Formulas are like steps to solve an equation but in cemistry a formula is 2 or more elements combined to make something
The balanced reaction is
2CH3OH (g) + 3O2 (g) ⟶ 2CO2 (g) + 4H2O (g)
as per equation two moles of methanol gas will react with 3 moles of oxygen
one mole of gas occupies 22.4 L of volume
so the moles and volume goes in same ratio
it means two unit volume of methanol will react with three unit volume of oxygen
therefore 1L of methanol gas will react with 3 /2 L of oxygen
Or 18 L of methanol gas will react with 3 x 18 /2 = 27 L of oxygen
So here oxygen is limiting reagent
As per balanced equation
10L of oxygen will react with = 2 X 10 /3 L of methanol = 6.67 L of methanol gas to give 6.67 L of CO2 gas and 13.33 L of water gas
So overall there will be = 18 - 6.67 L of left out methanol = 11.33 L
And 6.67 L of CO2 + 13.33 L of water = 20 L
Total volume of gas = 11.33+ 20 = 31.33 L
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
p =
<em>p equals m over V</em>
hope this helps!!
