Chemistry, 18.09.2019 08:50 jm00120

The mixing of which pair of reactants will result in a precipitation reaction? the mixing of which pair of reactants will result in a precipitation reaction? nh4br(aq) + nh4i(aq) naclo4(aq) + (nh4)2s(aq) ni(no3)2(aq) + na2co3(aq) k2so4(aq) + cu(no3)2(aq)

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NH4Br and NH4I are both soluble in water, and they don't react with each other. So this pair will not form precipitation.

NaClO4 and (NH4)2S are both soluble in water. While in water, they dissociate into Na+, ClO4-, NH4+ and S2- ions, which don't react with each other. So this pair will not form precipitation.

Ni(NO3)2 and Na2CO3 are both soluble in water. When the 2 solutions combine, NiCO3 forms. NiCO3 is not soluble in water, so it will precipitate out from the solution. This pair will result in precipitation.

K2SO4 and Cu(NO3)2 are both soluble in water. While in water, they dissociate into K+, SO42-, Cu2+ and NO3- ions, which don't react with each other. So this pair will not form precipitation.

Answer from: natcooper6204

Reaction between Ni(NO₃)₂ (aq) + Na₂CO₃ (aq) will result in a precipitation reaction

Further explanation

There are two types of chemical reactions that may occur.

namely single-replacement reactions and double-replacement reactions.

Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

To predict whether this reaction can occur or not is one of them, the precipitation reaction. A precipitation reaction occurs if two ionic compounds which are dissolved reacted to produce one of the products of the ion compound does not dissolve. Formation of these precipitating compounds that cause reactions can occur

Solubility Rules:

1. soluble compound

All compounds of Li +, Na +, K +, Rb +, Cs +, and NH4 +

All compounds of NO₃⁻ and C₂H₃O₂⁻

Compounds of Cl−, Br−, I− except Ag⁺, Hg₂²⁺, Pb²⁺

Compounds of SO₄²⁻ except Hg₂²⁺, Pb²⁺, Sr²⁺, Ba²⁺

2. insoluble compounds

Compounds of CO₃²⁻ and PO₄³⁻ except for Compounds of Li +, Na +, K +, Rb +, Cs +, and NH₄ +

Compounds of OH− except Compounds of Li +, Na +, K +, Rb +, Cs +, NH₄⁺, Sr²⁺, and Ba²⁺

We use the Solubility Rules to determine whether a precipitation reaction will occur or not

NH₄Br and NH₄I, have the same cation, so there will be no reaction between the two (Double-Replacement reactions occur if there is an ion exchange between two ion compounds in the reactant)

The reaction of NaClO₄ (aq) + (NH₄)₂S (aq) ---> Na₂S (aq)+ NH₄ClO₄(aq)

Both products include soluble compounds

Ni(NO₃)₂ (aq) + Na₂CO₃ (aq) ---> NiCO₃(s)+ NaNO₃ (aq)

NiCO₃ is an insoluble compound

K₂SO₄ (aq) + Cu(NO₃)₂ (aq) ---> KNO₃ + CuSO₄

Both products include soluble compounds

Learn more

A double-replacement equation

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the reducing agent

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substance loses electrons in a chemical reaction

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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 :

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

Answer from: Quest

The answers is bad germs in their

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