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What is the total number of grams of NaI(s) needed to make 1 L of a 0.010M solution?
To make a 0.010M NaI solution, you'll need 0.010 moles of NaI per liter. The molar mass of NaI is 149.89 g/mol. Therefore, to calculate the grams needed, you would multiply the molar mass by the number of moles, which gives you 1.499 g of NaI needed per liter of solution.
How do you calculate the number of moles of NaI in 50.0mL of a 0.400M solution?
To calculate the number of moles of NaI in 50.0mL of a 0.400M solution, you first need to convert the volume to liters by dividing by 1000 (since 1L = 1000mL). Then, use the formula moles = Molarity x Volume (in liters) to find the number of moles of NaI. In this case, it would be 0.400 mol/L x 0.050 L = 0.020 moles of NaI.
Is sodium iodide soluble?
Yes, sodium iodide (NaI) is highly soluble in water. It dissolves readily to form a clear solution.
What is the freezing point of a solution that contains 0.550 moles of NaI in 615 g of water?
To calculate the freezing point depression, you first need to find the molality of the solution using the moles of solute and mass of solvent. Then, use the molality to find the freezing point depression constant of water. Finally, apply the formula ΔTf = Kf * molality to find the freezing point depression.
What Color is NaI?
Sodium iodide (NaI) is typically a white or colorless solid.
How many moles of NaI would you need to prepare 2L of a 2M NaI solution?
599.6
What is the total number of grams of NaI(s) needed to make 1 L of a 0.010M solution?
To make a 0.010M NaI solution, you'll need 0.010 moles of NaI per liter. The molar mass of NaI is 149.89 g/mol. Therefore, to calculate the grams needed, you would multiply the molar mass by the number of moles, which gives you 1.499 g of NaI needed per liter of solution.
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 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.
What aqueous solution would form a precipitate when mixed with NaI?
Silver nitrate for example: AgI(s) silver iodide
What is the mass of sodium iodide in 100ml of 2.63 M of NaI?
By definition, a 2.63 molar solution of NaI contains 2.63 moles, or in this instance more properly formula units, of NaI per liter of solution. The gram atomic formula mass of NaI is 149.89 and 100 ml is 0.100 liter. Therefore, the required answer is (0.100)(149.89)(2.63) or 39.4 grams of NaI, to the justified number of significant digits.
What is the solution for NaI plus PbC2H3O2?
Sodium Iodide is the solute which is created in the reaction and Sodium actetate solution is created. NaI + PbC2H3O2 ---> PbI (Plumbum Iodide) + NaC2H3O2 (Sodium Acetate) solution.
What is the solution of CG PET 2009?
GAND MARAO BHOSDI WALO. ANSWER KAHIN NAI MILEGA.
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 does nai nai mean in Mandarin?
"Nai nai" in Mandarin Chinese means paternal grandmother.
How do you calculate the number of moles of NaI in 50.0mL of a 0.400M solution?
To calculate the number of moles of NaI in 50.0mL of a 0.400M solution, you first need to convert the volume to liters by dividing by 1000 (since 1L = 1000mL). Then, use the formula moles = Molarity x Volume (in liters) to find the number of moles of NaI. In this case, it would be 0.400 mol/L x 0.050 L = 0.020 moles of NaI.
