Answer:
0.027 moles of carbon dioxide produced by the reaction of 0.085 mol of oxygen.
Explanation:
Moles of oxygen gas = 0.085 mol
According to reaction, 50 moles of oxygen gives 16 moles of carbon dioxide gas, then 0.085 moles of oxygen gas will give:
of carbon dioxide
0.027 moles of carbon dioxide produced by the reaction of 0.085 mol of oxygen.
chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulae, wherein the reactant entities are given on the left-hand side and the product entities on the right-hand side with a plus sign between the entities in both the reactants and the products and an arrow that points towards the products, and shows the direction of the reaction.[1] The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers.
Explanation:
<h3>CARRYINGTOLEARN:)</h3>
Mass spectrum of Dodecane will give following information.
1 ) Molecular Peak or Parent Peak:
The Parent peak will appear at m/z = 170. The intensity of this peak will be very weak.
2) Fragments:
Usually the fragments of such long chain alkanes appear with spacing of 14 amu, Hence, the peaks in dodecane will be as follow,
170 - 156 - 142 - 128 - 114 - 100 - 86 - 72 - 58 - 44 - 30 - 16
3) Base Peak:
Most probably the Base peak will appear at m/z = 57. This peak is due to the formation of tertiary butyl cation as the intensity mainly depends upon the stability of cation. So this cation might form due to rearrangment giving the intensity of 100%.
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
