As he began to teach inorganic chemistry, Mendeleev could not find a textbook that met his needs. Since he had already published a textbook on organic chemistry in 1861 that had been awarded the prestigious Demidov Prize, he set out to write another one. The result was Osnovy khimii (1868–71; The Principles of Chemistry), which became a classic, running through many editions and many translations. When Mendeleev began to compose the chapter on the halogen elements (chlorine and its analogs) at the end of the first volume, he compared the properties of this group of elements to those of the group of alkali metals such as sodium. Within these two groups of dissimilar elements, he discovered similarities in the progression of atomic weights, and he wondered if other groups of elements exhibited similar properties. After studying the alkaline earths, Mendeleev established that the order of atomic weights could be used not only to arrange the elements within each group but also to arrange the groups themselves. Thus, in his effort to make sense of the extensive knowledge that already existed of the chemical and physical properties of the chemical elements and their compounds, Mendeleev discovered the periodic law.
Answer:
Nickel(III) oxide is the chemical name of the compound represented by the formula Ni2O3?
Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
<h3>How to calculate mass of substances?</h3>
The mass of a substance can be calculated using the following steps:
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
1 mole of Cu react with 2 moles of AgNO3
- Molar mass of AgNO3 = 169.87 g/mol
- Molar mass of Cu = 63.5g/mol
moles of AgNO3 = 262g/169.87g/mol = 1.54mol
1.54 moles of AgNO3 will react with 0.77 moles of Cu.
mass of Cu = 0.77 × 63.5 = 48.97g
Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
Learn more about mass at: brainly.com/question/6876669
<span>the solvent, hope this helps</span>
