The density of dry air measured at 25 degrees Celsius is 1.19x10 to the negative 3rd power. What is the volume of 50.0g of air?
d33z n_u_ts in yo Face D1_ick Sucker!
mass = volume x density.
Substitute for mass and density; solve for volume. Make sure the units of density (you don't list that) agrees with the unit you want in the answer.
can u explain that a little more because i still don't understand
50/.00119=42016.8067227mL>>then you have to round to 2 significant digits b/c 50 has 2 sig figs which is the least # of sig figs
the answe iri sf
Deez nuts in yo face
To find the volume of 50.0g of air, we can use the formula:
Density = mass / volume
Rearranging the formula, we get:
Volume = mass / density
Given:
Mass of air = 50.0g
Density of air = 1.19 x 10^-3 g/cm^3
To solve for volume, substitute the values into the formula:
Volume = 50.0g / (1.19 x 10^-3 g/cm^3)
First, let's convert the density from g/cm^3 to g/m^3. Since there are 100 cm in 1 m, the conversion factor is 1/100^3 = 1/1,000,000.
Volume = 50.0g / (1.19 x 10^-3 g/cm^3) * (1 g/m^3 / 1.19 x 10^-3 g/cm^3)
= 50.0g / (1.19 x 10^-3) x (1 x 10^6) m^3
= 50.0g / 1.19 x 10^-3 x 10^6 m^3
Simplifying further:
Volume = 50.0g / (1.19 x 10^-3 x 10^6) m^3
= (50.0g / 1.19) x (1/10^-3) x (1/10^6) m^3
= 42.0168 x 1 x 1 x 10^6 m^3
= 42.0168 x 10^6 m^3
= 4.20168 x 10^7 m^3
So the volume of 50.0g of air is approximately 4.20168 x 10^7 m^3.
50/.00119=42016.8067227mL>>then you have to round to 2 significant digits b/c 50 has 2 sig figs which is the least # of sig figs, so its 42,000mL
sma
There isn't anything to explain.
mass = volume x density. That's the formula. That's it. End of story. You read the problem, substitute the numbers, solve the equation.
mass in the problem = 50g
density in the problem = 1.19E-3. You don't list the units. I will assume for purposes of showing you that this is 1.18E-3 g/mL. If it is some other unit you will need to adjust that.
Substitute.
mass = volume x density
50 = volume(in mL) x 1.19E-3 g/mL
volume = 50/1.19E-3 = about 42000 mL.
If you will explain what trouble you are having perhaps I can focus on a better answer.