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2 years ago

HYDROGEN PEROXIDE WITH CONCENTRATION OF

Chemistry

1 answer:

stiv31 [10]2 years ago

4 0

Answer:

(a) To calculate the oxygen gas produced in litters at STP when the compound undergoes complete decomposition. First note the equation at decomposition below

2H2O2(aq) → 2H2O(l) + O2(g)

Then

(3.0 g/100 ml)(10 ml) = 0.3 g H2O2

divide by Mass of H2O2 which is 34 g/mole,

0.3/34 = 0.00882353

this will arise to 8.8 x 10-3 molesH2O2

then multiply the moles H2O2 by: (1 mole O2/2 mole H2O2)= 4.4 x 10-3moles O2

To calculate the volume (in L) of Oxygen 02, we have to use the ideal gas equation:

VO2 = nO2RT/P;

n = 4.4 x10-3moles, P = 1 atm, R - 0.0826 L-atm/mol-K andT = 273K

V02 = 99.22 X 10∧-3

(b) divide volume of O2 from (a) by 10 ml (0.010L - initialvolume of H2O2)

9.92 X10∧-3

Explanation:

a) To calculate the oxygen gas produced in litters at STP when the compound undergoes complete decomposition. First note the equation at decomposition below

2H2O2(aq) → 2H2O(l) + O2(g)

Then

(3.0 g/100 ml)(10 ml) = 0.3 g H2O2

divide by Mass of H2O2 which is 34 g/mole,

0.3/34 = 0.00882353

this will arise to 8.8 x 10-3 molesH2O2

then multiply the moles H2O2 by: (1 mole O2/2 mole H2O2)= 4.4 x 10-3moles O2

To calculate the volume (in L) of Oxygen 02, we have to use the ideal gas equation:

VO2 = nO2RT/P;

n = 4.4 x10-3moles, P = 1 atm, R - 0.0826 L-atm/mol-K andT = 273K

V02 = 99.22 X 10∧-3

(b) divide volume of O2 from (a) by 10 ml (0.010L - initialvolume of H2O2)

9.92 X10∧-3

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The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

bagirrra123 [75]

Answer:

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

Explanation:

The steps of the Ostwald process:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)$

$3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)$

Combinning the equations:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2$

$(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3$

$4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) + 4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)$

Simplifying:

$4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow 14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)$

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

4 0

3 years ago

What is the molarity of a solution if 300.0 ml of it contains 16.8 g of kno3? [formula weight: kno3, 101.11 amu]?

AVprozaik [17]

The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.

Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M

6 0

3 years ago

Read 2 more answers

What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?

Vlad1618 [11]

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

$\rm \Delta H_{rxn} = \rm q_{rxn}$