50 points!

how much heat energy is needed to convert 10.75g of ice at -22 celsius to steam at 118 celsius?

A heat energy equal to 6300 J is required to heat 10.75 g of ice to steam.

Given:

Amount of transferred energy = Q =?

Mass of ice (water) = m = 10.75 g

Initial temperature of ice =

Final temperature of steam =

Formula for Heat capacity is given by

Q = m×c×Δt ........................................(1)

where:

Q = Heat capacity of the substance (in J)

m=mass of the substance being heated in grams(g)

c = the specific heat of the substance in J/(g.°C)

Δt = Change in temperature (in °C)

Δt = (Final temperature - Initial temperature) =

Specific heat of water is c = 4.186 J /g. °C

Substituting these in equation (1), we get

Q = m×c×Δt =

Required heat energy is equal to 6299.93 Joules ≅ 6300 Joules.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-