You need to have more information, please list the
"following".
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Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.
What do you mean? Is this for anyone stuck on the same question cause I don’t think that’s how it works lol
Answer:
340 J·K⁻¹mol⁻¹
Explanation:
I looks like the standard entropy increases approximately linearly with the number of C atoms.
I plotted S° vs the number of C atoms and got the graph shown below.
It appears that S° for four carbon atoms should be about 340 J·K⁻¹mol⁻¹.·
Answer:
CaSO₄ (calcium suflate) is the precipitate formed
Explanation:
We can think the reactants:
BeSO₄ → Beryllium sulfate
Ca(OH)₂ → Calcium hydroxide
The reaction is:
Be₂SO₄ + Ca(OH)₂ → CaSO₄ ↓ + Be(OH)₂
We call it as a double-replacement reaction because two ions exchange places from 2 compounds to form two new compounds.
Sulfates can always make precipitate with the elements from group 2, Ca, Ba and Mg.
Hydroxides from group 2 are solubles, so we complete states:
BeSO₄ (aq) + Ca(OH)₂ (aq) → CaSO₄ ↓ (s) + Be(OH)₂ (aq)
