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Nesterboy [21]
3 years ago
11

Please help!!!! will mark as brainliest!!!!

Chemistry
1 answer:
nordsb [41]3 years ago
5 0

c because it acts as it its the spine of a flower to help keep it up


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Use atomic properties to explain the reduction of a less active metal by a more active one:(b) in the molten state. Give a speci
Bad White [126]

The reduction of a less active metal by a more active one is called metal displacement reactions. For example:

Fe + CuSO4 → FeSO4 + Cu

<h3>What is metal displacement reaction? </h3>

Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.

As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.

Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.

Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:

Fe + CuSO4 → FeSO4 + Cu

learn more about metal displacement reaction:

brainly.com/question/11777638

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Solid waste is eliminated from the body through the esophagus agree or disagree
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Which substance can be broken down by chemical means?
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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
Brilliant_brown [7]

Answer :  The rate of reaction is,

Rate=4.77\times 10^{-19}M/s

The appearance of NO_3 is, 4.77\times 10^{-19}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)

The rate law expression will be:

Rate=k[NO_2][O_3]

Given:

Rate constant = k=1.69\times 10^{-4}M^{-1}s^{-1}

[NO_2] = 1.77\times 10^{-8}M

[O_3] = 1.59\times 10^{-7}M

Rate=k[NO_2][O_3]

Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})

Rate=4.77\times 10^{-19}M/s

The expression for rate of appearance of NO_3 :

\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}

As, \text{Rate of reaction}=4.77\times 10^{-19}M/s

So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

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