I think the answer is transparent and translucent
<span>NaCl is poster-compound for ionic bonding. The bonds in NaCl have about 70% ionic character, making the bond highly polar. its overstatement to state that there is actual ion in NaCl with +1 and -1 charge but actual charge of Na and Cl is +1 and -1 ion, since Nacl exist as a network of highly charged particle and not discrete molecule, NaCl particle does not exhibit intermolecular forces.
Water molecule on other hand exhibit London dispersion force, keesom force, and hydrogen bonding.
The polar water molecule are attracted to the polarized Na and Cl atoms. This is what allow NaCl(s) to dissolve and ionize in water. Therefore type of attraction in NaCl is ion-dipole attraction.</span>
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This is all i know
First of all, let's write the equation of the reaction.
CH₃COOH + NaOH ⇒ CH₃COONa + H₂O
The formula to be used here is CaVa/CbVb = na/nb
where Ca is the concentration of the acid (?)
Cb is the concentration of the base (0.175M)
Va is the volume of the acid (25 mL)
Vb is the volume of the base (37.5 mL)
na is the number of moles of acid (1)
nb is the number of moles of base (1)
Ca × 25/0.175 × 37.5 = 1/1
Ca = 0.175 × 37.5 × 1 /25 ×1
Ca = 0.263 M
The concentration of the acid in the sample was 0.263 M