if the average concentration of PCBs in the body tissue of a human is 4.0 ppm, what mass of the PCB is present in a 64 kg person? is this a significant amount in regards to the health of the person?
What is the concentration of a solution of sodium nitrate that contains 35.0 g of the solute in a 1.5 L solution?
A typical household ammonia solution has a concentration o 1.24 M. What volume of this solution would contain .500 mol of NH3? What volume of this solution would contain 10.0 g of NH3?
What volume of concentrated 17.8 M sulfuric acid would a laboratory technician need to make 2.00 L of a .200 mol/L solution by dilution of the original, concentrated solution?
Fair warning. Many of us, me in particular, shy away from posts containing more than one question. It takes too long to answer a single question and some of us have just time enough some times to knock off one small question. In this case the answers were quick and I did them all but I suggest that you limit posts to one question in the future. So you post multiple times; if that is so please use the same screen name. It helps us keep the questions straight. We don't mind multiple posts by the same person.
1 ppm = (g solute/g solvent) x 1E6
1 ppm = (g PCB/g solvent) x 1E6
4 ppm = (g PCB/64,000 g) x 1E6
Solve for g PCB. You can look in your references as to the toxic effects of this amount.
Find mols NaNO3 = grams/molar mass
Then M = mols/1.5 L
NH3.
M = mols/L
1.24 = 0.5/L
Solve for L of 1.24 M needed for a 0.5 mol.
H2SO4.
c1v1 = c2v2
17.8*v1 = 0.200 x 2
solve for v1 in L.
To find the mass of PCB present in a 64 kg person, we can use the average concentration given in parts per million (ppm). "Parts per million" means that for every 1 million parts of a substance, there are 4 parts of PCB.
Step 1: Convert the concentration from parts per million to a decimal fraction.
4 ppm = 4/1,000,000 = 0.000004
Step 2: Multiply the concentration by the person's mass to find the mass of PCB present.
Mass of PCB = Concentration x Mass of person
Mass of PCB = 0.000004 x 64 kg = 0.000256 kg
The mass of PCB present in a 64 kg person is 0.000256 kg.
Regarding the significance of this amount to the person's health, it would depend on various factors such as exposure duration, individual sensitivity, and the toxicity of PCBs. PCBs are known to be toxic and can have adverse health effects, including developmental and reproductive issues, immune system dysfunction, and potential carcinogenic effects. However, to determine if the amount is significant for health, it would require a comprehensive assessment by a medical professional considering the person's specific circumstances.
To find the concentration of a solution of sodium nitrate that contains 35.0 g of the solute in a 1.5 L solution, we can use the formula:
Concentration (in g/L) = Mass of solute / Volume of solution
Concentration = 35.0 g / 1.5 L = 23.3 g/L
The concentration of the sodium nitrate solution is 23.3 g/L.
To find the volume of a 1.24 M ammonia solution that contains 0.500 mol of NH3, we can use the formula:
Volume (in L) = Amount of substance (in mol) / Concentration (in mol/L)
Volume = 0.500 mol / 1.24 M = 0.403 L = 403 mL
The volume of the 1.24 M ammonia solution containing 0.500 mol of NH3 is 403 mL.
To find the volume of a 1.24 M ammonia solution that contains 10.0 g of NH3, we need to convert the mass to moles using the molar mass of ammonia (NH3).
Step 1: Calculate the number of moles using the formula:
Moles = Mass / Molar mass
Molar mass of NH3 = 14.01 g/mol (nitrogen) + 3 * 1.01 g/mol (hydrogen) = 17.04 g/mol
Moles = 10.0 g / 17.04 g/mol = 0.586 mol
Step 2: Use the formula to find the volume:
Volume = Moles / Concentration
Volume = 0.586 mol / 1.24 M = 0.472 L = 472 mL
The volume of the 1.24 M ammonia solution containing 10.0 g of NH3 is 472 mL.
To make a 2.00 L solution of sulfuric acid with a concentration of 0.200 mol/L by dilution, we will use the formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Given:
Initial concentration (C1) = 17.8 M
Final concentration (C2) = 0.200 mol/L
Final volume (V2) = 2.00 L
Rearranging the formula, we have:
V1 = (C2 * V2) / C1
V1 = (0.200 mol/L * 2.00 L) / 17.8 M
V1 ≈ 0.0225 L ≈ 22.5 mL
To make a 2.00 L solution of a 0.200 mol/L sulfuric acid by dilution, the laboratory technician would need to take approximately 22.5 mL of the concentrated 17.8 M sulfuric acid and dilute it to a final volume of 2.00 L.