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Aleksandr-060686 [28]
3 years ago
5

Hydrogen peroxide, H2O2, is common in many households and is used as a bleaching agent. It usually comes in a dark, opaque bottl

e. When exposed to heat or light, hydrogen peroxide will decompose according to the reaction:
2 H2O2→2 H2O + O2

Iron Nitrate has been shown to be an effective catalyst to accelerate the decomposition process, and sodium phosphate can inhibit the effects of this catalytic effect.

Naturally, the consumer does not want their hydrogen peroxide to decompose. Consider the factors affecting reaction rate that you learned about in this lesson. What should the consumer should do to prevent the decomposition of hydrogen peroxide?. For example, should the hydrogen peroxide be stored in a warm or cool location? Use scientific reasoning to support your response.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
8 0

Answer:

Hydrogen peroxide should be stored in

1) a cool environment

2) with amber bottles away from sunlight

3) with little drops of sodium phosphate

Explanation:

It has been confirmed that heat and light aids in the decomposition of hydrogen peroxide according to the equation; 2H2O2→2 H2O + O2.

This means that hydrogen peroxide must be stored in a cool place. This will reduce its rate of decomposition. Secondly, it should be stored in amber bottles away from light since light also aids in its decomposition.

Thirdly, drops of sodium phosphate may be added to prevent its catalytic decomposition during storage.

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How much energy would be released as an electron moved from the n=4 to the n=3 energy level?
Thepotemich [5.8K]

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

                                     E_n =  -13.6/n^{2}  eV

where n is the energy level

Therefore,

                   E_4 = -13.6/4^{2} \\E_3  = -13.6/3^{2}

Thus, E_4  =   -0.85eV

         E_3  =  -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

                                       E_4 -E_3

                                   =  -0.85 - ( -1.51)

                                   = 0.66eV

To know more about energy released in electron transition

brainly.com/question/8384785

#SPJ4

8 0
2 years ago
Which of the following is unchanged at the end of the CNO cycle?
Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon (_{6}^{12}\textrm{C}), nitrogen isotope (_{7}^{13}\textrm{N}), and oxygen isotope (_{8}^{15}\textrm{O}), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon (_{6}^{12}\textrm{C}) nucleus in the last step.

_{6}^{12}\textrm{C} → _{7}^{13}\textrm{N} → _{6}^{13}\textrm{C} → _{7}^{14}\textrm{N} → _{8}^{15}\textrm{O} → _{7}^{15}\textrm{N} → _{6}^{12}\textrm{C}

3 0
3 years ago
Scientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below. A pur
Julli [10]

Answer:

dude that doesnt maek sense

Explanation:

6 0
3 years ago
An atom that normally has __________ electrons in its outer (valence) shell tends not to form chemical bonds with other atoms.
goblinko [34]
8 valence shell electrons.
5 0
3 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
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