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3 years ago

H2O2

Chemistry

1 answer:

Stels [109]3 years ago

7 0

Answer:

Total Ionic equation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)

Explanation:

Chemical equation:

H₂O₂(aq) + MgSO₃(aq) → MgSO₄(aq) + H₂O(l)

Balanced chemical equation:

H₂O₂(aq) + MgSO₃(aq) → MgSO₄(aq) + H₂O(l)

Total Ionic equation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)

Water can not split into ions because it is present is liquid form.

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3 years ago

Iron reacts with oxygen gas to produce iron (III) oxide. If there are 52 grams of

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Mass of Fe₂O₃ : 74.26 g

Volume of O₂ at STP = 15.624 L

<h3>Further explanation</h3>

Reaction

4Fe (s) + 3O₂ (g) ==> 2Fe₂O₃ (s)

mol Fe :

Ar Fe = 56 g/mol

$\tt \dfrac{52}{56}=0.93$

mol Fe₂O₃ :

$\tt \dfrac{2}{4}\times 0.93=0.465$

mass Fe₂O₃ :

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3 years ago

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3 years ago

A 25.0 mL solution of a monoprotic weak acid is titrated with 0.10 M NaOH. The equivalence point is reached after 38.52 mL of Na

Sveta_85 [38]

Answer:

0.154 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 25.0mL

M₂ = 0.10 M

V₂ = 38.52 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂

Putting the respective values ,

M₁ * 25.0mL = 0.10 M * 38.52 mL

M₁ = 0.154 M

8 0

4 years ago

The density of aluminium is 2.70g/cm^3 . If a cube of aluminum weights 13.5 grams, what is the lengths of the edge of the cube.

Harlamova29_29 [7]

Answer: -

1.71 cm

Explanation: -

Density of aluminium = 2.70 g / $cm^{3}$

Mass of aluminium = 13.5 g

We know density of a substance is the mass of the substance per unit volume.

Density = Mass / volume

Volume = Mass / density

Volume of aluminium = Mass of aluminium / Density of Aluminium

= 13.5 g / 2.70 g/ $cm^{3}$

= 5 $cm^{3}$

We know for a cube volume = $length^{3}$

Length of the cube = $Volume^{1/3}$

= $5 ^{1/3}$

=1.71 cm

Thus if the density of aluminium is 2.70g/ $cm^{3}$ and the cube of aluminum weights 13.5 grams, 1.71 cm is the length of the edge of the cube

5 0

3 years ago