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Aleks04 [339]
3 years ago
10

Two solid samples each contain sulfur, oxygen, and sodium, only. These samples have the same color, melting point, density, and

reaction with
an aqueous barium chloride solution. It can be concluded that the two samples are the same
(1) compound (3) mixture
(2) element (4) solution
Chemistry
1 answer:
katovenus [111]3 years ago
7 0
<span>The correct answer is 1. compound. There are three elements and not just one so it's not element. Mixtures wouldn't have the same properties that you described but these do so it's a compounds. Solution is not a solid thing but rather a liquid one. The two samples are therefore compounds.</span>
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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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Describe how phospholipids form a barrier between water inside the cell and water outside the cell
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<span>Phospholipids would have to form a phospholipid bilayer in order to achieve water on the outside and water inside. This is because the nonpolar tails of the phospholipids are facing each other in a water environment because they cannot interact with the water, only their own tails, while the phosphate heads of the molecule face the periphery of the tails and interact with water. Micelles are the simplest examples of these.</span>
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almond37 [142]

A. 0.02 mol of O2

B. 0.1 mol of CI2

C. 1 mol of N2

D. 2 mol of H2

Bolded answer is correct.

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Answer:

Only the H is unbalanced.

Explanation:

There are 4 H's and 2 of everything else

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