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3241004551 [841]
2 years ago
11

How do you convert kilograms to pounds?

Chemistry
1 answer:
Maru [420]2 years ago
4 0

Answer:  (1 Kilogram = 2.20462 pounds) . There are 2.2046226218 lb in 1 kilogram. To convert kilograms to pounds, multiply your figure by 2.205 for an approximate result. 1 kilogram is also equal to 2 lb and 3.27396195 oz. Working out a rough estimate in your head for converting to pounds and ounces may be tricky - remember that there are 16 ounces in a pound.

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What chemical reactions occur in a Nelson's cell? ​
Cerrena [4.2K]

Explanation:

The graphite anodes are suspended into the brine. During electrolysis, Cl ions are oxidized at the anode and chlorine gas goes out of the cell, while sodium ions are reduced at the mercury cathode forming sodium amalgam. ... Hydrogen gas is obtained as a by–product at the cathode.

8 0
1 year ago
When does boiling occur?
butalik [34]

Water boils at 100 Degrees Celsius or 212 degrees Fahrenheit

8 0
3 years ago
The decomposition of mercury (ii) oxide at high temperature, is it an endothermic or exothermic process? Write a chemical reacti
gulaghasi [49]

Answer:

that will be endothermic reaction...... as oxides of mercury decomposes and break s into simpler elements by absorbing energy

Explanation:

hope it helped u buddy

6 0
2 years ago
Read 2 more answers
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
Please explain your answer! <br><br> Thanks!
Yuri [45]
Hello Gary!
*Sorry if I'm late*

Your answer is going to be 65.
Element A (which is actually Zinc) has the atomic number of 65. (I remember having got memorize this also).
Pretty much the explanation is that the number of patrons is equivalent to the element's atomic number!

Hope this helps!
Have a nice day :D
8 0
3 years ago
Read 2 more answers
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