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marishachu [46]
2 years ago
6

List the spectator ions in the following reaction.

Chemistry
1 answer:
Rudik [331]2 years ago
4 0

Answer:

2Na⁺ (aq) and 2OH⁻(aq)

Explanation:

Spectator ions:

Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.

Given ionic equation:

Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)

In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq)  net ionic equation can be written as,

Net ionic equation:

Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)

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Mass of 2×10^21 number of atoms of an element is 0.4g. What is the mass of 0.5 mole of the element?
alexandr402 [8]

Answer:

66.7 g

Explanation:

Number of atoms = 2×10²¹

Mass of 2×10²¹ atoms = 0.4 g

Mass of 0.5 moles of that element = ?

Solution:

1 mole contain 6.022×10²³ atoms

2×10²¹ atoms × 1 mol / 6.022×10²³ atoms

0.33×10⁻² mol

0.003 mol

 0.003 mole have mass of 0.4 g

0.5 mol have mass 0.5/0.003×0.4 g = 66.7 g

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2 years ago
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Dimas [21]

Answer:

Not balance.

Explanation:

Chemical equation:

K + Cl₂    →    KCl

The given equation is not balance because there are one potassium and two chlorine atoms on left side of equation while on right side there are one potassium and one chlorine atom present.

Balance chemical equation:

2K + Cl₂    →    2KCl

the equation is balance now because there are two potassium and two chlorine atoms on left side of equation and two potassium and two chlorine atoms are also present on right side.

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The reaction of 2-chloropropane with sodium hydroxide can occur via both SN1 and SN2 mechanisms.
strojnjashka [21]

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Explanation:

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To summarize, this mechanism takes places in two separate steps. The mechanism is attached below.

Part 2: S_N2 reaction. This is a nucleophilic substitution reaction in which we have one step. Our nucleophile, hydroxide, attacks the carbon and then chlorine leaves simultaneously without an intermediate carbocation being formed.

The mechanism is attached as well.

(b) The rate determining step is the slow step. Formation of the carbocation has the greatest activation energy, so this is our rate determining step for S_N1. For S_N2, we only have one step, so the rate determining step is the attack of the nucleophile and the loss of the leaving group.

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Minchanka [31]

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Explanation:

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3 years ago
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