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1 year ago

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one mole of sodium chloride is added to 100 ml of water in beaker a. one mole of glucose, c6h12o6, is added to 100 ml of water i

n beaker b. how much more will the freezing point of water be lowered in beaker a than in beaker b?

1 answer:

Viefleur [7K]1 year ago

5 0

Twice as much more will the freezing point of water be lowered in beaker a than in beaker b.

<h3>What determines freezing point?</h3>

A liquid's freezing point rises if the intermolecular interactions between its molecules are strong. The freezing point, however, drops if the molecules of inter - molecular are minimal. The process through which a substance transforms from a liquid into a solid is known as freezing.

<h3>How significant is freezing point?</h3>

Freezing points play a big role in occupational safety. A chemical may perhaps turn harmful if held below its freezing point. A critical safety benchmark for assessing the effects of worker exposure to cold environments is the freezing point.

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What do the planets Earth and Mars have in common?

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I think the answer is B

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3 years ago

Can you please answer number 8 thanks

MAXImum [283]

A day is added to February every four tears as a corrective measure. The Earth does not exactly revolve around the sun for 365 days. There is a small offset that is corrected by adding a day to February. This happens every four years and it is called Leap year.

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3 years ago

Which subatomic particle can change the molecular geometry of a molecule?

emmainna [20.7K]

Answer:

Electrons

Explanation:

According to the Valence Shell Electron Pair Repulsion Theory (VSEPR), the geometry of a molecule depends on the number of electron pairs (regions of electron density) on the central atom of the molecule. Electron pairs on the valence shell of the molecule tend to position themselves as far apart in space as possible to minimize repulsion between them. Hence, the orientation of these electron pairs is the ultimate determinant of the observed geometry of a molecule.

Lone pairs of electrons cause more repulsion than bond pairs of electrons on the central atom of a molecule. Hence when the central atom of a molecule contains lone pairs of electrons, the molecular geometry is usually distorted from the expected geometry on the basis of VSEPR theory.

Hence, electrons are the subatomic particles which are responsible for any change in the observed molecular geometry of a molecule.

6 0

3 years ago

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

3 years ago

You are given the following three half-reactions:(1) Fe³⁺(aq) + e⁻ ⇄ Fe²⁺(aq) (2) Fe²⁺(aq) +2e⁻ ⇄ Fe(s) (3) Fe³⁺(aq) +3e⁻ ⇄ Fe(s

RideAnS [48]

The E°half-cell of the (3) is 0.33 volts .

Given,

(1) Fe3+(aq) +e =Fe2+(aq)

(2) Fe2+(aq) +2e = Fe(s)

(3) Fe3+(aq) +3e = Fe(s)

We know ,

E° half-cell of (1) is 0.77 volts .

E° half cell of ( 2) is -0.44 volts .

By resolving or adding equation 1 and 2 we will get (3) ,

Thus , the value of the E° half cell of (3) is given by ,

E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts

Hence the E° half-cell of ( 3 ) is 0.33 volts .

<h3>What is cell potential? </h3>

It is the difference between the electrode potentials of two half cells.

it also defined as the force which causes the flow of electrons from one electrode to the another and thus results in the flow of current.

Ecell = E( cathode) - E ( anode)

<h3>What is electrode potential ? </h3>

The tendency of the electrode to lose of gain electrons is called as electrode potential.

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4 0

2 years ago