Answer:
grams KI needed = 31.9 grams
Explanation:
? g KI + 500 g water => 6.0% KI solution
let x = grams KI needed.
Percent = part/total x 100% = (x/x+500)·100% = 6.0%
Solve for 'x' ...
x / x + 500 = 6/100 = 0.06 => x = 0.06(x + 500) => x = 0.06x + 30
x - 0.06x = 30 => 0.94x = 30 => x = 30/0.94 = 31.9 gms KI needed.
The molar mass is 286.94 g/mol
Answer:
9.45 × 10⁻³ M
Explanation:
Step 1: Given data
- Mass of calcium nitrate (solute): 155 mg (0.155 g)
- Volume of solution: 100. mL (0.100 L)
Step 2: Calculate the moles of solute
The molar mass of Ca(NO₃)₂ is 164.09 g/mol.
0.155 g × 1 mol/164.09 g = 9.45 × 10⁻⁴ mol
Step 3: Calculate the molarity of the solution
The molarity is equal to the moles of solute divided by the liters of solution.
M = 9.45 × 10⁻⁴ mol/0.100 L = 9.45 × 10⁻³ M
Answer: The pH of an aqueous solution of .25M acetic acid is 2.7
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.25 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.25\times 0.0084=0.0021](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.25%5Ctimes%200.0084%3D0.0021)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[0.0021]=2.7](https://tex.z-dn.net/?f=pH%3D-log%5B0.0021%5D%3D2.7)
Thus pH is 2.7
The correct answer is A. A <span> sample of sodium hydroxide (NaOH) has a mass of 160.0 g which is equal to 4 moles of NaOH. TO calculate this, we use the relation of the molar mass given.
160.0 g NaOh ( 1 mol NaOH / 40.0 g NaOH ) = 4.000 mol NaOH</span>
