Paper chromatography is a tool frequently used in analytical chemistry to separate a homogeneous mixture into its components.
In paper chromatography the stationary phase is water which drenches the paper and the mobile phase is a mixture polar organic solvent and water. The various components of the homogeneous mixture is separated based on their relative solubility in either the mobile or stationary phase.
The molar concentration of the KI_3 solution is 0.0833 mol/L.
<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
<em>Step 2</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
<em>Step 3</em>. Calculate the <em>molar concentration of the I_3^(-)</em>
<em>c</em> = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
Answer:
11.29 kJ (to 4 sf)
Explanation:
Here, ammonia is the limiting reactant.
For the reaction to occur once, 4 moles of ammonia must be consumed.
Thus means that the reaction can occur 0.1579/4=0.039475 times.
Therefore, (286)(0.039475) = 11.29 kJ (to 4 sf)
For a proton, the charge is represented by +e
For an electron, the charge is represented by -e
The charge of an electron [e] is equal to the - 1.602 * 10^-19 Coulombs
For an ionized atom, the net charge is equal to the sum of charges in the proton and the electron in that atom.
From the question, we are told that, the iron atom has 26 protons and 7 electrons.
The net charge Q = 26 [ +e] + 7 [- e] = +19 e
+19 e = + 19 [1.602 * 10^-19 C] = + 3.04 * 10^-18 C.