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oee [108]
3 years ago
8

Solid iron(II) hydroxide decomposes to form solid iron(II) oxide and liquid water. Write the balanced chemical equation for the

reaction described. Phases are optional.
Chemistry
1 answer:
Bess [88]3 years ago
5 0

Answer:

Fe(OH)_2_(_s_)~->~FeO_(_s_)~+~H_2O_(_l_)

Explanation:

The <u>formula</u> for iron (II) hydroxide is Fe(OH)_2

The <u>formula</u> for iron(II) oxide is FeO

The <u>formula</u> for liquid water is H_2O

The <u>reaction</u> would be:

Fe(OH)_2_(_s_)~->~FeO_(_s_)~+~H_2O_(_l_)

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Explanation:

The heat absorbed by the aluminum in this case is:

q = m x C x ΔT     m= q/ (C x ΔT)

q= 9.86 J

C = 0.90 J/g-K

ΔT = ( 30.5 ºC - 23.2 ºC ) = 7.3 ºC = 7.3 K (this is a range of temperature)

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Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
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Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

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Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

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0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

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