Answer:
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of NaOH = x
Molar mass of NaOH = 40 g/mol
Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)
Molarity of the solution of NaOH = 0.250 M
Solving for x:
x = 4.5 g
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
The solubility KI is 50 g in 100 g of H₂O at 20 °C. if 110 grams of ki are added to 200 grams of H₂O <u>the </u><u>solution </u><u>will be </u><u>saturated</u><u>.</u>
<h3>What is solubility?</h3>
Solubility is a condition where the solute is fully dissolved in the solvent. When fully mixed with the solvent.
Given that 50 g of KI is added to 100 g of water at 20 °C it means 100 g of water can dissolve a maximum of 50 g of KCl.
1 g of water will dissolve an quantity of 0.5 g of KCl.
To assay for 200 g of water: 200 g of water can disintegrate a maximum of (0.5) x 200 g of KCl.
The maximum amount of KCl that will dissolve is 100 g
Actualised amount dissolved = 110 g
when Amount dissolved > Maximum solubility limit
110 g > 100 g
Thus, the solution is saturated.
To learn more about solubility, refer to the below link:
brainly.com/question/8591226
#SPJ4
