Answer:
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Explanation:
Complete question
Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation
Equilibrium equation between the undissociated acid and the dissociated ions
HF(aq)⇌H+(aq)+F−(aq)
Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−
NaOH(aq)→Na+(aq)+OH−(aq)
Hydroxide anions and the hydrogen cations will neutralize each other to produce water.
H+(aq)+OH−(aq)→H2O(l)
On combining both the equation, we get –
HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)
The Final equation is
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:
<h3>Van't Hoff factor:</h3>
In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.
In such a way, for calcium iodide, we write its ionization equation as shown below:

Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.
Learn more about the van't Hoff factor: brainly.com/question/23764376
The answer is (4) amino acid. This molecule has one carboxyl and one amidogen linked at the same carbon atom. This is the property of amino acid. So this is an amino acid.
Answer:
thick, insulating fur
Explanation:
If an animal lives in a freezing climate, it makes sense logically that the animal would adapt and develop a layer of thick fur to keep its body insulated and maintain homeostasis.
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Answer:
Iodine
Explanation:
It's in the same group as chlorine.