<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>
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
Answer:
8.66 g of Al₂O₃ will be produced
Explanation:
4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
This is the reaction.
Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.
4.6 g / 26.98 g/mol = 0.170 moles
Ratio is 4:2.
4 moles of aluminum can produce 2 moles of Al₂O₃
0.170 moles of Al, may produce (0.170 .2)/ 4 = 0.085 moles
Let's convert the moles of Al₂O₃ to mass.
0.085 mol . 101.96 g/mol = 8.66 g
Answer:
B. hydrogen
Explanation:
that's because hydrogen is an element.
Answer:
the region of negative charge surrounding an atomic nucleus that is associated with an atomic orbital. The region is defined mathematically, describing a region with a high probability of containing electrons.
Explanation:
