<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
Explanation:
B. HCIO these is the answer
Option D
A precipitate is the term for a solid that forms when two solutions are mixed
<u>Explanation:</u>
A solid set from a couple of solutions is termed a precipitate. A precipitate is an unsolved solid that makes when a pair of solutions are fused and react chemically. Unsolvable means that the solid will not melt. When the effect transpires in a liquid solution, the solid developed is denominated the 'precipitate'.
The substance that generates the solid to make is termed the 'precipitant'. Seldom the development of a precipitate symbolizes the existence of a chemical reaction. Precipitation may additionally transpire immediately from a supersaturated solution.
Answer:
-2
Explanation:
7 x 1 - 2 x 1 + 1 x 1 + 3C = 0 (no charge)
6 + 3C = 0
C = -2
