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Blizzard [7]
2 years ago
14

Which net ionic equation best represents the reaction that occurs when an aqueous solution of barium chloride is mixed with an a

queous solution of lithium sulfate?
Chemistry
1 answer:
vovangra [49]2 years ago
6 0

Answer : The net ionic equation will be:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The balanced molecular equation will be,

BaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+BaSO_4(s)

The complete ionic equation in separated aqueous solution will be,

Ba^{2+}(aq)+2Cl^-(aq)+2Li^+(aq)+SO_4^{2-}(aq)\rightarrow 2Li^+(aq)+2Cl^-(aq)+BaSO_4(s)

In this equation the species, Li^+\text{ and }Cl^{-} are the spectator ions.

By removing the spectator ions , we get the net ionic equation.

The net ionic equation will be:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

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Which of the following are cations? Check all that apply.
Lostsunrise [7]

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a

barium

b

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e

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Explanation:

4 0
3 years ago
A 2.17 gm sample barium reacted completely with water what is the equation for the reaction how many milliliters of dry H2 evole
vaieri [72.5K]

Answer:

400 mL

Explanation:

Given data:

Mass of barium = 2.17 g

Pressure = 748 mmHg (748/760 = 0.98 atm)

Temperature = 21 °C ( 273+ 21 = 294k)

Milliliters of H₂ evolved = ?

Solution:

chemical equation:

Ba + 2H₂O →  Ba(OH)₂ + H₂

Number of moles of barium:

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Now we  will compare the moles of barium with H₂.

                       Ba        :       H₂

                         1         :         1

                  0.016        :     0.016

Milliliters of H₂:

PV = nRT

V = nRT/P

V = 0.016 mol ×  0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm

V = 0.39 atm. L/0.98 atm

V = 0.4 L

L to mL

0.4 × 1000 = 400 mL

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