Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.
Explanation:
In the elemental form oxidation state is zero. Here chlorine is present in elemental form so oxidation state is zero.
Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.
If electron is gained oxidation number becomes negative.
If electron is lost then oxidation number is positive.
If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.
Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.
Answer:
[N2] = 0.3633M
[H2] = 1.090M
[NH3] = 0.2734M
Explanation:
Based on the reaction of the problem, Kc is defined as:
Kc = 0.159 = [NH3]² / [N2] [H2]³
<em>Where [] are the equilibrium concentrations.</em>
The initial concentrations of the reactants is:
N2 = 1.00mol / 2.00L = 0.500M
H2 = 3.00mol / 2.00L = 1.50M
When the equilibrium is reached, the concentrations are:
[N2] = 0.500M - X
[H2] = 1.50M - 3X
[NH3] = 2X
<em>Where X is reaction quotient</em>
Replacing in the Kc equation:
0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³
0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴
0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²
0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0
Solving for X:
X = 0.1367. Right solution.
X = 1.8286. False solution. Produce negative concentrations
Replacing:
[N2] = 0.500M - 0.1367M
[H2] = 1.50M - 3*0.1367M
[NH3] = 2*0.1367M
The equilibrium concentrations are:
<h3>[N2] = 0.3633M</h3><h3>[H2] = 1.090M</h3><h3>[NH3] = 0.2734M</h3>
Volume = nRT/P
n = number of particles (moles)
R = universal gas constant (0.0821)
T = temperature (Kelvin)
P = pressure (atm)
(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He
Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K
Now we can set up the equation for volume: (3mol)(0.0821)(373.15)/1.2atm = 76.6 L of Helium gas
Answer:
storm runoff
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