What is the answer slam sms

The osmotic pressure of an aqueoussolution of urea at 300 K is

den301095 [7]

Answer: The freezing point of solution is -0.09°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=imRT$

where,

$\pi$ = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the solution = 300 K

Putting values in above equation, we get:

$120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=i\times K_f\times m$

where,

i = Vant hoff factor = 1 (for non-electrolytes)

$K_f$ = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

$\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

The equation used to calculate freezing point of solution is:

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

where,

$\Delta T$ = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC$

Hence, the freezing point of solution is -0.09°C

4 0

3 years ago

An experiment produced 0.10 g CO2, with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is t

Pachacha [2.7K]

The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.

3 0

3 years ago

Zinc is a pure substance. How could you make a mixture from a 1.0-g zinc disc?

anygoal [31]

Answer:

D)Coat the disc with another pure substance, such as copper.

Explanation:

To make a mixture of the zinc, simply coat it with another pure substance such as copper. This will result in a mixture.

What then is a mixture?

They are impure substances with indefinite composition.
They retain their physical properties
their constituents reacts differently to changed conditions
They can easily be separated into constituents by physical methods.

We can see that coating the copper on the zinc does not alter the physical properties of the metal.

Most times, the copper will wear off and leave behind the zinc.

The metal coated can easily be removed and this signifies just a physical mix.

8 0

3 years ago

How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed?

spayn [35]

The balanced chemical reaction is written as:

2Al2O3 = 4Al + 3O2

We are given the amount of aluminum oxide used in the reaction. This will be the starting point for the calculations. We do as follows:

26.5 mol Al2O3 ( 3 mol O2 / 2 mol Al2O3 ) = 39.75 mol O2 produced

7 0

3 years ago

What volume (ml) of 3.0 m naoh is required to react with 0.8024-g copper(ii) nitrate? what mass of copper(ii) hydroxide will for

babunello [35]

The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂

Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction

Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction

7 0

4 years ago

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