Answer:
<u>5 moles S x (36.02 g S/mole S) = 180.1 grams of S</u>
Explanation:
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
, is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x
. <u>The mole is correlated to the atomic mass of that element.</u> The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. <u>But it can also be read as 36.02 grams/mole.</u>
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<u>This means that 36.02 grams of S contains 1 mole (6.02x</u>
<u>) of S atoms</u>.
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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) = <u>180.1 grams of S</u>
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. [<u>Note how the units cancel to leave just moles</u>]
Answer:
The volume of water in water bath is 1,011 Liters.
Explanation:
Length of the water bath, L = 1.85 m
Width of the water bath, W= 0.810 m
Height of the water bath ,H= 0.740 m
Height of the water in water bath, h= 0.740 m - 2.57 inches
1 m = 39.37 inch

Volume of the water in bath = L × W × h



The volume of water in water bath is 1,011 Liters.
Answer:
0.21mol Ar (g)
Explanation:
To convert from litres to moles at STP we must divide the amount of litres by 22.4.
4.7 / 22.4 = 0.21mol Ar (g)
Can i get a list of possible answers plz?
Explanation:
It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.
Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.
So, vapor pressure of mixture = 1% vapor pressure of pure water
Therefore,
=
= 27.785 Pa
Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.