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59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example.

professor190 [17]

Answer:

Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.

H2O2 act as oxidizing agent in acidic medium.

Explanation:

Example : 2FeSO4 +H2SO4 +H2O2 —>

(ferrous sulphate)

Fe2(SO4)3 +2H2O

(ferric sulphate)

3 0

2 years ago

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of

timama [110]

<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

$HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)$

$KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)$

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, $A^-$ ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

If the solution has high hydrogen ion concentration, then the pH will be low.
If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of $H^+$ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

8 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

What is most likely oxidation state of aluminum (AI)?

blagie [28]

Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.

7 0

3 years ago

How did the atmosphere get its oxygen

riadik2000 [5.3K]

Because super tiny organisms called cyanobactierea (alge) they conducted photosynthesis and they us sunshine water and Carbon D to make Oxygen

7 0

3 years ago

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