Consider the phase change of ice to liquid water if the molar heat of fusion for water is 6.01 kj mol-1 how many joules are required to melt 1 gram of ice at 273 k

The correct answer is 333.8 J

Explanation:

Ice (water in solid state) melts at 273 K (0ºC). During the phase change from solid to liquid, water absorbs 6.01 KJ per mol until it melts completely. To calculate the heat absorbed, we first have to convert the mass of water (1 g) to mol by using the molecular weight of water (H₂O):

Molecular weight of water= 2 x molar mass H + 1 x molar mass O

Molecular weight of water= 2 x 1 g/mol + 16 g/mol= 18 g/mol

mol ice= = 0.055 mol

heat= 0.055 mol x 6.01 KJ/mol= 0.3338 KJ

Finally, we convert KJ to J (1 KJ= 1000 J so we multiply by 1000):

0.3338 KJ x 1000 J/1 KJ= 333.8 J

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.

wher is the rest of the question