Answer:
Explanation:
In this case, we have to remember the <u>relationship between the Ka value and the pH</u>. We can use the general reaction for any acid with his Ka value expression:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
In the Ka expression, we have a<u> proportional relationship</u> between Ka and the concentration of 
. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the <u>smaller Ka value</u> in this case .
I hope helps!
<h2>
Answer: Mg²⁺ & N³⁻</h2>
<h3>
Explanation:</h3>
Magnesium (group IIA) loses two electrons to become a cation. This is the most energetically favorable way for Magnesium to obtain a valence electron that is stable (octet). When an atom loses electrons it has a positive charge that matches the number of electrons lost.
Nitrogen (group VA) gains three electrons to become an anion. This is the most energetically favorable way for Nitrogen to obtain a valence electron that is stable (octet). When an atom gains electrons it has a negative charge that matches the number of electrons gained.
Answer: the density is 997 kg
Explanation:
The answer is cold water because the water is cold and it’s not using thermal energy
Explanation:
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by
E=hf=hcλ(energy of a photon)E=hf=hcλ(energy of a photon),
where E is the energy of a single photon and c is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is h = 4.14 × 10−15 eV · s.
Since many wavelengths are stated in nanometers (nm), it is also useful to know that hc = 1240 eV · nm.
These will make many calculations a little easier.
All EM radiation is composed of photons. Figure 1 shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and γ rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.
