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3 years ago

Why is copper (i) sulfate Cu2So4^-2? not copper (ii)?

2 answers:

const2013 [10]3 years ago

5 0

Copper (I) oxidation state is 1 Cu2So4
copper (II) oxidation state is +2 CuSo4
copper (i) also give up one electron so you need two of them to react with the sulfate ion (which has charge of -2)
and also all metallic ions have an multiple oxidation levels corresponding to the number of electrons they can exchange or loose
Hope this helps

ArbitrLikvidat [17]3 years ago

5 0

<h2>Answer</h2>

This is because Copper (I) has +1 charge and Copper (II) has +2 charge.

<h2>Explanation</h2>

The chemical formula of an ionic compound consists of cation first, followed by anion. Both cations and anions are incolved to make compound to be nutral or charge entity for example Cu2SO4 is cuprous compound contain Cu (I) where two Cu give only one electron as

Cu is cation and SO4 is anion

SO4 has a valency of -2

So, Cu2 has a valency of +2

But in Cu (II), The single copper gives two electron to SO4 and form CuSO4 compound named as copper sulphate.

These metalls are called transition metals who can change their oxidation states

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Answer:

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2 years ago

The difference in interchain stability between the polysaccharides glycogen and cellulose is due to: Group of answer choices the

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3 years ago

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3 years ago

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mariarad [96]

The balanced reaction is 3 Ca ( s ) + N 2 ( g ) → Ca 3 N 2 ( s ).

Explanation:

A chemical equation is said to be balanced when the total number of atoms present on the reactants side is equal to the total number of atoms present on the product side.

The unbalanced chemical equation is as follows,

Ca ( s ) + N 2 ( g ) → Ca 3 N 2 ( s )

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Calcium has 1 atom on the reactant and 3 on the products side. To balance the reaction we need to multiply the calcium atom by 3 on the reactants side.

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3 years ago

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

pH = 10.05

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = 0.20

I hope it helps!

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3 years ago