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Novosadov [1.4K]
3 years ago
14

I need help asap!!!!

Chemistry
2 answers:
ddd [48]3 years ago
7 0

Answer:

24.64 L of O2 will be produced

Explanation:

Step 1: Data given

Mass of hydrogen peroxide = 75.0 grams

Molar mass of hydrogen peroxide (H2O2) = 34.015 g/mol

Step 2: The balanced equation

2H2O2 → 2H2O + O2

Step 3: Calculate moles H2O2

Moles H2O2 = 75.0 grams / 34.015 g/mol

Moles H2O2 = 2.20 moles

Step 4: Calculate moles O2

For 2 moles H2O2 we'll have 2 moles H2O and 1 mol O2

For 2.20 moles H2O2 we'll have 2.20/2 = 1.10 moles O2

Step 5: Calculate volume of O2

For 1 mol we have 22.4 L

For 1.10 moles we have 1.10 * 22.4 L = 24.64 L

24.64 L of O2 will be produced

Butoxors [25]3 years ago
6 0

Answer:

2H₂O₂ →  2H₂O + O₂

24.7 L are the liters of formed oxygen.

Explanation:

We state the reaction:

2H₂O₂ →  2H₂O + O₂

2 moles of peroxide decompose to 2 moles of water and 1 mol of oxygen gas.

We convert the mass to moles: 75 g . 1 mol / 34 g = 2.20 moles

As ratio is 2:1, per 2.20 moles of peroxide I would produce the half of moles, of O₂ → 2.20 /2 = 1.10 moles

We convert the moles to mass → 1.10 mol . 32 g / 1 mol = 35.3 g

Let's use oxygen's density to find out the volume

δ O₂ = 1.429 g/L    (mass/volume)

35.3 g . 1L / 1.429g = 24.7 L

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Ymorist [56]

Answer:

I and III

Explanation:

6 0
3 years ago
The discovery of protons and electrons most directly refuted or replace the idea that
Effectus [21]

Answer:

The atom resembles plum pudding

Explanation:

The discovery of the electron in 1897 by J. J Thomson and the proton in 1917 by Rutherford most directly refuted or replace the idea that the atom resembles plum pudding.

8 0
3 years ago
The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
3 years ago
In the titration of HCl with NaOH, the equivalence point is determined
kondaur [170]

Answer:

In the titration of HCl with NaOH, the equivalence point is determined from the point where the phenolphthalein turns pink and then remains pink on swirling.

Explanation:

The equivalence point is the point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl). Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask.

Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. It remains colorless throughout the range of acidic pH levels, but it begins to turn pink at a pH level of 8.3 and continues to a bright purple in stronger alkalines.

It will appear pink in basic solutions and clear in acidic solutions.

The more NaOH added, the more pink it will be. (Until pH≈ 10)

In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above 13.0 pH

a. from the point where the pink phenolphthalein turns colorless and then remains colorless on swirling.

⇒ the more colorless it turns, the more acid the solution. (More HCl than NaOH)

b. from the point where the phenolphthalein turns pink and then remains pink on swirling.

The equivalence point is the point where phenolphtalein turns pink and remains pink ( Between ph 8.3 and 10). (

Although, when there is hydrogen ions are in excess, the solution remains colorless. This begins slowely after ph= 10 and can be noticed around ph = 12-13

c. from the point where the pink phenolphthalein first turns colorless and then the pink reappears on swirling.

Phenolphthalein is colorless in acid solutions (HCl), and will only turn pink when adding a base like NaOH

d. from the point where the colorless phenolphthalein first turns pink and then disappears on swirling

Phenolphthalein is colorless in acid or neutral solutions. Once adding NaOH, the solution will turn pink. The point where the solution turns pink, and stays pink after swirling is called the equivalence point. When the pink color disappears on swirling, it means it's close to the equivalence point but not yet.

3 0
3 years ago
Write the balanced standard combustion reaction for the c5h7on3s2
AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0
3 years ago
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