In the following experiment, a coffee-cup calorimeter containing 100 ml of h2o is used. the initial temperature of the calorimeter is 23.0 ∘c. if 3.00 g of cacl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? the heat of solution δhsoln of cacl2 is −82.8 kj/mol. assume that the specific heat of the solution form

28.3 °C

Explanation:

We assume that the specific heat of the solution is equal to the specific heat of water (4.184 Jg⁻¹ °C⁻¹).

First, we find the heat released by the dissolution of CaCl₂. The grams will  be converted to moles using the molar mass (110.986 g/mol), then multiplied by the molar heat of solution:

(3.00 g)(mol/110.986 g)(-82.8 kJ/mol) = -2.23812 kJ

The negative sign indicates that heat is released. Extra significant figures are included to avoid round-off errors.

The amount of heat released by the CaCl₂ dissolution is equal to the heat absorbed by the water. The equation is rearranged to solve for Δt, the temperature change of the water.

Q = mcΔt ⇒ Δt  = Q/(mc)

Δt = (2.23812 kJ)(1000 J/kJ) / (100 mL)(1 g/mL)(4.184 Jg⁻¹ °C⁻¹) = 5.3 °C

We can then calculated the final temperature t₂ of the solution:

Δt = t₂ - t₁

t₂ = Δt + t₁ = 5.3 °C + 23.0 °C = 28.3 °C

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

First, we will convert 3.00 g of CaCl₂ to moles using its molar mass (110.98 g/mol).

The heat of solution (ΔHsoln) of CaCl₂ is −82.8 kJ/mol. The heat released by the solution of 0.0270 moles is:

According to the law of conservation of energy, the sum of the heat released by the solution (Qs) and the heat absorbed by the calorimeter (Qc) is zero.

Assuming the density of water is 1 g/mL, we have 100 mL (100 g) of water and 3.00 g of CaCl₂. The mass of the solution (m) is:

Finally, we can calculate the final temperature of the system using the following expression.

where,

c: specific heat of the solution (same as water 4.18 J/g.°C)

T₁ and T₂: initial and final temperature

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

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0.11 moles of the gas are present in the sample of dry gas.

Explanation:

Data given:

mass of the gas  = 2.1025 grams

volume of the gas = 2.850 litres

temperature =  22 degrees (273.15+22) = 295.15 K

Pressure = 740 mm Hg or 0.973 atm

moles of the gas =?

R = 0.08206 atmL/Mole K

From the ideal gas law the number of moles can be calculated in the sample of dry gas. Number of moles will be determined by the pressure exerted, volume and temperature of the gas.

The formula:

PV = nRT

n =

putting the values in the above equation:

n =

  = 0.11 moles

0.11 moles of the dry gas is present in the sample given.