Chemistry, 20.12.2020 01:30 odinmorton

The first part of a balance chemical with Michelle and see you plus 2AGNO3 in order for this equation to be balanced how many oxygen atoms must be present in the products

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Answer from: 100154281

the number of oxygen atoms that must be present in the products is 6 oxygen atoms.

Explanation:

Chemical equations can be defined as a way of summarizing a chemical reaction in symbols and formulas . In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. This obeys the law of conservation of energy as matter cannot be created nor destroyed.

When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced.

A simple example is illustrated below:

The chemical reaction between chlorine atom and hydrogen which are diatomic

H2 + Cl2 > HCl

(Reactants)> (products)

In describing the equation, there are two atoms of hydrogen reacting with two atoms of chlorine to give one molecule of hydrogen chloride containing one atom each of the reactants. We can fix this by including the coefficient 2 on the product side:

H2 + Cl2 > 2HCl

Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.

Therefore, the number of oxygen atoms that must be present in the products is 6 oxygen atoms in order for the equation to be balanced.

Answer from: Quest

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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