increase in collision rates causes the rate of reaction to increase.
