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Yes,Na represents an element. It is the symbol for sodium.Its atomic number is 11.

Na stands for sodium. It locates in the 3rd period because it has electrons in three different energy levels. It is an alkali metal as it has only one electron in outermost shell.

Yes, Na is an element in the 1st group. It is an alkaline metal. It is highly reactive.

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What is the name of the compound with the formula Nai?

NaI = Sodium iodide


How many grams of NaI would be used to produce a 2.0 M solution with a volume of 1.00 L?

To calculate the grams of NaI needed to make a 2.0 M solution with a volume of 1.00 L, you first need to determine the molar mass of NaI, which is approximately 149.89 g/mol. Using the formula Molarity (M) = moles of solute / liters of solution, you can rearrange the equation to solve for moles of solute, which in this case is 2.0 mol. Finally, you can convert moles of NaI to grams by multiplying by the molar mass, resulting in approximately 299.78 grams of NaI needed.


What volume of 0.380 M NaI would be required to oxidize 45.0 ml of 0.500 M KMnO4 in acidic solution?

To determine the volume of 0.380 M NaI needed to oxidize 45.0 mL of 0.500 M KMnO4 in acidic solution, we first calculate the moles of KMnO4: [ \text{Moles of KMnO4} = 0.500 , \text{M} \times 0.045 , \text{L} = 0.0225 , \text{moles} ] In acidic conditions, the balanced reaction shows that 1 mole of KMnO4 reacts with 5 moles of I⁻ (from NaI). Therefore, the moles of NaI required are: [ \text{Moles of NaI} = 5 \times 0.0225 = 0.1125 , \text{moles} ] Now, using the concentration of NaI, we can find the volume needed: [ \text{Volume of NaI} = \frac{0.1125 , \text{moles}}{0.380 , \text{M}} \approx 0.296 , \text{L} \text{ or } 296 , \text{mL} ] Thus, approximately 296 mL of 0.380 M NaI is required.


How would NaI interact with water?

NaI (sodium iodide) is highly soluble in water and will readily dissociate into Na+ and I- ions when mixed with water. The ions will interact with water molecules through ion-dipole interactions, forming a homogenous solution of NaI in water.


What is the molarity of a solution prepared by dissolving 10.7 g NaI in 0.250 L?

To find the molarity, we first need to calculate the number of moles of NaI using its molar mass (149.89 g/mol). Then, we divide the moles of NaI by the volume of the solution in liters (0.250 L). This gives us the molarity, which would be around 1.43 M.