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serg [7]
3 years ago
5

Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO₃) i

nto solid calcium oxide and gaseous carbon dioxide. Suppose 23.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 380.0 °C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

CaCO3 (s)  → CaO (s) + CO2 (g)

The mass of carbonate that must have reacted was 43.03 grams

Explanation:

CaCO3  → CaO + CO2

Relation between reactant and product is 1:1

Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.

P . V = n . R . T

1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K

(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n

0.43 moles = n

0.43 moles of CO2, were produced from 0.43 moles of CaCO3.

Molar weight of CaCO3 = 100.08 g/m

Mass = Molar weight . moles

Mass = 100.08 g/m 0.43 m = 43.03 g

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Read 2 more answers
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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