Consider the total ionic equation below.

Chemistry

1 answer:

LenaWriter [7]3 years ago

7 0

Answer:

The spectator ions in this equation are :

CrO²⁻₄ and Ba²⁺

Explanation:

Spectator ions : These are the ions which are present in the solution but do not take part in chemical reaction.They are present as such in both reactant as well as product .

In this reaction,

1.H+ and OH- are combined and converted into water (H2O) : Changed to H2O

So , H+ and OH- are not Spectator ions

2.CrO²⁻₄ and Ba²⁺ are unchanged . Present as such in product and reactants

CrO²⁻₄ and Ba²⁺are Spectator ions

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Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$