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4 years ago

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Rank the following aqueous solutions from highest to lowest freezing point: 0.1 m FeCl3, 0.30 m glucose (C6H12O6), 0.15 m CaCl2.

Assume complete dissociation.

1 answer:

Tresset [83]4 years ago

8 0

Answer:

Explanation: See images below for explanation

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You make an experiment to determine the percent mass of carbon in sugar. You find out that there is 46.63g of carbon in 75.00g o

Aleksandr [31]

Taking into account the definition of percentage composition, the percent composition of carbon in this sample is 62.17%.

<h3>Percentage composition</h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

Then, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percentageby mass= \frac{mass of solute}{mass of solution}x100$

<h3>This case</h3>

In this case, you know:

mass of solute= 46.63 g
mass of solution= 75 g

Replacing:

$percentageby mass= \frac{46.63 g}{75 g}x100$

Solving:

<u><em>percentage by mass= 62.17 %</em></u>

Finally, the percent composition of carbon in this sample is 62.17%.

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3 years ago

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago