Use Hess's law to calculate AG° rxn using the following information.

Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 degrees celcius. I

snow_lady [41]

Answer:

c. 8.1 L

Explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

$3O_2\rightarrow 2O_3$

3 moles of oxygen gas on reaction gives 2 moles of ozone

Also,

1 mole of oxygen gas on reaction gives 2/3 moles of ozone

So,

0.50 mole of oxygen gas on reaction gives $\frac{2}{3}\times 0.50$ moles of ozone

Moles of ozone = 0.3333 mol

Pressure = 1 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25.0 + 273.15) K = 298.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K

⇒V = 8.1 L

6 0

2 years ago

Please balance the equation, putting the correct coefficient in each box.

Mariulka [41]

Answer: $Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced equation will be:

$Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3$

5 0

3 years ago

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

4 0

3 years ago

Which statement best describes a typical difference that could be found between the "Analysis" and "Conclusion" sections of a

IgorC [24]

The statement “Only the “Conclusion” section discusses whether the original hypothesis was supported, and both sections suggest further research”, best describes the difference between analysis and conclusion.

Answer: Option 4

<u>Explanation: </u>

In research, we do experiments and derive the results. Then, those results were analyzed by us. In this analysis part, we compare our results with the related results published elsewhere. Also, we correlate the similarities and point out the differences between our analysis and other reported results.

In conclusion part, we have to check hypothesis or it supported. And, we summarise our analysis and figure out the further research need to be done on that to improvise our research. So, the final statement is the correct option which best describes the difference between analysis and conclusion.

3 0

3 years ago

In which color coded area would you find an element that behaves both like a metal and a non-metal?

Kay [80]

what elements are each color over? can you list one from each color so i can answer it

4 0

2 years ago