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xz_007 [3.2K]
3 years ago
14

Explain why crude oil is a fossil fuel​

Chemistry
2 answers:
Step2247 [10]3 years ago
6 0
Cause crude oil is a fossil fuel which is also yo mama
Alecsey [184]3 years ago
5 0
They were formed from fossils
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BRAINLIEST AND POINTS!!
umka21 [38]

Answer:

There are several ways that scientists communicate our results, including written reports and scientific journal publications, and by giving presentations to our colleagues and the public. One popular venue for scientists to present to colleagues is at scientific conferences.

Explanation:

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3 years ago
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4. Identify the different parts to this word problem. A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is
Dafna1 [17]

<u>Given:</u>

Mass of the ball = 35.2 kg

Momentum =  218 kg m/s

<u>To determine:</u>

The velocity of the ball

<u>Explanation:</u>

Momentum (p) of an object is the product of its mass  (m) and velocity (v)

p = m*v

v = p/m = 218 kg.ms-1/35.2 kg = 6.19 m/s

Ans: The velocity of the ball is 6.19 m/s

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3 years ago
Help me out plz, I will mark you the brainiest.
liberstina [14]

Answer : The correct option is (B).

Explanation :

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The given compound has 4 carbon atoms and also saturated compound because there is no double or triple bond present.

As we know that there are two structural isomers of butane.

(1) n-butane

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Thus, the correct option is (B).

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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

       

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3 years ago
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Answer:

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