Calculate the final temperature of 202 mL of water initially at 32 degrees celsius upon absorption of 19 kJ of heat

The  Final Temperature of water = 54.48°C.

Explanation:

Given,

Volume of water = 202 mL.

Initial Temperature T₁ = 32°C

Heat energy absorbed = 19 kJ = 19000 Joules

Final Temperature T₂ = ?

Final temperature can be calculated using the formula,

Q = (mass) (ΔT) (Cp)

Here  ΔT = T₂ - T₁

Cp = specific heat of water = 4.184 J/g°C.

For finding the mass, we use the formula,

Density = Mass/Volume

Since, Density of water =  1 g/mL.

Mass of water = 202 grams.

Now Substituting all values in the formula Q = (mass) (ΔT) (Cp)

19000 = (202)(T₂ - T₁)(4.184)

Solving, we get,

(T₂ - T₁) = 22.48°C

Therefore,  T₂ = 22.48°C + 32°C = 54.48°C.

Therefore, Final Temperature of water = 54.48°C.

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

there are two significant figures.