The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
<h2> ║∧║······→Hello.←······║∧║</h2>
Your answer should be:
N2 + O2 → 2NO
<h2>····················································································</h2>
I hope this answered your question!
Thanks for spending time reading this :)
(<em>A brainliest would be appreciated!)</em>
⊕If this answer doesn't answer your question or you are displeased by it please tell me in the comments I would like to know. And if I can, I will remove it.⊕
<h2>Have a wonderful day</h2>
Answer:
Mass = 182.4 g
Explanation:
Given data:
Number of moles of Al₂O₃ = 3.80 mol
Mass of oxygen required = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Now we will compare the moles of aluminum oxide and oxygen.
Al₂O₃ : O₂
2 : 3
3.80 : 3/2×3.80 = 5.7
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 5.7 mol × 32 g/mol
Mass = 182.4 g
Answer:
b 13kg/m³ is the answer to this question
3 ethyl, 4 methylheptane. The compound is named by first identifying the longest carbon chain in the structure. in this case the chain has seven carbon atoms thus the prefix hept-.
Next you identify the substituent groups attached to the long carbon chain and name them from the lowest value of the integer assigned to the carbon atoms from either side. From the right, the ethyl group is attached to carbon number 3 while from the left, the methyl group is attached to carbon number 4. We therefore start with the right and name the attached groups first, including the carbon atoms to which they are attached.
Then we also take into consideration the highest number of bonds between the carbon atoms which is one from the question. Thus the suffix -ane is added if a maximum of one bond, -ene,if two bonds and -yne if three bonds.