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1 year ago

Which of the following does NOT demonstrate the law of conservation of matter?

Chemistry

1 answer:

Brrunno [24]1 year ago

8 0

The one that does not demonstrates the law of conservation of matter would be the third option: $2NO_2 + H_2O -- > HNO_3 + HNO_2$

<h3>What is law of conservation of matter? </h3>

It is a law that explains that matters are conserved during the course of chemical reactions. They can neither be destroyed nor created.

Thus, the number of moles of the species in a reaction remains the same before and after the reaction.

Thus, an equation that demonstrates the law of conservation of matter will be balanced in terms of the number of moles of species.

The only equation, in this case, is $2NO_2 + H_2O -- > HNO_3 + HNO_2$

There are 3 oxygen species on the reactant while 5 are present on the product side.

More on the law of conservation of matter can be found here: brainly.com/question/9434062

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At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

3 0

3 years ago

What compounds are classified as electrolytes?

Kitty [74]

1- KNO3 and H2SO4. electrolyte is a liquid that consists of ions, which decomposes during the process of electrolysis, electrolytes will dissolve in liquids, like water. Since KNO3 and H2SO4 both dissolve in water, they are the right answer

5 0

2 years ago

Find the molar enthalpy of formation for paraffin wax (C2H526)) given the following reaction

pogonyaev

Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ

Explanation:

The balanced chemical reaction is,

$C_{25}H_{52}(g)+38O_2(g)\rightarrow 25CO_2(g)+26H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_{25}H_{52}}\times \Delta H_{C_{25}H_{52}})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})]$

$\Delta H_{C_{25}H_{52}}=-2460.5kJ/mol$

Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ

3 0

2 years ago

A common car battery consists of six identical cells each of which carries out the reaction: Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 +

docker41 [41]

Answer:

The correct answer is 4.58 grams.

Explanation:

Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).

For the reaction, n * 96500 C = molar mass

1C = molar mass/n*96500 = Equivalent wt / 96500

w = Equivalent wt / 96500 * I * t

In the given reaction,

Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.

Now putting the values in the equation we get,

w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)

w = 4.58 gm.

6 0

3 years ago

What would a bond between potassium and chlorine be

Ivenika [448]

Answer:Potassium react with elemental chlorine, an electron transfers from a potassium atom to a chlorine atom, forming potassium ion and chloride ion. ... As the bond formation between ions, the bond will be ionic. This is because, this process of transfer of electrons from one atom (K) to another atom (Cl).

Explanation:

4 0

2 years ago