Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ
Explanation:
Using this formular, q = [mCpΔT] and = [nΔHfusion]
The energy that is needed in the different physical changes is thus:
The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:
q = [mCpΔT]
q = 52.0 x 2.09 x 10
q = 1.09 kJ
While from 0°C to 100°C is calculated as:
q = [mCpΔT]
q = 52.0 x 4.18 x 100
q = 21.74 kJ
And for fusion at 0°C is called Heat of fusion and would be given as:
q = n ΔHfusion
q = 52.0 / 18.02 x 6.02
q = 17.38 kJ
And that required for vaporization at 100°C is called Heat of vaporization and it's given as:
q = n ΔHvaporization
q = 52.0 / 18.02 x 40.7
q = 117.45 kJ
Add up all the energy gives 157.8 kJ
Explanation:
<em><u>2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2</u></em>
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<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
