Answer:
Depends how much water and the temperature of the water. To heat 1 mL of water by 1 degree C 1 cal of energy (4.184 Joules) is required. Assuming that the water is at 25 degrees C, to boil one litre (liter) of water you would require 75,000 cal or 313.8 kJ.
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
Water is often referred as a <span>universal solvent </span>because it is capable dissolving much more solutes as compared to any other solvent. This is because, water is a high polar molecule. In water, H has partial positive charge while O has partial negative charge.
Due to this, water favors dissociation of molecules into positively and negatively charged ions. Positively charge ions gets attracted towards oxygen i.e. negatively charges, while negatively charged ions get attracted towards positive end of water molecule.
However, it is worth nothing that, despite water being referred as universal solvent, many compounds are insoluble or partially soluble in water. For instance, most of the hydroxide displays poor solubility in water.
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:
Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L
㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days
Answer:
I know for a fact the correct answer is B
