A 2.9-L sample of paint that has a density of 4.65 g/mL is found to contain 26.7 g Pb3N2(s). How many grams of lead were in the paint sample?
You don't need the density or volume of the paint.
Convert 26.7 g Pb3N2 to mols moles = grams/molar mass.
Moles Pb must be 3x moles Pb3N2.
grams Pb = mols Pb x molar mass Pb.
Well, well, well, looks like we've stumbled into the colorful world of paints and densities! Let's calculate the number of grams of lead hiding in that sample.
First, we need to calculate the volume of the paint sample. Since we have the density (4.65 g/mL) and the sample volume (2.9 L), we can multiply these two fandangoes together:
Volume = Density x Sample volume
Volume = 4.65 g/mL x 2.9 L
But uh-oh, we have different units here! So let's convert liters into milliliters. There are 1000 mL in a liter, math magic time!
Volume = 4.65 g/mL x 2.9 L x 1000 mL/L
Now we can calculate the volume:
Volume = 13485 mL
Now we're getting somewhere!
Since we know that the sample contains 26.7 g of Pb3N2, we can go ahead and calculate the number of grams of lead it contains.
Pb3N2 is our painter's protagonist here, so we need to figure out the ratio of lead to the entire compound. Surprise, surprise, Pb3N2 consists of three lead atoms (Pb) per compound.
To calculate the number of grams of lead, we can multiply the molar mass of Pb by the number of moles of lead in Pb3N2, and then multiply that by Avogadro's number.
But hey, I'm just a jolly joke-telling bot, not a walking periodic table! So let's assume you know that the molar mass of lead (Pb) is approximately 207.2 g/mol.
Now, divide the mass of Pb3N2 (26.7 g) by the molar mass of Pb3N2 (554.2 g/mol) to get the number of moles of Pb3N2. Then multiply that by the ratio of lead atoms per Pb3N2 (3 Pb atoms). Now we're cooking, or should I say, painting!
Grams of Pb = (26.7 g / 554.2 g/mol) x (3 Pb atoms / 1 Pb3N2)
Now it's time for a drumroll and the grand finale:
Grams of Pb = 0.048 g
So, there you have it! In that 2.9-liter sample of paint, there are approximately 0.048 grams of lead. Happy painting, my friend!
To find the number of grams of lead (Pb) present in the paint sample, we need to calculate the volume occupied by the lead (Pb3N2) and then use its density.
Given:
Volume of paint sample = 2.9 L
Density of paint = 4.65 g/mL
Mass of Pb3N2(s) = 26.7 g
First, we need to find the volume occupied by the lead (Pb3N2) in the paint sample.
Volume = Mass / Density
Volume = 26.7 g / 4.65 g/mL
Calculating the volume:
Volume = 5.74 mL
Since we are given the density of the paint, we can now calculate the mass of lead (Pb) present in the paint sample.
Mass = Volume x Density
Mass = 5.74 mL x 4.65 g/mL
Calculating the mass:
Mass = 26.691 g
Therefore, there were approximately 26.691 grams of lead (Pb) in the paint sample.
To find the number of grams of lead in the paint sample, we need to use the given density and volume of the paint sample.
First, we need to calculate the mass of the paint sample. We can do this by multiplying the density by the volume:
Mass = Density × Volume
Given: Density = 4.65 g/mL, Volume = 2.9 L
Mass = 4.65 g/mL × 2.9 L
To calculate the mass, we need to convert the volume from liters (L) to milliliters (mL) because the density is given in grams per milliliter (g/mL). Since 1 L is equal to 1000 mL:
Mass = 4.65 g/mL × 2.9 L × 1000 mL/L
Mass = 13535 g
Now that we have the total mass of the paint sample, we can determine the mass of lead by dividing the mass of the lead compound (Pb3N2(s)) by the total mass of the paint sample:
Mass of lead = (Mass of Pb3N2 / Mass of paint) × Total mass of paint
Given: Mass of Pb3N2 = 26.7 g, Total mass of paint = 13535 g
Mass of lead = (26.7 g / 13535 g) × 13535 g
Mass of lead = 26.7 g
Therefore, there are 26.7 grams of lead in the paint sample.