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Goryan [66]
3 years ago
13

How many moles of carbon dioxide can be produced when 3.05 of calcium carbonate are heated​

Chemistry
1 answer:
vredina [299]3 years ago
3 0

Answer:

0.0305mol

Explanation:

CaCO3 ---> CaO + CO2

Mass of CaCO3 mol = 40 + 12 + (16 x 3) = 100g/mol

Number of CaCO3 moles heated = 3.05/100 = 0.0305 mol

One CaCO3 mol produces 1 mol CO2

Therefore 0.0305mol  of CO2 produced.

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Which of the following will occur when a solution of Pb(NO3)2(aq) is mixed with a solution of KI(aq)?
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When a solution of Pb(NO3)2(aq) is mixed with a solution of KI(aq), a precipitate of PbI₂ will form; K⁺ and NO₃⁻ are spectator ions.

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When an aqueous solution of lead nitrate (Pb(NO₃)₂ is mixed with aqueous solution of potassium iodide (KI), then there is a precipitate formation of lead iodide (PbI₂), and the potassium  (K⁺) ion and nitrate (NO₃⁻) ion acts as spectator ions that is ions do not involved in the reaction.

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The mass of an object is 100 mg. What is the object's mass in kg?
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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of
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Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and 7.2m^3 respectively.

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The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

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P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

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T_2 = final temperature of gas = 75^oC=273+75=348K

Now put all the given values in the above equation, we get:

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

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PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

Formula for new density will be:

\rho_2=\frac{P_2M}{RT_2}

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M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

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\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

\rho_2=0.2281g/L

The new density of carbon dioxide gas is 0.2281 g/L

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