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3 years ago

consider the following equilibrium:h2co3(aq) h2o(l) h3o (aq) hco3-1(aq).what is the correct equilibrium expression?

1 answer:

4 0

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:<span>

<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>

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The flotation process used in metallurgy involves Multiple Choice the roasting of sulfides. separation of gangue from ore. elect

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The flotation process used in metallurgy involves the separation of gangue from ore.

<h3>How does the flotation process work?</h3>

The various wettability qualities of a material's surface are the foundation of flotation operations. The basic principles of flotation are quite similar to those of a sink and float process, where the materials' relative densities to the medium in which they are deposited determine the basis of the separation.

<h3>What is the process of separating minerals from gangue known as?</h3>

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1 year ago

How many moles of FeCI3 are present in 345.0 g FeCI3

Sphinxa [80]

Answer:

$\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}$

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

Fe: 55.84 g/mol

Cl: 35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

$\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Multiply by the given number of grams, 345.0.

$345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Flip the fraction so the grams of FeCl₃ will cancel.

$345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}$

$345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }$

$\frac{345.0 \ mol \ FeCl_3}{162.19 }$

Divide.

$2.12713484 \ mol \ FeCl_3$

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

$\approx 2.127 \ mol \ FeCl_3$

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

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2 years ago

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nignag [31]

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3 years ago

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3 years ago