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2 years ago

1. An acetic acid solution that is 60.0% HC2H302 (by mass) contains

Chemistry

1 answer:

d1i1m1o1n [39]2 years ago

7 0

Answer:

A solution of acetic acid that is 60.0% HC₂H₃O₂ (by mass) indicates that it contains 60.0 g of acetic acid and 100.0 g of water.

Explanation:

A percentage is a way of expressing an amount as a fraction of 100. The mass percentage corresponds to physical units of the solutions and they allow to establish more precisely the concentration of the solutions and express them in terms of percentages.

Mass percentage indicates the amount in grams of solute per 100 grams of solution.

So a solution of acetic acid that is 60.0% HC₂H₃O₂ (by mass) indicates that it contains 60.0 g of acetic acid and 100.0 g of water.

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The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

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3 years ago

Help i just need people to answer the question today

yKpoI14uk [10]

Answer:

1) Hydrogen

2) Methane

3) Carbon

4) Structural isomer

5) Ethene also known as ethylene

6) Hydrocarbons are widely used as fuel

7) Crude oil

Explanation:

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3 years ago

I’m only having a problem with this number that’s all if anyone could help I would really appreciate it! Thank you!

Masja [62]

Answer:D.

Explanation:i used photo math. It is really helpful when I'm doing math homework etc...

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3 years ago

How did Bohr refine the model of the atom? How did Bohr refine the model of the atom? He developed electrochemistry. He discover

BartSMP [9]

Answer:

He developed the concept of concentric electron energy levels

Explanation:

Before Bohr's model, Rutherford's model was proposed. This model explains most of the properties of the atom but failed to explain the stability of the atom.

As per Rutherford's model, electrons revolve around the nucleus in the orbit.

But revolving electron in their orbit around nucleus would give up energy and so gradually move towards the nucleus and therefore, eventually collapse.

Bohr's proposed that the electrons around the nucleus move orbit of fixed energy called "stationary states". Electrons in these stationary states do not radiate energy.

Therefore, proposal of concentric electron energy levels refine the atomic models.

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3 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

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3 years ago