The periodic table has mass units for every element that can be correlated with the number of atoms of that element. The relationship is known as Avogadro's Number. This number, 6.02x $10^{23}$ , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x $10^{23}$ . The mole is correlated to the atomic mass of that element. The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. But it can also be read as 36.02 grams/mole.

This means that 36.02 grams of S contains 1 mole (6.02x $10^{23}$ ) of S atoms.

This relationship holds for all the elements. Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole. ^5.38 grams of Zn contains 1 mole of Zn atoms.

And so on.

5.0 moles of Sulfur would therefore contain:

(5.0 moles S)*(36.02 grams/mole S) = 180.1 grams of S

Note how the units cancel to leaves just grams. The units are extremely helpful in mole calculations to insure the correct mathematical operation is done. To find the number of moles in 70 g of S, for example, we would write:

(70g S)/(36.02 grams S/mole S) = 1.94 moles of S. [Note how the units cancel to leave just moles]

4 0

2 years ago

I 13 Suggest why this group is called Group 8.

yarga [219]

Answer:

because it's in group 8

Explanation:

3 0

3 years ago

3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.

igor_vitrenko [27]

Volume percent = Volume of solute
----------------------------------
Volume of the solution

   2 Volume of the solute
------- =   ------------------------------
100 250

Volume of the solute = 2 x 250
------------
100

= 5 mL.

Hope this helps!

4 0

3 years ago

Assuming that sea water is a 3.5 wt % solution of NaCl in water, calculate its osmotic pressure at 20°C. The density of a 3.5% N

olga nikolaevna [1]

Answer:

π = 14.824 atm

Explanation:

wt % = ( w NaCL / w sea water ) * 100 = 3.5 %

assuming w sea water = 100 g = 0.1 Kg

⇒ w NaCl = 3.5 g

osmotic pressure ( π ):

π = C NaCl * R * T

∴ T = 20 °C + 273 = 293 K

∴ C ≡ mol/L

∴ density sea water = 1.03 Kg/L....from literature

⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln

⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol

⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M

⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K

⇒ π = 14.824 atm

7 0

3 years ago