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:
3= Lithium (Li) = [He] 2s1
6= Carbon (C) = [He] 2s2 2p2
8=Oxygen (O)= [He] 2s2 2p4
13=Aluminium (Al)= [Ne] 3s2 3p1
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The balanced equation is:
Then proceed with the following equations.
The answer is
.
Answer:
51207 torr is the new pressure of the gas
Explanation:
We can solve this question using combined gas law that states:
P1V1T2 = P2V2T1
<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em> </em>
Computing the values of the problem:
P1 = 710torr
V1 = 5.0x10²mL
T1 = 273.15 + 30°C = 303.15K
P2 = ?
V2 = 25mL
T2 = 273.15 + 820°C = 1093.15K
Replacing:
710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K
3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K
P2 = 51207 torr is the new pressure of the gas
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles
Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules
