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
It mean it consisted of 1 g of lead and 0.077 g of O2.
divide these numbers by molar mas.
1/82=0.012 Pb /0.004 = 3
0.077/16= 0.004 O /0.004 =1
Pb3O
Explanation:
Initial Pressure = 24 lb in-2
Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)
Final Pressure = ?
Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)
No Change in Volume.
From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P1T1 = P2T2
P2 = P1T1 / T2
P2 = 24 * 268 / 308 = 20.88 lb in-2
There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.
