How to write a compound that contains one aluminum atom for every three chlorine atoms?

How is burning magnesium different than burning methane

olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

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6 0

3 years ago

Determine the mass of a ball with a wavelength of 3.45 × 10-34 m and a velocity of 6.55 m/s

Yuki888 [10]

293 grams The formula for the wavelength of a massive particle is Î» = h/p where Î» = wavelength h = Plank constant (6.626070040Ă—10^â’34 J*s) p = momentum (mass times velocity) So let's solve for momentum and from there get the mass Î» = h/p Î»p = h p = h/Î» Substitute known values and solve p = 6.626070040Ă—10^â’34 J*s/3.45Ă—10^-34 m p = 1.92 J*s/m Since momentum is the product of mass and velocity, we have p = M * V p/V = M So substitute again, and solve. p/V = M 1.92 J*s/m / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 0.293 kg = M So the mass is 293 grams

6 0

3 years ago

Read 2 more answers

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade

egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

Explanation:

Assuming Pressure is constant.

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

$\frac{300}{273} = \frac{V_2}{288}$

$V_2 = \frac{288*300}{273} \\V_2 = 316.5$

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

5 0

2 years ago

At a certain temperature, the vapor pressure of pure benzene ‍ is atm. A solution was prepared by dissolving g of a nondissociat

erastova [34]

Answer:

Molar mass of solute: 300g/mol

Explanation:

Vapor pressure of pure benzene: 0.930 atm

Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm

It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:

$P_{sln} = X_{solvent}P_{solvent}^0$

Moles in 78.11g of benzene are:

78.11g benzene × (1mol / 78.11g) = 1 mol benzene

Now, mole fraction replacing in Raoult's law is:

0.900atm / 0.930atm = 0.9677 = moles solvent / total moles.

As mole of solvent is 1:

0.9677× total moles = 1 mole benzene.

Total moles:

1.033 total moles. Moles of solute are:

1.033 moles - 1.000 moles = 0.0333 moles.

As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:

10.0g / 0.033moles = 300g/mol

8 0

3 years ago

What are the various ways in which an atom of an element can achieve the noble gas configuration ?

azamat

Answer:

<h3>Atoms attain noble gas configuration by obtaining or donating and sharing of electrons present in their outermost shell. </h3>

7 0

2 years ago