Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2
Answer:
k = ![\frac{[HOCl]^2}{[Cl]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BCl%5D%5E2%7D)
Explanation:
The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.
Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium
The global reaction is:
2 HgO (s) + H₂O (l) +2 Cl₂ (g) ⇌ 2 HOCl (aq) + HgO⋅HgCl₂ (s)
Thus, equilibrium-constant expression is:
<em>k =
</em>
You don't include HgO nor HgO⋅HgCl₂ because are pure solids nor water because is pure liquid.
I hope it helps!
62 g of water are vaporized and the energy required is 2.4 kJ/g
So 62g x 2.4 kJ/g = 148.8 kJ or 148,800 Joules
Q = mCΔT
Q is energy in joules, m is mass of water, C is the specific heat, delta T is change in temp
148,800 = m(4.18)(41 - 19) = 1618g or 1.6 kg of water
Answer:
The heat of combustion of magnesium metal is 24.76 kJ/gram
Explanation:
Step 1: Data given
Mass of magnesium sample = 0.1946 grams
Molar mass of magnesium = 24.3 g/mol
bomb calorimeter that has a heat capacity of 1349 J/°C
Mass of water = 500 grams
Temperature change = 1.40 °C
Step 2: Calculated heat released
Q = (1349 J/°C * 1.40 °C) + (500 grams * 4.184 J/g°C * 1.40 °C)
Q =4817.4 J = 4.82 kJ
Step 3: Calculate the heat given off by the burning Mg, in kJ/g
4817.4 J / 0.1946 grams = 24755.4 J/ gram = 24.76 kJ/ gram
The heat of combustion of magnesium metal is 24.76 kJ/gram