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Alika [10]
3 years ago
9

A compound is made up of iron and oxygen, only. The ratio of iron ions to oxide ions is 2:3 in this compound. The IUPAC name for

this compound is
(1) triiron dioxide (3) iron(III) oxide
(2) iron(II) oxide (4) iron trioxide
Chemistry
2 answers:
never [62]3 years ago
6 0

Answer: The correct option is 3.

Explanation: We are given a compound which is made up of iron and oxygen only. The ratio of the two are given as:

\frac{\text{iron ions}}{\text{oxide ions}}=\frac{2}{3}

This means that, number of iron ions are 2

Number of oxide ions are 3

From the above information, the formula becomes : Fe_2O_3

The valency of iron = 3

Valency of oxide = 2

This compound is named as iron (III) oxide.

Hence, the correct option is 3

nasty-shy [4]3 years ago
4 0
A compund is made up of iron and oxygen, only. The ratio of iron ions to oxide irons is 2:3 in this compund. The IUPAC name for this compound is<em> Iron (III) Oxide. </em>
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Which is a property of covalent compounds?
ira [324]

Answer:

Explanation:

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6 0
3 years ago
Potassium permanganate(KMnO4) reacts with oxalic acid (H2C2O4) in aqueous sulfuric acid according to: 2KMnO4 5H2C2O4 3H2SO4-&gt;
Andru [333]

Answer:

57.3mL of 0.250M KMnO4 are required

Explanation:

3.225g of oxalic acid are:

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Based on the reaction, 5 moles of oxalic acid react with 2 moles of KMnO4, thus, for a complete reaction of oxalic acid you need:

0.03582mol of oxalic acid × (2mol KMnO4 / 5mol Oxalic Acid) = <em>0.01433mol of KMnO4</em>

<em />

If concentration of KMnO4 is 0.250M, liters in 0.01433mol are:

0.01433mol × (1L /0.250mol) = 0.0573L ≡<em> 57.3mL of 0.250M KMnO4 are required</em>

<em></em>

I hope it helps!

5 0
3 years ago
a container contains 5L of nitrogen gas at 25°C. what will be it's volume if the temperature increases by 35°C keeping the press
Olin [163]

Answer:

5.168 L

Explanation:

Applying Charles Law,

V/T = V'/T'..................... Equation 1

Where V = Initial Volume, T = Initial Temperature in Kelvin, V' = Final Volume, T' = Final Temperature in Kelvin,

Make V' the subject of the equation,

V' = VT'/T................. Equation 2

From the question,

Given: V = 5L, T = 25°C = (25+273) = 298K, T' = 35°C = (35+273) = 308K

Substitute these values into equation 2

V' = (5×308)/298

V' = 5.168 L

4 0
3 years ago
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