Answer:
H2(g)+I2(s)→2HI(s)
Explanation:
Hello there!
In this case, according to the given information and unbalanced chemical reaction, we infer it must be balanced in agreement with the law of conservation of mass because the reactants side has two hydrogen and iodine atoms whereas the products side has just one. In such a way, by placing a 2 on HI, we obtain the following balanced reaction:
H2(g)+I2(s)→2HI(s)
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<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:
Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:
Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:
By plugging in the moles and molarity, we obtain:
Which in mL is:
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Answer:
hmmm
Explanation:
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