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Answer: 1Ca + 2O + 2H
Explanation:
Ca is 1 since there is no subscript
O and H each have 2 because the subscript 2 is outside the parenthesis so you multiply their subscript (1) by 2
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%
Answer:

Explanation:
By definition one <em>half-life</em> is the time to reduce the initial concentration to half.
For a <em>second order reaction </em>the rate law equations are:
![\dfrac{d[B]}{dt}=-k[B]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D-k%5BB%5D%5E2)
![\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BB%5D%7D%3D%5Cdfrac%7B1%7D%7B%5BB%5D_0%7D%2Bkt)
The <em>half-life</em> equation is:
![t_{1/2}=\dfrac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%7D)
Thus, substitute the<em> rate constant</em>
and the <em>half-life </em>time <em>224s</em> to find [A]₀:
![224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}](https://tex.z-dn.net/?f=224s%3D%5Cdfrac%7B1%7D%7B1.30%5Ctimes10%5E%7B-3%7DM%5E%7B-1%7D%5Ccdot%20s%5E%7B-1%7D%5BA%5D_0%7D)
![[A]_o=0.291M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.291M)