3.8 mole Cl2
Explanation:
3.8molesC6H5Cl x 1moleCl2/ 1mole C6H5Cl
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>
<u></u>
<u>This means that 36.02 grams of S contains 1 mole (6.02x</u>
<u>) of S atoms</u>.
<u></u>
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:
6.31g/mol
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
Mole (n) = mass (m)/molar mass (Mm)
* Mm = m/n
Also, density (p) = mass (m) ÷ volume (V)
PV = nRT
Since n = M/Mm
PV = M/Mm. RT
PV × Mm = m × RT
Divide both sides by V
P × Mm = m/V × RT
Since p = m/V
P × Mm = p × RT
Mm = p × RT/P
Mm = 0.249 × 0.0821 × 293/0.95
Mm = 5.989 ÷ 0.95
Mm = 6.31g/mol
Answer:
The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)
Explanation:
Let's think all the situation.
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
Initially 0.20 - -
Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.
React x x/2 x/2
Because the ratio is 2:1, in the reaction I have the half of moles.
So in equilibrium I will have
(0.20 - x) x/2 x/2
Notice that I have the concentration in equilibrium so:
0.20 - x = 0.060
x = 0.14
So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)
Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).
As we have a volume of 2L, the values must be /2
Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²
Kc = (0.07/2 . 0.07/2) / (0.060/2)²
Kc = 1.225x10⁻³ / 9x10⁻⁴
Kc = 1.36