B .pure water it’s made up of oxygen and hydrogen “H2O”
<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>
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Answer:
we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
Explanation:
when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate
Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, 
Explanation:
The reaction equation for given reaction is as follows.
Here, 1 mole of reacts with 3 moles of 
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide 
is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of is calculated as follows.
Now, moles of .given by 0.5 mol of is calculated as follows.
As molar mass of is 2.016 g/mol. Therefore, mass of is calculated as follows.
Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, .
