Explanation:
Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
Writing the information from the ICE Table in Equation form yields
6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Manipulating the equation to get everything on one side yields
0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this information is plugged into the quadratic formula to give
x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
The quadratic formula yields that x=0.013745 and x=-0.014405
However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86
Answer: First of all planning is done to set the goals to how to test the target. then scanning is done. after that target is uncovered to see how the target exploits by the use of web application attacks. then analysis is done an tests of the penetration tests are compiled.
Explanation:
M(dextrose) = 50 g.
V(solution) = 1 L.
n(dextrose) = 50 g ÷ 180 g/mol.
n(dextrose) = 0,27 mol.
Osmotic concentration (osmolarity)<span> is a measure of how many </span><span>osmoles of particles of solute</span><span> it contains </span>per liter.
The osmolarity = n(dextrose) ÷ V(solution).
The osmolarity = 0,27 mol ÷ 1 L.
The osmolarity = 0,27 mol/L · 1000 mmol/m.
The osmolarity (dextrose) = 270 mosm/L.
The osmolarity (dextrose monohydrate) = 50 g÷197 g/mol·1000 =254mosm/L
molecule- when two or more of the same atoms of an element chemically join together
ex: O2
compund- when the two types of atoms are different
ex: H2O
