Answer:
A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m.
Explanation:
because in the right side there is 2 hydrogen( H) and 2 chloride (Cl) to balace them we have to replace 2.
The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.
First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.
We can notice that once the NaCl is diluted in water, all what you can see is a clear solution. You cannot see the separate particles of NaCl in water.
..............> observation I
Now, we will heat this solution until it boils and water starts evaporating. We will place a cold surface above the steam coming out from the boiling solution.
What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II
Based on this, we managed to use boiling, evaporation and condensation (physical methods) to restore the components of the solution separately.
.............>conclusion
Based on observation I, observation II and the conclusion. we were able to prove that NaCl solution is a homogeneous mixture.
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
If i’m correct it’s b, bouyance force.
