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3 years ago

Write the net ionic reaction for the following double replacement reaction:

Chemistry

1 answer:

Schach [20]3 years ago

6 0

Net ionic equation

Cu²⁺(aq)+S²⁻(aq)⇒CuS(s)

<h3>Further explanation</h3>

Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

Reaction

CuSO₄(aq)+Na₂S(aq)⇒CuS(s)+Na₂SO₄

ionic equation

Cu²⁺(aq)+SO₄²⁻(aq)+2Na⁺(aq)+S²⁻(aq)⇒CuS(s)+2Na⁺(aq+SO₄²⁻(aq)

spectator ions : 2Na⁺ and SO₄²⁻

Net ionic equation

Cu²⁺(aq)+S²⁻(aq)⇒CuS(s)

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The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th

Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $T_2$

$K_1$ = rate of reaction at $T_1$

$Ea$ = activation energy of the reaction

R = gas constant = 8.314 J/K mol

$E_a=300 kJ/mol=300,000 J/mol$

$T_2=625^oC=898.15 K,T_1=525^oC=798.15 K$

$\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]$

$\log (\frac{K_2}{K_1})=2.185666$

$K_2=153.344\times K_1$

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

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3 years ago

Which transformation represents light energy transforming into chemical energy?

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3 years ago

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3 years ago

Which formula is an empirical formula?

poizon [28]

<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. $NH_{3}$

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_{2}O_{4}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_{2}O_{4}=2(NO_{2})$

B. $NH_{3}$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_{3}H_{6}$