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How many grams of NaI would be used to produce a 2.0 M solution with a volume of 1.00 L?
To prepare a 2.0 M solution of NaI with a volume of 1.00 L, you would need to use 284 grams of NaI. This is calculated by multiplying the molarity by the volume in liters and the molar mass of NaI (149.89 g/mol).
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 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.
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.
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.
How many grams of NaI would be used to produce a 2.0 M solution with a volume of 1.00 L?
To prepare a 2.0 M solution of NaI with a volume of 1.00 L, you would need to use 284 grams of NaI. This is calculated by multiplying the molarity by the volume in liters and the molar mass of NaI (149.89 g/mol).
Which solution of NaI would be most dissociated?
The most dissociated solution of NaI would be the one with the highest concentration, as the greater the concentration of ions in solution, the greater the likelihood of dissociation. Additionally, higher temperatures also promote greater dissociation in solutions.
What does je nai's plus un mean?
I don't have one anymore
Name the compound formed from Na plus and I- ions?
NaI is sodium iodide
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 is the solution of CG PET 2009?
GAND MARAO BHOSDI WALO. ANSWER KAHIN NAI MILEGA.
What does nai nai mean in Mandarin?
"Nai nai" in Mandarin Chinese means paternal grandmother.
What aqueous solution would form a precipitate when mixed with NaI?
Silver nitrate for example: AgI(s) silver iodide
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.
