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Ahat [919]
3 years ago
14

What is always true of an object with a lit of mass ?

Chemistry
2 answers:
yawa3891 [41]3 years ago
4 0
A greater force is required
sertanlavr [38]3 years ago
3 0

Such an object makes a larger dent in the fabric of space-time than an object with little mass. (It has a greater gravitational attraction than less massive objects)

A greater force is required to accelerate such an object than a less massive object

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A rectangular solid has the following measurements 6 cm, 10 cm, and 5 cm. what is the volume of the solid?
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300cm cubed ( have a great night! )
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When a pot of water has been boiling vigorously for several minutes, the bubbles that emerge from the liquid water are composed
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Well, water is composed H2 gas and O2 gas in a 2:1 ratio, and the bubbles that form at the bottom of the pan is just the water changing from a liquid to a gas, and im pretty sure its water vapor that comes out of the pan. I hope that helped??? :)
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✞Why is propane stored in household tanks but natural gas is not?♡♡♡​
Klio2033 [76]

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In order to keep propane a liquid at room temperature (70° F or 21° C), it has to be held in a tank at a pressure of about 850 kPa. ... Household metal tanks cannot withstand this pressure. In short, natural gas is not stored in household tanks because the symmetry of its molecule makes it hard to liquify.

4 0
3 years ago
If an atom of Bromine (Br) becomes an ion, which of the following does it most likely form?
lawyer [7]

Answer:

An atom of Bromine (Br) forms an ion and becomes Br⁻

Explanation:

  • Atoms of elements gain or lose electron(s) to attain a stable configuration and form ions.
  • When an atom gains electron(s) it forms a negatively charged ion called an anion.
  • For example, Bromine is a halogen and its atom requires to gain one electron to attain stability and form a bromine ion (Br⁻).
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5 0
3 years ago
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
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