For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et

Alexxx [7]

Answer: The vapor pressure of solution is 459.17 mmHg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For testosterone:

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol$

For diethyl ether:

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}$

$\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095$

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

$p^s$ = vapor pressure of the solution = ?

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

$\chi_{\text{solute}}$ = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

$\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg$

Hence, the vapor pressure of solution is 459.17 mmHg

7 0

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!

4 0

3 years ago

Why do you have to have a certain number of reactants on one side?

Debora [2.8K]

Answer:

EWQEQW

Explanation:

WEQWE

3 0

3 years ago

Ruby has observed that plants in her garden vary in height. She wants to investigate whether a plant species (species A) grows f

Agata [3.3K]

Answer: Tens place.

Step-by-step explanation: hmm wait one moment

7 0

1 year ago

Read 2 more answers

At 700 K, the reaction below has an Kp value of 54. An equilibrium mixture at this temperature was found to contain 0.933 atm of

kati45 [8]

Answer:

See explanation below

Explanation:

In this case, we have the equilibrium reaction which is:

H₂ + I₂ <------> 2HI Kp = 54

Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:

Kp = PpHI² / PpH₂ * PpI₂

Solving for the partial pressure of iodine:

PpI₂ = PpHI² / PpH₂ * Kp

Replacing the given values, we have:

PpI₂ = (2.1)² / 0.933 * 54

PpI₂ = 4.41 / 50.382

PpI₂ = 0.088 atm

4 0

4 years ago