<h3>
Answer:</h3>
Anion present- Iodide ion (I⁻)
Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)
<h3>
Explanation:</h3>
In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.
Additionally we need to know the color of the precipitates.
Some of insoluble salts of silver and their color include;
- Silver chloride (AgCl) - white color
- Silver bromide (AgBr)- Pale cream color
- Silver Iodide (AgI) - Yellow color
- Silver hydroxide (Ag(OH)- Brown color
With that information we can identify the precipitate of silver formed and identify the anion present in the sample.
- The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
- Therefore, the anion that was present in the sample was iodide ion (I⁻).
- Thus, the corresponding net ionic equation will be;
Ag⁺(aq) + I⁻(aq) → AgI(s)
Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
Answer:
600K
Explanation:
PV=nRT
T=PV/nR
= 1.6atm* 15.0L/ 0.5mol*0.0821LatmK^-1mol^-1
=600K
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Decay is a type of degradation reaction and thus is considered a first order reaction. thus the formula goes like this.
rate constant= 0.693/half life
so here...
rate constant= 0.693/1620 year^-1
