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enyata [817]
3 years ago
5

Which combustion equation violates the law of conservation of matter? A) CH4 + 2O2 -> CO2 + 2H2O B) C3H8 + 6O2 -> 3CO2 + 4

H2O C) 2C6H14 + 19O2 -> 12CO2 + 14H2O D) 2C8H18 + 25O2 -> 16CO2 + 18H2O
Chemistry
2 answers:
Dmitriy789 [7]3 years ago
5 0

Answer : The combustion equation violates the law of conservation of matter is, (B) C_3H_8+6O_2\rightarrow 3CO_2+4H_2O

Explanation :

According to the law of conservation of matter, the number of atoms of each elements of reactant side must be equal to the product side in the chemical reaction. Or we can say that the mass of reactant side must be equal to the mass of product side in the chemical reaction.

(A) CH_4+2O_2\rightarrow CO_2+2H_2O

This combustion equation follow the law of conservation of matter.

(B) C_3H_8+6O_2\rightarrow 3CO_2+4H_2O

This combustion equation does not follow the law of conservation of matter because in this reaction, the atoms of carbon and hydrogen elements are equal on both the sides but the atoms of oxygen element on reactant side are not equal to the product side.

(C) 2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

This combustion equation follow the law of conservation of matter.

(D) 2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

This combustion equation follow the law of conservation of matter.

Hence, the combustion equation violates the law of conservation of matter is, equation (B)

tankabanditka [31]3 years ago
4 0

B - C3H8 + 6O2 -> 3CO2 + 4H2O

THis is because Oxygen is not balanced on both sides, thus defying the law of conservation of matter.

On the reactants side of the equation, there are 12 Oxygen atoms, however, on the products side, there are only 10 Oxygen atoms.

Hope this helps!

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HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into
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<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M

<u>Explanation:</u>

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

                       2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>               1.0

<u>At eqllm:</u>        1.0-2x          x           x

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We can determine the empirical formula by first converting each of the grams to moles. remember to do this, first, we need the molar mass of the molecules which can be calculated by adding the mass of the atoms from the periodic table. 

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Hydrogen: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H

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Now let's convert each mass to moles.

C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol

now that we have the moles of each atom, we need to divide them by the smallest value to find the ration. If you do not get the whole number, you need to multiply until to get a whole number.

C:  0.54 mol / 0.18 mol = 3
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empirical formula--> C₃H₃O



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7 grams of H2 is only about 3.5 moles. 
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