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3 years ago

Calculate the number of moles of caco3 (calcium carbonate, or limestone) in a 20.0g sample of this substance

1 answer:

6 0

First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.

molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃

Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃

<em>Answer: 0.2 moles</em>

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II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

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Вывести формулу оксигенсодержащего соединения, если массовая доля карбона в нем 52,2%, а гидрогена 13%

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2 years ago

Sodium hydrogen carbonate and citric acid<br><br> Word equation

timofeeve [1]

Answer:

the right answer is:

Explanation:

How to Balance: NaHCO3 + HC2H3O2 = NaC2H3O2 + CO2 + H2O|

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2 years ago

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A mixture of 9.22 moles of A, 10.11 moles of B, and 27.83 moles of C is placed in a one-liter

AysviL [449]

The equilibrium constant is 1.3 considering the reaction as written in the question.

<h3>Equilibrium in chemical reactions</h3>

In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;

A (g) + 2B (g) ⇌ 3C (g).

The initial concentration of each specie is;

A - 9.22 M
B - 10.11 M
C - 27.83 M

The equilibrium concentration of B is 18.32 M

We now have to set up the ICE table as follows;

A (g) + 2B (g) ⇌ 3C (g)

I 9.22 10.11 27.83

C -x -x +x

E 9.22 - x 10.11 - x 27.83 + x

The equilibrium concentration of B is 18.32 M hence;

10.11 - x = 18.32

x = 10.11 - 18.32 = -8.21

Hence;

Equilibrium concentration of A = 9.22 - (-8.21) = 17.43

Equilibrium concentration of C = 27.83 + (-8.21) = 19.62

Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2

K = 1.3

Learn more about equilibrium constant: brainly.com/question/17960050

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2 years ago