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Vedmedyk [2.9K]
3 years ago
12

Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are

Chemistry
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

Decomposition of aluminium oxide forms  aluminium atoms and  oxygen atoms.

Explanation:

<u>Decomposition reaction:</u>

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

AB\rightarrow A+B

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

Al_{2}O_{3}\rightarrow Al+O_{2}

The balanced equation is as follows.

2Al_{2}O_{3}\rightarrow 4Al+3O_{2}

Therefore, Decomposition of aluminium oxide forms aluminium atoms and  oxygen atoms.

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Indicate if the following statements are TRUE (T) or FALSE (F):
Aleks [24]

Answer:

a) T

b) T

c) F

d) F

e) T

f) T

g) T

h) F

I) F

j) F

k) F

l) F

Explanation:

The w/v concentration is obtained from, mass/volume. Hence;

%w/v= 50/1000= 5%

In the %w/w we have;

25g/100 g = 25% w/w

In combustion reaction, energy is given out hence it is exothermic.

Neutralization reaction yields a salt and water

% by mass of carbon is obtained from;

8× 12/114 × 100 = 84.1%

All the ionic substances mentioned have very low solubility in water.

One mole of a substance contains the Avogadro's number of each atom in the compound.

There are two iron atoms so one mole contains 2× 55.85 g of iron.

Some sulphates such as BaSO4 are insoluble in water.

Halides are soluble in water hence NaI is soluble in water.

The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.

The equation represents a decomposition of calcium carbonate as written.

3 0
3 years ago
A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)
Natalija [7]

Answer:

2%

Explanation:

.98 is 98% of one and therefore they are missing 2%

4 0
3 years ago
Read 2 more answers
How would having too much sample in the melting point tube most likely affect the melting point measurement? Select the correct
oksano4ka [1.4K]

Answer:

2-4 mm height of capillary tube.

Explanation:

Sample should be around 2-4 mm in height.

It should be packed well so that it does not have air packets that caues the lowering of melting point.

If you take greater amount, then there will be needed more heat, resulting a wide range of melting point.

7 0
3 years ago
If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?
GenaCL600 [577]
K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K          +            10KNO₃  ⇒  6K₂O + N₂
2mol        :            10mol
3,97mol   :           1,21mol
                             limiting reagent

KNO₃ is limiting reagent

5 0
3 years ago
A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced.  The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = \frac{1.085 X 100}{1.45} =74.8 %.

7 0
2 years ago
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