Compound

A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 d

DENIUS [597]

Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

where $k_2$ = rate constant at temp $T_2$

$k_1$ = rate constant at temp $T_1$

$E_a$ = activation energy

R= gas constant

$T_1$ = temperature = $20^0C=(20+273)K=293K$

$T_2$ = temperature = $30^0C=(30+273)K=303K$

$ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]$

$ln \frac{k_{2}}{k_{1}}=1.15$

$\frac{k_{2}}{k_{1}}=3$

Thus rate increases 3 times on raising the temperature from 20degree to 30 degree

3 0

3 years ago

Help plz ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!

ddd [48]

Honestly I don’t even know

7 0

3 years ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

3 years ago

A 1.00 liter solution contains 0.24 M hypochlorous acid and 0.31 M sodium hypochlorite. If 0.160 moles of potassium hydroxide ar

ryzh [129]

Explanation:

When OH- (as in potassium hydroxide) is added, it reacts with the acid (HOCl) to reduce the amount of HOCl and increase the concentration of sodium hypochlorite.

Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.

This occurs as follows:

HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)

since water is formed, this maintains the pH. Thus ...

A. The number of moles of HClO will decrease. - TRUE

B. The number of moles of ClO- will increase. - TRUE

C. The equilibrium concentration of H3O+ will remain the same. - TRUE

D. The pH will decrease. - FALSE

E. The ratio of [HClO] / [ClO-] will decrease. -TRUE. It will decrease as HClO goes down and ClO- goes up.

5 0

3 years ago

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

neonofarm [45]

Answer: The total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

Explanation:

Given: Mass = 46 g

Initial temperature = $-25^{o}C$

Final temperature = $100^{o}C$

Specific heat capacity of ice = 2.05 $J/g^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = m \times C \times (T_{2} - T_{1})\\= 46 g \times 2.05 J/g^{o}C \times (100 - (-25))^{o}C\\= 11787.5 J (1 J = 0.001 kJ)\\= 11.787 kJ$

Thus, we can conclude that the total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

7 0

3 years ago