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Nadusha1986 [10]
2 years ago
12

?Al + ?H2SO4 → ?Al2(SO4)3 +3H2

Chemistry
1 answer:
AfilCa [17]2 years ago
5 0

The answer is: 27 grams of aluminium.

Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

n(H₂) = 1.5 mol; amount of hydrogen.

Form chemical reaction: n(Al) : n(H₂) = 2 : 3.

n(Al) = 2 · 1.5 mol ÷ 3.

n(Al) = 1.0 mol; amount of aluminium.

m(Al) = n(Al) · M(Al).

m(Al) = 1 mol · 27 g/mol.

m(Al) = 27 g; mass of aluminium.

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The oxides of nitrogen are very important ingredients in determining urban air pollution. Name each of the following compounds.(
deff fn [24]

Answer:

(a) N₂O - Dinitrogen monoxide or Dinitrogen (I) oxide

(b) NO - Nitrogen monoxide or Nitrogen (II) oxide

(c) NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide

(d) N₂O₅ - Dinitrogen pentoxide

(e) N₂O₄ - Dinitrogen tetraoxide.

Explanation:

To name each of the compounds,

(a) N2O; It is properly written as N₂O. N₂O is a colorless, sweet tasting gas also known as "laughing gas".

N₂O - Dinitrogen monoxide OR Dinitrogen (I) oxide.

(b) NO; NO is a colorless gas.

NO - Nitrogen monoxide OR Nitrogen (II) oxide.

(c) NO2; It is properly written as NO₂

NO₂ is a reddish-brown gas.

NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide.

(d) N2O5; It is properly written as N₂O₅

N₂O₅ is a white solid.

N₂O₅ - Dinitrogen pentoxide.

(e) N2O4; It is properly written as N₂O₄;

N₂O₄ is a red-brown liquid with an unpleasant smell.

N₂O₄ - Dinitrogen tetraoxide.

6 0
3 years ago
What are the kingdoms of eukaryotes
Talja [164]
Protists,fungi,plantar and animalia
4 0
3 years ago
Abbey always gets zits on her chin and she can't figure out
lesya [120]

Experiment has to do with the examination of the relationship between the dependent variable and the independent variable.

<h3>What is an experiment?</h3>

The term experiment has to do with the examination of the relationship between the dependent variable and the independent variable. In this case, the independent variable is the variable that we can manipulate while the dependent variable the one that changes when we manipulate the independent variable.

Let us now identify the variables in the experiment;

independent variable; the type of product used

constant  variable; length of time that the product is used

Hypothesis; The Zits could be cleared from her face by a skin care product

The control group is the washing with water

The experimental group is the washing with a product

The constant is the face on which the product s used.

Learn more about experiment;brainly.com/question/11256472

#SPJ1

Missing parts;

Abbey always gets zits on her chin and she can’t figure out the best method to get rid of them. She decides to test several products to see which clears her face up fastest: washing with soap and water, washing with an acne face wash from the pharmacy, and washing her face with organic oils. For one week she just washes her face with water. The next week she washes every day with soap and water. The following week she uses the acne wash, and the last week she uses the organic oils. Each week she records the number of zits she has on her chin at the end of the day. What is the independent variable? What is the constant variable? What is the hypothesis? What is the control group? What is the experimental group? What is the constant?

6 0
1 year ago
Will Upvote!
Greeley [361]
I would think energy has been transferred <span />
4 0
3 years ago
Read 2 more answers
Use the half-reaction method to balance the following equation in acidic solution. It is not necessary to include any phases of
yawa3891 [41]

Answer:

2H⁺ + 2MnO4⁻ + 3CN⁻ → 2MnO₂  + H₂O + 3CNO⁻

Explanation:

CN⁻   +   MnO4⁻  →  CNO⁻   +   MnO₂

In right side permangante, with Mn element has +7 as oxidation number.

In the MnO₂, Mn acts with +4.

This is the half reaction of reduction, where the Mn has gained 3 electrons.

in right side, cianide with C element has +2 as oxidation number. In the anion cianate, C acts with +4.

The oxidation number has increased in this half reaction. It's oxidation where the C, has lost 2 elecontrons

( 4H⁺ + MnO4⁻ + 3e⁻  → MnO₂  + 2H₂O ) .2

(H₂O  +  CN⁻  →  CNO⁻  + 2e⁻  + 2H⁺) .3

I have to add water, to ballance the amount of oxygens and protons to ballance H, in the opposite side

To ballance the half reactions, I have to multiply x2 (reduction) and x3 (oxidation)  so I can cancel, the electrons.

2H⁺  +  2MnO4⁻ + 3CN⁻ → 2MnO₂  + H₂O + 3CNO⁻

The electrons are now cancelled, and I can also modify water. In reactant side I have 3H₂O and in product side, I have 4H₂O so, H₂O in right side is gone so I finally obtained 1 H₂O on left. I had 8H⁺ on right, and 6H⁺ on left, so finally I obtained 2H⁺ on right, and the protons of product side are gone.

6 0
3 years ago
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