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Lerok [7]
3 years ago
5

What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are

solid calcium oxide and carbon dioxide gas). Report your answer to 4 sig figs
Chemistry
1 answer:
damaskus [11]3 years ago
7 0

Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

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Use the Lewis model to determine the formula for the compound that forms from each pair of atoms.
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Answer:

KI

Explanation:

K is 1+  & I is 1- so we need 1 of each to balance out the compound.

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. If this same atom with 22 protons and 19 electrons were to gain 3 electrons, the net charge on the atom would be
frez [133]

Answer: No charge (0)

Explanation:

The atom has a proton of 22 and electron number of 19. This means the element has lost 3 electrons. Therefore, the net charge is +3.

So, if this atom gains 3 more electrons, the net charge would be zero (neutral).

The total number of electron would now be 22 which is the same as the proton number.

This implies the atom is now in an unreacted state or ground state.

3 0
3 years ago
(7) A 955 gram piece of tungsten metal is heated from 18°C to 100°C. The amount of heat absorbed
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Answer:

The answer to your question is: Cp = 0.129 J/g°C

Explanation:

Data

mass = 955 g of tungsten

T1 = 18°C

T2 = 100°C

Q = 10180 Joules

Cp = ?

Formula

Q = mCpΔT

Clear Cp from the equation

Cp = Q / mΔT

Substitution

Cp = 10180 / (955(100-18))

Cp = 10180 / (955(82))

Cp = 10180 / 78310

Cp = 0.129 J/g°C

3 0
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What conversion factor is needed to calculate the number of atoms in 8.6 moles of Aluminum?
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Explanation:

7 0
2 years ago
Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate
Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of K_3PO_4 = 150 g

Mass of Al_2(CO_3)_3 = 90 g

Molar mass of K_3PO_4 = 212.27 g/mole

Molar mass of Al_2(CO_3)_3 = 233.99 g/mole

Molar mass of K_2CO_3 = 138.205 g/mole

First we have to calculate the moles of K_3PO_4 and Al_2(CO_3)_3

\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles

\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles

The given balanced reaction is,

2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4

From the given reaction, we conclude that

2 moles of K_3PO_4 react with 1 mole of Al_2(CO_3)_3

0.7066 moles of K_3PO_4 react with \frac{1}{2}\times 0.7066=0.3533 moles of Al_2(CO_3)_3

But the moles of Al_2(CO_3)_3 is, 0.3846 moles.

So, Al_2(CO_3)_3 is an excess reagent and K_3PO_4 is a limiting reagent.

Now we have to calculate the moles of K_2CO_3.

As, 2 moles of K_3PO_4 react to give 3 moles of K_2CO_3

So, 0.7066 moles of K_3PO_4 react to give \frac{3}{2}\times 0.7066=1.0599 moles of K_2CO_3

Now we have to calculate the mass of K_2CO_3.

\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3

\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100

\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%

Therefore, the % yield of potassium carbonate is, 85.33%

6 0
3 years ago
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