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>]
b the awnser is b its a compound of carbon C and 2 oxygon atoms O sub 2
Answer:
D
Explanation:
Scientists always perform tests and observe and measure in the physical world to prove their points or answer their questions.
Answer:
pOH= 14.248
[H+]=1.77 M
[OH-]=5.65 x10^-15M
Explanation:
pH+pOH= 14
pOH= 14-pH
pOH=14-(-0.248)
pOH= 14.248
[H+]=10^-pH= 10^-(-0.248)=1.77 M
[OH-]=10^-pOH= 10^-14.248=5.65 x10^-15M
The balanced chemical reaction for the complete combustion of C4H10 is shown below:
C4H10 + (3/2)O2 --> 4CO2 + 5H2O
The enthalpy of formation are listed below:
C4H10: -2876.9 kJ/mol
O2: none (because it is pure substance)
CO2: -393.5 kJ/mol
H2O: -285.8 kJ/mol
The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.
ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
= -<em>126.1 kJ</em>
Thus, the enthalpy of combustion of the carbon is -126.1 kJ.
