During an experiment, a student adds 0.339 g of calcium metal to 100.0 ml of 2.05 m hcl. the student observes a temperature increase of 11.0 °c for the solution. assuming the solution's final volume is 100.0 ml , the density is 1.00 g/ml , and the specific heat is 4.184 j/(g⋅°c) , calculate the heat of the reaction, δhrxn . ca(s) 2h (aq)⟶ca2 (aq) h2(g)

The heat of the reaction is 17.94 J.

Explanation:

mass of Ca (m) = 0.339 g

increase in temperature (∆T)= 11°C

specific heat (s)= 4.184 J/(g. °C)

Now, to calculate the total heat we will have to multiply the mass with specific heat and increase in temperature.

∆H = m × s × ∆T

      = 0.339 × 4.184 × 11

      = 17.94 J

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

the formula mass of a molecule or a compound is the sum of the atomic weights of the atoms in its empirical formula

(that's what i know about formula mass)