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3 years ago

Which substance is a compound iron or nitrogen oxide

Chemistry

2 answers:

den301095 [7]3 years ago

6 0

Answer:

Nitrogen oxide is compound.

Explanation:

Compound is defined as chemical substance made up two or more than two different elements. For example : NaOH is the compound made up of 1 sodium atom, 1 oxygen atom and 1 hydrogen atom.

Now, according to the question, out iron and nitrogen oxide. Nitrogen dioxide (NO)is a compound because its made up of one nitrogen atom and 1 oxygen atom.

Where as iron is not a compound because it is just composed of iron atoms only.

Zanzabum3 years ago

4 0

I think it might be Nitrogen dioxide, but please check behind

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The melting point of the alkali metals (increased/decreases) from lithium to francium

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3 years ago

What mass (in mg) does 2.63 moles of nickel have? 3.56 × 105 mg 1.54 × 105 mg 2.23 × 104 mg 129 mg?

jarptica [38.1K]

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3 years ago

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kobusy [5.1K]

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4 0

4 years ago

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This type of fatty acid has at least one double bond in the carbon-to-carbon chain and also contains protective qualities

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8 0

3 years ago

The equilibrium constant for the reaction 2NO(g)+Br2(g)⥫⥬==2NOBr(g) is Kc=1.3×10−2 at 1000 K. At this temperature does the equil

o-na [289]

Answer :

The equilibrium favors NO and $Br_2$ .

(1) The value of equilibrium constant for this reaction is, 76.9

(2) The value of equilibrium constant for this reaction is, 8.77

Explanation:

The given chemical equation is:

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

The value of equilibrium constant for the above equation is $K_c=1.3\times 10^{-2}$ .

The value of $K_c$ that means equilibrium lies to the left side. Thus, the equilibrium favors NO and $Br_2$ .

We need to calculate the equilibrium constant for the given equation of above chemical equation, which is:

(1) $2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_1}=\frac{1}{K_c}$

$K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9$

Thus, the value of equilibrium constant for this reaction is, 76.9

(2) $NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_2}=(\frac{1}{K_c})^{1/2}$

$K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77$

Thus, the value of equilibrium constant for this reaction is, 8.77

7 0

3 years ago

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