Chemistry, 17.12.2019 19:31 eamccoy1

Which equation represents a conservation of atoms? group of answer choices latex: 4fe\: +3o_2\: \longrightarrow\: 2fe_2o_3 4 f e + 3 o 2 ⟶ 2 f e 2 o 3 latex: 2fe\: +2o_2\: \longrightarrow\: fe_2o_3 2 f e + 2 o 2 ⟶ f e 2 o 3 latex: 4fe\: +2o_2\: \longrightarrow\: 2fe_2o_3 4 f e + 2 o 2 ⟶ 2 f e 2 o 3 latex: 2fe\: +\: 3o_2\: \longrightarrow\: fe_2o_3

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Answer from: jailenevazquez755

Answer : The equation represents a conservation of atoms is:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Explanation :

Law of conservation of matter : It states that matter can neither be created nor be destroyed but it can only be transformed from one form to another form.

The balanced chemical reaction always follow the law of conservation of matter.

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

(1) The given chemical reaction is,

$4Fe+3O_2\rightarrow 2Fe_2O_3$

This reaction is a balanced chemical reaction in which number of atoms of iron and oxygen are equal on the both side of the reaction. So, this reaction obey the law of conservation of matter.

(2) The given chemical reaction is,

$2Fe+2O_2\rightarrow 2Fe_2O_3$

This reaction is an unbalanced chemical reaction in which number of atoms of iron and oxygen are not equal on the both side of the reaction. So, this reaction does not obey the law of conservation of matter.

(3) The given chemical reaction is,

$4Fe+2O_2\rightarrow 2Fe_2O_3$

This reaction is an unbalanced chemical reaction in which number of atoms of oxygen are not equal on the both side of the reaction. So, this reaction does not obey the law of conservation of matter.

Hence, the equation represents a conservation of atoms is:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

Idon't know hope someone you

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