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3 years ago

A beaker contains 20.0 ml of a 50.0 g/L solution. What will the new concentration be if you add 40.0 mL of water?

Chemistry

1 answer:

Ronch [10]3 years ago

8 0

Answer:

moles SO42- = 0.0500 L x 0.20 M=0.010

moles Ba2+ = 0.0500 L x 0.10 M = 0.0050

Ba2+ + SO42- = BaSO4 (s)

moles SO42- in excess = 0.010 - 0.0050=0.0050

total volume = 100 mL = 0.100 L

[SO42-]= 0.0050/0.100= 0.050 M

[Na2SO4] = 2 /2 = 1 M

moles Na2SO4 = 2 M x 0.500 L = 0.500

mass Na2SO4 = 0.500 x 142 g/mol=71.0 g

moles MgBr2 = 46 /184 =0.25

moles Br- = 0.25 x 2 = 0.50

[Br-]= 0.50 / 0.50 L = 1 M

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AveGali [126]

Answer: The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.

Explanation:

The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

The net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:

$Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(s)$

The ionic equation follows:

$Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow PbS(s)+2Na^+(aq)+2NO_3^-(aq)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+S^{2-}(aq)\rightarrow PbS(s)$

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