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3 years ago

One mole of copper(ii) sulfate, cuso4, contains ________ o atoms.

Chemistry

2 answers:

Naddik [55]3 years ago

8 0

Answer : One mole of copper(II)sulfate contains $24.088\times 10^{23}$ number of oxygen atoms.

Solution :

As we know that, 1 mole of contains $6.022\times 10^{23}$ number of atoms.

In $CuSO_4$ , there are 1 number of copper atom, 1 number of sulfur atom and 4 number of oxygen atoms.

As 1 mole of copper(II)sulfate contains $4\times 6.022\times 10^{23}=24.088\times 10^{23}$ number of oxygen atoms.

Therefore, one mole of copper(II)sulfate contains $24.088\times 10^{23}$ number of oxygen atoms.

Ierofanga [76]3 years ago

6 0

According to avogadro constant, the number of units in one mole of any substance contain 6.022 x10 ^23 atoms

therefore the number of o atoms in one mole of CuSO4 = 6.022 x 10 ^ 23

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What group of elements easily gains one valence electron

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3 years ago

The first step in the free radical mechanism for the preparation of polyethylene is:

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Answer:

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

Explanation:

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2. Chain propagation

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(see the attached image)

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<u><em>The sequence of Polymerisations are:</em></u>

b. heating an organic peroxide to break the O-O bond

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3 years ago

What is the total number of electrons in As-3?

V125BC [204]

Answer:

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Explanation:

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2 years ago

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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