Answer:
b. double bonds, triple bonds, carbon atoms,and hydrogen atoms.
Explanation:
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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^-.
Strong acids, strong bases, and salts are strong electrolyte.
Answer:
Explanation:
If an antacid has sodium hydrogen carbonate/Calcium carbonate, it reacts with HCl as shown
NaHCO₃+ HCl → NaCl + CO₂+ H₂O
Antacid acid salt gas water
CaCO₃+ 2HCl → CaCl₂+ CO₂+ H₂O
Antacid acid salt gas water
The formation of gas CO₂ is shown by brisk effervescence when the antacid (sodium hydrogen carbonate/calcium carbonate) reacts with HCl (acid). So CO₂ is the additional product formed and its formation is supported by observation of brisk effervescence as HCl is added to the antacid.
It easier to remove electrons from a large element(bottom of the periodic table) because there further away from the nucleus.
