When calcium carbonate is heated, it produces calcium oxide and carbon dioxide. The equation for the reaction is CaCO3(s) CaO(s)
1 answer:
When calcium carbonate is heated, it produces calcium oxide and carbon dioxide. The equation for the reaction is CaCO3(s) CaO(s) CO2(g). How many grams of calcium carbonate (molar mass = 100 g/mol) need to decompose to produce 44.5 g of Calcium oxide.
Answer: 79.5 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, contains avogadro's number of particles and weighs equal to the molecular mass of the substance.
According to stoichiometry :
1 mole of CaO is produced by 1 mole of
That means 56 g of CaO is produced by 100 g of
Thus 44.5 g of CaO is produced by = of
Thus 79.5 g of need to decompose to produce 44.5 g of Calcium oxide.
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Answer:
\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}
Explanation:
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The suitable differential equation for this case is:
As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:
Of course, .
Now, since the volume of a cone is and the ratio
or
, the volume becomes:
We proceed to its differentiation:
Then, we compute
Finally, at h=2:
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