HEL 50. How many joules of heat must be absorbed by 0.375 kg of water to raise its temperature by 45.0 to 70.0°C. Specific heat of water is 4.184 J/gC or 1.0 cal/gC
51. What mass of water can be heated from 25.0 to 50.0°C by the adding of 2825 Joules?(Cp of H2O=4.184 J/g °C)

  50.  54,915 J

  51.  27.0 g

Explanation:

The dimensional analysis tells you how to compute these. Multiplying specific heat, J/(g·°C), by the product of mass and temperature change will give you energy.

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50.  (375 g)((70 -45) °C)(4.184 J/(g·°C)) = 54,915 J

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51. M((50 -25)°C)(4.184 J/(g·°C)) = 2825 J

  M = 2825/104.6 g ≈ 27.0 g

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

i think its 100.

hope this !