Dissolving 4.02 g of cacl2 in enough water to make 349 ml of solution causes the temperature of the solution to increase by 3.91 oc. assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 j/goc and 1.00 g/cm3, respectively) calculate δh per mole of cacl2 (in kj) for the reaction under the above conditions.

ΔH per mole of CaCl2 is 160.3 kJ

Explanation:

Step 1: The balanced equation

CaCl2 (aq)→  Ca2+  +  2Cl-(aq)

Step 2: Data given

Molar mass of CaCl2 =110.98 g/mole

mass of CaCl2 = 4.02 grams

volume of the solution = 349 mL = 0.349 L

Temperature increases with 3.91 °C

Specific heat of the solution = specific hea of water = 4.18 J/°C*g

Density solution = density water = 1g/cm³

Step 3: Calculate moles of CaCl2

Numer of moles of CaCl2 = mass of CaCl2 / Molar mass of CaCl2

Number of moles = 4.02 grams /110.98 g/mole= 0.036 moles

Step 4 = Calculate amount of heat added

Q = m*c*ΔT

with m = the mass 349 grams + 4.02 grams = 353.02 grams

c = 4.18 J/°C*g

ΔT = 3.91 °C

Q = 353.02*4.18*3.91 = 5769.69 J

Step 5: Calculate ΔH per mole

5769.69 J /0.036 moles = 160269.1 J = 160.3 kJ

ΔH per mole of CaCl2 is 160.3 kJ

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

it is a good conductor

explanation: