The answer is C) [Ar]4s2 3d6
Explanation:
passing hydrogen through the anode of a fuel cell and oxygen through the cathode.
hopes this helps
In order to obtain solid NaCl, the student should do a few steps.
First, he/she should do filtration. Pass the mixture through a filter paper, where all the sand should be filtered out already because they're not dissolved in the solution plus they're too small to pass through the filter paper.
Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
<span>Group 1 can be characterized as atoms that have 1 electron in their valence shell. This is valuable when dealing with these questions, because the loss or gain of valence electrons is what defines ionic relationships. When group 1 elements form ionic bonds with other atoms, they are extremely likely to lose their valence electron, since the nucleus has a weaker pull on it than, say, a chlorine atom has on its 7 valence electrons. The weaker pull between the nucleus and the valence electron of group 1 elements means that the radius is high, since the electron is more free to move with less pull on it. This also means that the first ionization energy is low, since it takes relatively little energy for that electron to be pulled away to another atom.</span>
