Suppose that 0.88g of water condenses on a 75.0-g block of iron that is initially at 22 degrees celsius. if the heat released during condensation goes only to warming the block, what is the final temperature (in degrees celsius) of the iron block? assume a constant enthalpy of vaporization for water of 44.0 kj/mol.

The final temperature (in degrees Celsius) of the iron block is 85.82 °C

Explanation:

Determine the heat emitted by the condensing water.

qw = 0.88 g H2O    

qw = -2.1492 kJ

Determine the heat absorbed by the iron block

qw = - qw = - (-2.1492 kJ) = 2.1492 kJ

Look up the specific heat for iron. Use the specific heat equation to solve for the final temperature of the iron block.

q = mcΔT

ΔT = q / mc

Temp f = (q / mc) + Temp i

Temp f = 2.1492 x 10^3 J / 75 g (0.449 J/g.°C) + 22 °C

Temp f = 85.82 °C

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.