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deff fn [24]
3 years ago
10

Which of the following chemical equations proves that the law of conservation of mass is in effect?

Chemistry
1 answer:
bogdanovich [222]3 years ago
5 0

Answer:

2KCl + F₂   →     2KF + Cl₂

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

2KCl + F₂   →     2KF + Cl₂

In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.

All other options are incorrect because mass is not conserved.

Mg₂ + LiBr ---> LiMg + Br

In this equation mass of magnesium is more on reactant side.

Na +O₂ ---> Na₂O

In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.

H₂O ---> H₂ + O₂

In this equation there is less oxygen on reactant side while more oxygen on product side.

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5 0
3 years ago
Can u please help me with these 2 answers!
Lelu [443]
5 is 2 I’m not sure about 4 though....
5 0
2 years ago
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
A sample of gas at 1.10 atm has a volume of 326 mL. What is the new volume if the pressure is changed to 1.90 atm?
Brut [27]

Answer

For this we use ideal gas equation which is:

P1V1 = P2V2

P1 = 1.10 atm

V1 = 326 ml

P2 = 1.90

V2 = ?

By rearranging the ideal gas equation:

V2 =  P1V1 ÷ P2

V2 = 1.10 × 326 ÷1.90

V2 = 358.6 ÷ 1.90

V2 = 188.7 ml

8 0
3 years ago
What is the major underlying principle of chromatography?.
KatRina [158]
Separation will be achieved if one component adheres to the stationary phase more than the other component does.
6 0
2 years ago
Read 2 more answers
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