Answer:
Potassium cation = K⁺²
Explanation:
The metal cation in K₂SO₄ is K⁺². While the anion is SO₄²⁻.
All the metals have tendency to lose the electrons and form cation. In given compound the metal is potassium so it should form the cation. The overall compound is neutral.
The charge on sulfate is -2. While the oxidation state of potassium is +1. So in order to make compound overall neutral there should be two potassium cation so that potassium becomes +2 and cancel the -2 charge on sulfate and make the charge on compound zero.
2K⁺² , SO₄²⁻
K₂SO₄
The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
One side will burn to a crisp and the other side will freeze... and after some time the earth would be pulled into the sun and cooked
Answer:
For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.
Explanation:
If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
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