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3 years ago

The nature of zinc powder and Cobalt (II )oxide is heated the following reaction occurs ;

1 answer:

8 0

1) The metal which reduces the other compound is the one higher in the reactivity. So in this case it is $\mathrm{Zn}$ .

2) The substance which brings about reduction while itself getting oxidised (that is losing electrons) is called a reducing agent. Here, $\mathrm{Zn}$ is the reducing agent and reduces Cobalt Oxide to Cobalt while itself getting oxidised to Zinc oxide.

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(GIVING BRAINLIEST!!!)What is the correct formula for the ionic compound that forms when oppositely- charged ions of magnesium a

allochka39001 [22]

Pretty confident that MgO is the answer

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A base

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When excess potassium hydroxide and H2SeO3 react whats the net ionic reaction for the acid-base rxn

Bad White [126]

The required net ionic equation is; 2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

The molecular reaction equation is;

H2SeO3(aq) + 2KOH(aq) -----> K2SeO3(aq) + 2H2O(l)

The complete ionic equation is;

2H^+(aq) + SeO3^2-(aq) + 2K^+(aq) + 2OH^-(aq)-----> 2K^+(aq) + SeO3^2-(aq) + 2H2O(l)

Net ionic equation;

2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

We can clearly see that this is a neutralization reaction hence water is the product of the net ionic equation.

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4 0

2 years ago

(only reply if you know the answer or get reported) Please solve this ty

kolezko [41]

Answer:

7) 50 8)4

Explanation:

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4 0

3 years ago

Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1

timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0

3 years ago