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3 years ago

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By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?

1 answer:

34kurt3 years ago

5 0

Answer:

By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?.

Explanation:

The freezing point of a pure solvent decreases further by adding a nonvolatile solute.

This is called depression in freezing point.

When an ionic solute is dissolved then the depression in the freezing point is proportional to the number of ions present in the solution.

In aluminum sulfate, there are five ions formed as shown below:

$Al_2(SO_4)_3(aq)->2Al^3^+(aq)+3SO_4^2^-(aq)$

But sucrose is a covalent compound and it does not undergo dissociation.

Hence, aluminum sulfate decreases the freezing point of water by five times compared to sucrose.

Explanation:

You might be interested in

Which beakers in the model contain solid insoluble substances?

ladessa [460]

Although the models are not provided, I was able to find them and the beakers with solid present in them are:
1C
2A
2C
3A
3C

This is determined by the fact that the beakers all have a piece of closely packed substance laying at the bottom. This closely packed lattice is characteristic of solid substances, and the fact that they exist in the solution in the solid states indicates that they are insoluble.

7 0

3 years ago

A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What

NeTakaya

Answer:

Approximately $1.854\; \rm mol\cdot L^{-1}$ .

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of $\rm Sr$ , $\rm O$ , and $\rm H$ on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

$\rm Sr$ : $87.62$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Calculate the formula mass of $\rm Sr(OH)_2$ :

$M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}$ .

<h3>Number of moles of strontium hydroxide in the solution</h3>

$M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}$ means that each mole of $\rm Sr(OH)_2$ formula units have a mass of $121.634\; \rm g$ .

The question states that there are $10.60\; \rm g$ of $\rm Sr(OH)_2$ in this solution.

How many moles of $\rm Sr(OH)_2$ formula units would that be?

$\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}$ .

<h3>Molarity of this strontium hydroxide solution</h3>

There are $8.71467\times 10^{-2}\; \rm mol$ of $\rm Sr(OH)_2$ formula units in this $47\; \rm mL$ solution. Convert the unit of volume to liter:

$V = 47\; \rm mL = 0.047\; \rm L$ .

The molarity of a solution measures its molar concentration. For this solution:

$\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}$ .

(Rounded to four significant figures.)

5 0

3 years ago

fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she

enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0

3 years ago

Science question down below

aliya0001 [1]

Option 1/A (It is the first one)

6 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

b)

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago