H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Hope that helps!

4 0

3 years ago

125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an init

Sveta_85 [38]

The heat released by the substance in the calorimeter is equal to the heat absorbed by water which results to the decrease and increase in temperature, respectively.
We use m Cp ΔT to balance the heat involved

(m Cp ΔT) subs in calorimeter = (m Cp ΔT) water
125 g * Cp * (97.0-23.5 ) C = 250 g *(4.18 J/C g)* (23.5-20)
Cp = 0.398 J/Cg

Answer is B

4 0

3 years ago

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