Chemistry, 24.09.2019 04:30 rachel693

What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed? pbk2 pbso4 h2o none of the above k(no3)2?

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The formula of the precipitate that will be formed is PbSO4.
When lead ll nitrate react with potassium sulphate, one of the product formed is lead sulphate. The reaction is a double decomposition precipitation reaction; PbSO4 is the precipitate formed.

Answer from: jrfranckowiak

$\rm PbSO_4$ will precipitate in the given reaction.

Explanation:

The reaction takes place between lead nitrate $\rm Pb(NO_3)_2$ and potassium sulphate $\rm K_2SO_4$ .

Now, the reaction will be a double displacement reaction in which the ions will exchange between lead nitrate $\rm Pb(NO_3)_2$ and potassium sulphate $\rm K_2SO_4$ .

So, the chemical reaction can be written as,

$\rm {Pb(NO_3)_2+K_2SO_4}\rightarrow PbSO_4+2KNO_3$

In the above equation, $\rm PbSO_4$ is a precipitate which will be settled down in the bottom and $\rm KNO_3$ will form an aqueous solution.

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Answer from: Quest

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

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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