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aniked [119]
3 years ago
11

Try to make only one molecule of water in the simulation. Is it possible?

Chemistry
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.

Explanation:

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An aqueous solution of methylamine (ch3nh2) has a ph of 10.68. how many grams of methylamine are there in 100.0 ml of the soluti
ruslelena [56]

Answer:

3.4 mg of methylamine

Explanation:

To do this, we need to write the overall reaction of the methylamine in solution. This is because all aqueous solution has a pH, and this means that the solutions can be dissociated into it's respective ions. For the case of the methylamine:

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻     Kb = 3.7x10⁻⁴

Now, we want to know how many grams of methylamine we have in 100 mL of this solution. This is actually pretty easy to solve, we just need to write an ICE chart, and from there, calculate the initial concentration of the methylamine. Then, we can calculate the moles and finally the mass.

First, let's write the ICE chart.

       CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻     Kb = 3.7x10⁻⁴

i)            x                                      0            0

e)          x - y                                  y            y

Now, let's write the expression for the Kb:

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]

We can get the concentrations of the products, because we already know the value of the pH. from there, we calculate the value of pOH and then, the OH⁻:

The pOH:

pOH = 14 - pH

pOH = 14 - 10.68 =  3.32

The [OH⁻]:

[OH⁻] = 10^(-pOH)

[OH⁻] = 10^(-3.32) = 4.79x10⁻⁴ M

With this concentration, we replace it in the expression of Kb, and then, solve for the concentration of methylamine:

3.7*10⁻⁴ = (4.79*10⁻⁴)² / x - 4.79*10⁻⁴

3.7*10⁻⁴(x - 4.79*10⁻⁴) = 2.29*10⁻⁷

3.7*10⁻⁴x - 1.77*10⁻⁷ = 2.29*10⁻⁷

x = 2.29*10⁻⁷ + 1.77*10⁻⁷ / 3.7*10⁻⁴

x = [CH₃NH₂] = 1.097*10⁻³ M

With this concentration, we calculate the moles in 100 mL:

n = 1.097x10⁻³ * 0.100 = 1.097x10⁻⁴ moles

Finally to get the mass, we need to molar mass of methylamine which is 31.05 g/mol so the mass:

m = 1.097x10⁻⁴ * 31.05

<h2>m = 0.0034 g or 3.4 mg of Methylamine</h2>
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