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3 years ago

A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

2 answers:

6 0

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.
m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.

n = 8.45 g / 232 g/mol = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality
Calculate the mass of the solvent in kilogram given the volume and its density.
m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.
The total weight of the solution is equal to 302.5 g.

weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. Weight percent in ppm.
weight percent of solute x (10000 ppm/1%)
= (2.79%)(10000 ppm/1%) = 27933.88 ppm

tigry1 [53]3 years ago

6 0

Answer : The molarity, molality, weight percent and ppm are, 0.0987 mole/L, 0.0839 mole/Kg, 2.79% and 27933.8843 respectively.

Solution : Given,

Mass of solute $(C_{12}H_{22}O_{11})$ = 8.45 g

Molar mass of solute $(C_{12}H_{22}O_{11})$ = 342.3 g/mole

Volume of solvent (water) = volume of solution = 250 ml

Density of solution = $1.21g/cm^3=1.21g/ml$

Calculation for molarity

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$Molarity=\frac{8.45g\times 1000}{342.3g/mole\times 250ml}=0.0987mole/L$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=Density\times Volume=1.21g/ml\times 250ml=302.5g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=302.5-8.45=294.05g$

Calculation for molality

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}$

$Molality=\frac{8.45g}\times 1000}{342.3g/mole\times 249.05g}=0.0839mole/Kg$

Calculation for weight percent

$weight\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{8.45g}{302.5g}\times 100=2.79\%$

Calculation for ppm

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6=\frac{8.45g}{302.5g}\times 10^6=27933.8843$

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When the surface water temperature is still well below the water boiling temperature, some bubbles at the bottom tend to float u

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Answer:

The reasons why the seemingly floating bubbles disappear was that they tend to loss their latent heat to the water molecules at the surface water.

Explanation:

Heat energy has a considerable effect on the velocity of molecules including water. The water molecules below the container will receive much more heat energy than those above it. This heat energy in the form of specific heat capacity and latent heat that result in the increase in the speed of individual molecules of water and finally to the escape of the molecules to a colder region of the container, in this case the upper region. At the collision of the bottom water to the surface water, they tend to exchange their heat content, the hotter molecules will lose their heat to the cold ones. When the formerly hot molecules encounter this, it will result in lowering the temperature and consequentially to the reduction of their movement, once in the form of bubble, now become ordinary water. This convectional transfer of heat energy will continue until the whole system has a uniform temperature depending on the consistency of the heat source.

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3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

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3 years ago

Wine goes bad soon after opening because the ethanol ch3ch2oh in it reacts with oxygen gas o2 from the air to form water h2o and

lilavasa [31]

The balanced chemical equation of the reaction described above is,

C2H6O + O2 --> H2O + C2H4O2

If we have 3.84 g of oxygen, we divide by its molar mass.

n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
n = 0.11 moles O2

Using ratio and proportion,

number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
= 0.11 moles C2H6

Then, we multiply the calculated value to its molar mass, 46 grams /mol.
mass of ethanol = (0.11 mol) x (46 grams / mol)
= 5.06 grams

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3 years ago

Two compounds are standing at the same temperature. Compound "A" is evaporating more slowly than compound "B." According to the

frosja888 [35]

Answer:

Option D) Compound B may have a lower molecular weight.

Explanation:

Compound A and B are standing at the same temperature yet compound A is evaporating more slowly than compound B.

This simply indicates that compound B have a lower molecular weight than compound A.

This can further be seen when gasoline and kerosene are placed under same temperature. The gasoline will evaporate faster than kerosene because the molecular weight of the gasoline is low when compared to that of the kerosene.

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3 years ago

4. Which class of matter is the least evenly

lys-0071 [83]

Answer:

Pure iron sulfide is homogeneous (uniform in appearance and properties), shows constant composition (a consistent ratio of iron to sulfur throughout any sample of it, large or small), consists of molecules all of one type, is no longer separable into two separate substances without another chemical reaction, and is .

Explanation:

Mixtures in two or more phases are heterogeneous mixtures. ... The exception would be solutions that contain another phase of matter. For example, you can make a homogeneous solution of sugar and water, but if there are crystals in the solution, it becomes a heterogeneous mixture.

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3 years ago