<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>
Answer:
= 15.51 mL
Explanation:
Here's is the reaction:
2HgO(s) ⇒ 2 Hg(s)+O₂(g)
In this reaction 2mol HgO = 1mol O₂
The molecular weight of HgO = 216.59g
so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO
= 0.0138511 molHgO
The amount of Oxygen follows:
0.0138511 molHgOx1/2= 0.00692555 mol O₂
Now, volume of 1 any gas = 22400mL
so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂
= 15.513232mL O₂
The sub-atomic particles of an atom are the proton, electron and the neutron. An electron has a charge of -1 and a smaller
mass than a proton. Proton has the same mass with the neutron. The ratio
between the mass of a proton and an electron is about 2000. An electron has an
equal value but negative charge with the proton.