Answer:
120 mol Mg
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
120 moles H₂
<u>Step 2: Identify Conversions</u>
RxN: 3 mol H₂ = 3 mol Mg
<u>Step 3: Stoichiometry</u>
<u />
= 120 mol Mg
Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as 
The equation for the ionization of the is given as:
By stoichiometry of the reaction:
1 mole of 
gives 1 mole of 
and 2 mole of 
When the solubility of is S moles/liter, then the solubility of will be S moles\liter and solubility of will be 2S moles/liter.
![K_{sp}=[Ca^{2+}][OH^{-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BOH%5E%7B-%7D%5D%5E2)
![6.5\times 10^{-6}=[S][2S]^2](https://tex.z-dn.net/?f=6.5%5Ctimes%2010%5E%7B-6%7D%3D%5BS%5D%5B2S%5D%5E2)
Thus solubility of this compound in pure water is 0.012 M
Answer:
28%
Explanation:
concentration=(solute mass/solution mass)*100
C=concentration
md-solute mass
ms-solution mass
the solution mass is equal to the sum of the solute and solvent mass
ms=3.82+1.49=5.31lb
c=1.49/5.31 * 100=0.28*100=28%
For every 3 Fe there is only 1 Fe3O4 producing.
So, ratio would be: Fe3O4 / Fe = 1 / 3
In short, Your Answer would be: 1 : 3
Hope this helps!
Answer:
47%
Explanation:
MgCO₃ (s) → MgO (s) + CO₂ (g)
CaCO₃ (s) → CaO (s) + CO₂ (g)
_______________________
MgCO₃ (s) + CaCO₃ (s) → MgO (s) + CaO (s) + 2CO₂ (g)
MgO: 40.3044 g/mol
MgCO₃: 84.3139 g/mol
CaO:56.0774 g/mol
CaCO₃: 100.0869 g/mol
CO₂: 44.01 g/mol
15.42 g is the total mass of dolomite.
7.85 g is the sum of MgO and CaO produced.
This means that 7.57 g of CO₂ were produced.
44.01 g CO₂_____ 1 mol
7.57 g CO₂ _____ x
x = 0.172 mol CO₂
Considering the global reaction, we had 0.172/2 = 0.086 mol of MgCO₃ in the original sample.
1 mol MgCO₃ _______ 84.3139 g
0.086 mol MgCO₃ ___ y
y = 7.25 g
15.42 g dolomite ______ 100%
7.25 g MgCO₃ ________ z
z = 47%
