Water (2230 g ) is heated until it just begins to boil. If the water absorbs 5.73×105 J of heat in the process, what was the initial temperature of the water?

Explanation:

To calculate the initial temperature of the water:

where,

q = heat absorbed =

= specific heat of water =

m = mass of water = 2230 g

= final temperature of water =

= initial temperature of metal = ?

Now put all the given values in the above formula, we get:

Thus, the initial temperature of the water is

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2