Chemistry, 19.02.2020 08:54 springcoates

Separation of mixture by gravity of sodium carbonat into calcium chloride

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By combining aqueous solutions of calcium chloride and sodium carbonate, a solid calcium carbonate is produced. The CaCO 3 is trapped by the paper which forms as a precipitate, but the NaCl passes through the filter paper with the water.

Explanation:

In gravity separation, a combination of two incompatible liquids can be separated by using a separating funnel. The operation is based on the variation within the density of the liquids. The heavier liquid that settles down is drained out first from the funnel into the beaker, and then the lighter liquid is drained out into another beaker. Filtration is used to separate a substance from a mixture because one is insoluble and the other is soluble. By combining aqueous solutions of calcium chloride and sodium carbonate, solid calcium carbonate is produced.

CaCl 2 (aq) + Na 2 CO 3 (aq) → CaCO 3 (s) + 2NaCl(aq)

A filter paper is inserted in a funnel and the mixture may be poured through a funnel. The NaCl is passed through the filter paper with the water but CaCO3 is trapped by the paper which forms the precipitate.

Answer from: Quest

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

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.