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netineya [11]
2 years ago
12

For the following pairs of ions, use the concept that a chemical compound must have a net charge of zero to predict the formula

of the simplest compound that the ions are most likely to form. Write the formula of the metal first and follow by the non metal

Chemistry
1 answer:
Dafna11 [192]2 years ago
4 0

\\ \sf\longmapsto FeP

\\ \sf\longmapsto Fe_2S_3

\\ \sf\longmapsto FeCl_2

\\ \sf\longmapsto MgCl_2

\\ \sf\longmapsto MgO

\\ \sf\longmapsto Mg_3N_2

\\ \sf\longmapsto Na_3P

\\ \sf\longmapsto Na_2S

\\ \sf\longmapsto CoO

\\ \sf\longmapsto Co_2S_3

\\ \sf\longmapsto AlCl_3

\\ \sf\longmapsto CsBr

\\ \sf\longmapsto Ti_2O_4

\\ \sf\longmapsto Ag_2S

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What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are
damaskus [11]

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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