Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
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>]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
