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3 years ago

Is Fe2(CO3)3 soluble?

Chemistry

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

Yes it’s a soluble because the carbonate should dissolve in any strong acid with much effervescence

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Fe(s) + CuSO4(aq) ⇒ Cu(s) + FeSO4(aq)

Alexxandr [17]

Answer:

The answer to your question is CuSO₄

Explanation:

To answer your question just remember the following information

- A chemical reaction is divided into sections

reactants and products

reactants on the left side of the reaction

products on the right side of the reaction

- All the symbols have a meaning

(s) means that that compound is in solid phase

(aq) means that that compound is dissolved in solution.

Then, the answer is CuSO₄

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Veseljchak [2.6K]

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3 years ago

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lyudmila [28]

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3 years ago

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kati45 [8]

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3 years ago

Read 2 more answers

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.

6 0

3 years ago