What happens to the copper sulphate crystals when they are heated that causes them to change color?

Chemistry

1 answer:

mote1985 [20]3 years ago

7 0

Answer:

Heating the CuSO4. 5H2O crystals causes then to loose the water of crystallisation that is the 5H2O part. It becomes anhydrous copper sulphate. Its colour changes to white from blue.

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By titration, it is found that 47.5 mL of 0.146 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentrati

OleMash [197]

Answer:

The correct answer is 0.277 M HCl

Explanation:

NaOH is a strong base and HCl is a strong acid. NaOH reacts with HCl to form a salt (NaCl) and water via a neutralization reaction, as follows:

NaOH + HCl → NaCl + H₂O

At the equivalence point, the moles of NaOH added is equal to the moles of HCl present in the solution to be titrated. Moreover, the moles can be calculated as the product of the concentration of base or acid (Cb or Ca) and the volume (Vb and Va), as follows:

moles NaOH = moles HCl

Cb x Vb = Ca x Va

Thus, we calculate the concentration of HCl (Ca), as follows:

Ca = (Cb x Vb)/Va = (0.146 M x 47.5 mL)/(25.0 mL)= 0.277 M

6 0

3 years ago

When H2SO4 is added to PbI2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If

Nookie1986 [14]

Answer:

$n_{I^-}=3.11x10^{-4}molI^-$

Explanation:

Hello,

In this case, the undergoing chemical reaction is shown below:

$H_2SO_4(aq)+PbI_2(aq)\rightarrow PbSO_4(s)+2HI(aq)$

Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:

$n_{I^-}=0.0471gPbSO_4*\frac{1molPbSO_4}{303.26gPbSO_4}*\frac{1molPbI_2}{1molPbSO_4}*\frac{2molI^-}{1molPbI_2} \\n_{I^-}=3.11x10^{-4}molI^-$