At STP, P = 1 atm, and T = 0 C
Thus, PV = nRT => V = nR(273). We will use this later...
if you have 35.4 Ca, and the molar mass of Ca is 40.08, you get .883 moles Ca. Thus, since it takes 2 moles of Ca to form a reaction, you only need half the moles of Ca of O2. Thus, n(O2) = .883/2
Tie this back to the first equation and you get
V = .442 * <span>0.082057(which is R) * 273 = 9.9 L</span>
Becuase you can't change it back to the shiny nail that it was at first So so therefore its a chemical and chemicals can not be changed back into its regular state.
Answer:
This metal has a specific heat of 0.9845J/ g °C
Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)
Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)
- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4
Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C
This metal has a specific heat of 0.9845J/ g °C