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oee [108]
4 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]4 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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If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost
dimulka [17.4K]
For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.


4 0
4 years ago
PLS HELP ASAP. list 5 chemicals that contribute to ozone layer depletion?​
Rashid [163]

Answer:chlorofluorocarbons (CFCs)

halon.

carbon tetrachloride (CCl4)

methyl chloroform (CH3CCl3)

hydrobromofluorocarbons (HBFCs)

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Explanation:

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3 0
3 years ago
What is a substanceeas​
Mumz [18]

Answer:

substance*? a particular kind of matter with uniform properties.

Explanation:

5 0
3 years ago
What is the correct name of the compound Pb(NO3)2?
Margaret [11]

Answer:

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4 0
2 years ago
If the temperature of an ideal gas is raised from 100◦C to 200◦C, while the pressure remains constant, the volume 1. increases b
mel-nik [20]

Answer:

3. doubles

Explanation:

for an ideal gas behavior, the relationship between volume and temperature is given by Charles law

Charles law states that the volume of a given mass of gas is directly proportional to its temperature provided that pressure remains constant. Mathematically, this is represented as

V ∝ T

V=KT

K = V/T

where V is the volume of the gas

T is the Temperature

k represents the constant of proportionality

For initial and final conditions of a gas,

\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }

where 1 and 2 represent initial and final conditions respectively

therefore, T₁ = 100 and T₂ = 200

\frac{V_{1} }{100} = \frac{V_{2} }{200}

200 × V₁ = 100 × V₂

divide both sides by 100

2V₁ = V₂

final volume,V₂ = 2V₁

there the volume doubles

3 0
3 years ago
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