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Ostrovityanka [42]
3 years ago
15

*2Al(OH), + 3H,SO, Al(SO4)2 + 6H2O reaction type

Chemistry
1 answer:
Tamiku [17]3 years ago
3 0
Substitution because you switch some of the other atoms when you get to the other side
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8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a
sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

                  Volume percentage = \frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100

                                                    = \frac{88.5}{760} \times 100

                                                    = 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

6 0
3 years ago
Help me and explain steps
Pachacha [2.7K]

Answer:

<u>2</u> Fe + <u>3</u> H2SO4 —> Fe2(SO4)3 + <u>3</u> H2

I hope I helped you^_^

3 0
2 years ago
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is
lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
  • 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.6 mol CH₄ * \frac{2molH_2O}{1molCH_4} = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

  • 1.2 mol * 18 g/mol = 21.6 g
4 0
2 years ago
Carbon exists in nature as three isotopes. Given the atomic masses and relative abundance of the isotopes, what is the average a
Ivan

The most abundant carbon isotope is carbon-12.

The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.

the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.

6 0
3 years ago
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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