Yes, K+ is<span> a </span>potassium<span> ion, and Mg</span>2+ is<span> a magnesium ion. But However, when non-metallic elements gain the </span>electrons<span> to form anions, Yes the end of their name </span>is<span> changed to “-ide.” and yes the example, a fluorine </span>atom<span> gains </span>one electron<span> to </span>become<span> a yes fluoride ion (F</span>-<span>) sooo yeessyes</span>
Answer:
most likely that (2) the replicated experiment was performed incorrectly.
Why, u ask? u dare question me:
1- The initial experiment invalidness cannot be proven.
2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>
3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.
4- Already knowing the data and errors would increase the precision of the replicated experiment.
5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.
happy learning!
Option C: elements produce spectra with only few distinct lines.
The spectra are not continuos and are different for every element. This permits to identify elements.
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
- <em>initial volume of gas in the container, V₁ = 2.65 L</em>
- <em>initial number of moles of gas, n₁ = 0.12 mol</em>
- <em>additional concentration, n = 0.182 mol</em>
The total number of moles of gas in the container is calculated as follows;

The final volume of gas in the container is calculated as follows;

Thus, the final volume of the methane gas in the container is 6.67 L.
Learn more here:brainly.com/question/21912477
Nitrogen=2, Hydrogen=8, Carbon=1, Oxygen=3
Hydrogen=4, Carbon=2, Oxygen=2
Iron=1, Nitrogen=2, Oxygen=6