An 18.7 g sample of platinum metal increases in temperature from 21.2C to 23.5C when 5.7 J of heat are added. What is the specific heat of platinum?

132.53 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁) Equation 1

Where Q = amount of heat, c = specific heat capacity of platinum metal, m = mass of platinum metal, t₁ = initial temperature, t₂ = Final temperature.

Make c the subeject of the equation

c = Q/m(t₂-t₁) Equation 2

From the question,

Given: Q = 5.7 J. m = 18.7 g = 0.0187 kg, t₁ = 21.2°C, t₂ = 23.5°C

Substitute these values into equation 2

c = 5.7/[0.0187×(23.5-21.2)]

c = 5.7/0.04301

c = 132.53 J/kg.°C

- decrease the number of plants and animals living in the water.

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