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Hunter-Best [27]
2 years ago
6

Some amount of hydrogen peroxide (H2O2) breaks down to produce 3 molecules of oxygen (O2) and 6 molecules of water (H2O). How ma

ny atoms of hydrogen are there?
Chemistry
2 answers:
cricket20 [7]2 years ago
8 0

Answer:

The answer to the question is ;

By counting the number of hydrogen in the decomposed hydrogen peroxide, there are a total 12 atoms of hydrogen in the reaction.

Explanation:

To solve the question, we note that

6H₂O₂ (l) → 6H₂O (l) + 3O₂ (g)

From the stoichiometry of the reaction, as it can be seen by balancing the equation that 6 moles of hydrogen peroxide liquid decomposes to produce 6 moles of liquid water molecule (H₂O) and 3 moles of oxygen gas molecules O₂.

Therefore, taking  count of hydrogen atoms in the reactant, there are 12 atoms of hydrogen atoms in the 6 molecules of hydrogen peroxide before the decomposition took place.

soldier1979 [14.2K]2 years ago
3 0

Answer:

Atoms_H=12Atoms

Explanation:

Hello,

In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:

Atoms_H=6moleculesH_2O*\frac{1molH_2O}{6.022x10^{23}moleculesH_2O}\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}AtomsH}{1molH} \\Atoms_H=12Atoms

Best regards.

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The difference in an area with high concentration and an area with low concentration is called.
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Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
drek231 [11]

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

6 0
3 years ago
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