Answer: option B. there are 35 mg of lead in 1.0 L of this solution
Explanation:
1ppm = 1mg/L
Therefore 35ppm = 35mg/L
So, the solution contains 35mg of lead in 1L of the solution
Answer: D is the answer since it is the product of this equation
Explanation: HOPE I AM RIGHT AND IT HELPS!!!
need more explanation feel free to comment in the comment box
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
