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3 years ago

When iron forms an ion with a plus three charge, the formula of a compound with iron (Fe3+) and sulfur (S) is Fe2S3, right?

Chemistry

2 answers:

LekaFEV [45]3 years ago

6 0

Answer: Yes the formula of the compound formed is $Fe_2S_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

If iron is having an oxidation state of +3 called as $Fe^{3+}$ cation. Sulphur is an anion with oxidation state of -2 and forms sulphide ion $S^{2-}$ . Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Fe_2S_3$ .

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

Setler79 [48]3 years ago

4 0

Yes

2Fe(3+) + 3S(2-) ----> Fe2S3

:-) ;-)

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Formula of a copper (II)sulfate hydrate lab

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Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

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Explanation:

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3 years ago

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3 years ago

What is the volume of 9.5 g fluorine gas, F2, at STP?

Tju [1.3M]

Answer:

5.6L

Explanation:

At STP, the pressure and temperature of an ideal gas is

P = 1 atm

T = 273.15k

Volume =?

Mass = 9.5g

From ideal gas equation,

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant =0.082J/mol.K

T = temperature of the ideal gas

Number of moles = mass / molar mass

Molar mass of F2 = 37.99g/mol

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3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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2 years ago