If this same quantity of energy were transferred to 2.5 kg of water at it's boiling point what fraction of the water would be vaporized

Fraction of water that can be vaporized = 0.0961 or 9.61%

Note : The question is incomplete. The complete question is given below:

A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body.

If this same quantity of energy were transferred to 2.5 kg of water at its boiling point, what  fraction of the water would be vaporized?

Explanation:

Energy released by a serving of Cheeze-Its = 130 kcal

Since 1 kcal = 4.18 kJ, 130 kcal = 130 * 4.18 kJ = 543.4 kJ

Heat of vaporization (evaporating or condensing) Hv, of water = 2260 J/g

From formula, quantity of heat, Q = mass  * heat of vaporization

mass of water = 2.5 kg = 2500 g

Q = 2500 g * 2260 J/g

Q  = 5650000 J = 5650 kJ

Therefore 2.5 kg of water requires 5650 kJ of heat for complete vaporization

Fraction of water that can be vaporized by 543.4 kJ energy = 543.4/5650

Fraction of water that can be vaporized = 0.0961 or 9.61%

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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