Ice at 0.0°c is placed in a styrofoam cup containing 352 g of a soft drink at 20.0°c. the specific heat of the drink is about the same as that of water. some ice remains after the ice and soft drink reach an equilibrium temperature of 0.0°c. determine the mass of ice that has melted. ignore the heat capacity of the cup. (hint: it takes 334 j to melt 1 g of ice at 0.0°c.)

88.23 g of ice

Given;

Mass of the soft drink = 352 g

Initial temperature of soft drink = 0°C

Initial temperature of ice = 0°C

Final temperature of the mixture = 0°C

We are required to determine the mass of ice that melted;

Quantity of heat = Mass × specific heat × change in temperature

Q = m × c ×Δt

The specific heat capacity of soft drink = 4.186 J/g°C

Δt = 20 °C

Therefore;

Heat = 352 g × 4.186 J/g°C × 20°C

        = 29,469.44 joules

Heat used to melt ice will be calculated by multiplying the mass of ice by the latent heat of fusion.

Latent heat of fusion = 334 J/g

Assuming the mass of ice is m

Then;

Heat = 334 m Joules

Step 3: Calculating the mass of ice

Since heat released by the soft drink is used to melt the ice

Then;

334 m Joules = 29,469.44 joules

       m = 29,469.44 joules ÷ 334

           = 88.232 g

           = 88.23 g

Therefore, the mass of ice that melted is 88.23 g

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

d

explanation: