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olya-2409 [2.1K]
3 years ago
7

rank the 4 gases (air,exhaled air, gas produced from decomposition of H202, gas from decomposition of NaHCO3) in order of increa

sing concentration of oxygen
Chemistry
1 answer:
nikitadnepr [17]3 years ago
7 0
Air has about 21% oxygen
Exhaled air has about 16% oxygen
The gas produced from decomposition of H₂O₂ is pure oxygen
The gas produced from decomposition of NaHCO₃ is pure carbon dioxide
Thus, the order is:
1) Gas produced from decomposition of H₂O₂
2) Air
3) Exhaled air
4) Gas produced from decomposition of NaHCO₃
You might be interested in
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
3 years ago
In which state of the following compounds does nitrogen have the most positive oxidation state? group of answer choices no2 nano
Amiraneli [1.4K]

The compound nitrogen have most positive oxidation state is NO₂. The correct option is b.

<h3>What is oxidation state?</h3>

The total number of electrons gained or lost by an atom in order to form a chemical bond with another atom.

The charge on an ion is equal to the sum of the oxidation states of all the atoms in the ion. A substance's more electronegative elements are given a negative oxidation state.

A positive oxidation state is assigned to the less electronegative element.

Thus, the correct option is b, NO₂.

Learn more about oxidation state

brainly.com/question/11313964

#SPJ4

4 0
2 years ago
The limiting reactant in a reaction a. is the reactant for which there is the least amount in grams b. is the reactant which has
Shalnov [3]

Actually the correct answer must be:

The limiting reactant in the reaction is the one which has the lowest ratio of moles available over coefficient in the balanced equation

 

This is because the actual mass or number of moles of the reactant does not directly dictate if it is a limiting reactant, this must be relative to the other reactants.

 

So the answer is:

e. none of the above

5 0
2 years ago
Can you please help me it’s not a hard question
aliya0001 [1]
The answer for acceleration is m/s^2
5 0
2 years ago
An object is traveling at a speed of 7500.0 centimeters per second. How many kilometers per day
ivann1987 [24]

Answer:

6,480 kilometers

Explanation:

1 day = 86,400 seconds

1 kilometer = 100,000 centimeters

Equation:

86,400 x 7,500 = 648,000,000

648,000,000 ÷ 100,000 = 6,480

Hope this helped :  )

7 0
2 years ago
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