If you are operating a power driven vessel that is underway in condition of restricted visibility, you are expected to do the following: sound prolonged blasts every two minutes. If the vessel is underway but is not moving, it is expected to sound two prolonged blast every two minutes. When one hears any of the signal above, one is expected to reduce speed to the minimum that is needed to keep on course.
Answer:
8.3 × 10³ mL
Explanation:
Step 1: Calculate the mass of water that contains 100 mg of Pb
The concentration of Pb in the sample is 0.0012% by mass, that is, there are 0.0012 g of Pb every 100 g of water. The mass of water that contains 100 mg (0.100 g) of Pb is:
0.100 g Pb × 100 g Water/0.0012 g Pb = 8.3 × 10³ g Water
Step 2: Calculate the volume corresponding to 8.3 × 10³ g of water
Since the solution is diluted, we will assume the density of the sample is equal than the density of water (1 g/mL).
8.3 × 10³ g × 1 mL/1 g = 8.3 × 10³ mL
The dissociation of both salts NaCl and CaCl₂ are as follows;
NaCl --> Na⁺ + Cl⁻
CaCl₂ --> Ca²⁺ + 2Cl⁻
the molar ratio of NaCl to Cl⁻ is 1:1
therefore number of NaCl moles is equal to number of Cl⁻ ions dissociated from NaCl
then number of Cl⁻ ion moles - 0.233 mol/L x 0.1000 L = 0.0233 mol
molar ratio of CaCl₂ to Cl⁻ ions is 1:2
1 mol of CaCl₂ gives out 2 mol of Cl⁻ ions.
number of CaCl₂ moles - 0.150 mol/L x 0.2500 L = 0.0375 mol
then the number of Cl⁻ ion moles - 0.0375 x 2 = 0.0750
total number of Cl⁻ ion moles = 0.0233 mol + 0.0750 mol = 0.0983 mol
volume of solution - 100.0 + 250.0 = 350.0 mL
concentration of Cl⁻ = 0.0983 mol / 0.3500 L = 0.281 M
concentration of Cl⁻⁻ is 0.281 M
The formula unit of compound made up from Pb4+ and oxygen is PbO2.
