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3 years ago

What are the similarities and differences for the properties of copper (II) sulphate and copper?

1 answer:

ivann1987 [24]3 years ago

3 0

It is to do with the ionisation of the atom. Copper is a metal, so it will lose electrons. When reacted with a non-metal, it will form an ionic bond.

In copper (I) sulphate, the copper ions have a charge of +1, ie they have lost ONE electron each.

Copper (I) sulphate has the chemical formula Cu2SO4. Each ionic bond involves two Cu+1 ions and a sulphate ion (SO4.

In copper (II) sulphate, the copper ions have a charge of +2, ie they have lost TWO electrons each.

Copper (II) sulphate has the chemical formula CuSO4. Each ionic bond involves a single Cu+2 ion and a sulphate ion (SO4).

So, really, it’s down to the chemical structure and the ionisation of the atom. Apart from the chemistry, copper (I) sulphate a very obscure chemical. Although, after a bit of googling, I have managed to find some info and vendors, it appears that this chemical is rarely seen and doesn’t have many practical uses.

Copper (II) sulphate, on the other hand, is incredibly common. It’s in every school chemistry lab. If someone says “copper sulphate” they will be talking about this chemical, not copper (I) sulphate. In pure form, it is a boring white powder, but when hydrated, it takes on it’s better known blue colour, with blue crystals and blue solution.

Hope this helps.

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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction

nekit [7.7K]

Answer : The volume of $NaOH$ required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?$

Putting values in above equation, we get:

$2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL$

Hence, the volume of $NaOH$ required to neutralize is, 340 mL

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4 years ago

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3 years ago

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3 years ago

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma

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Answer:

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Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

$\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})$

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

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4 years ago

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