The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.
The elements which are in between the metals and non-metals can be classified as semi-metals.
Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.
Answer:
390.85mL
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 780 torr
Initial volume (V1) = 400mL
Initial temperature (T1) = 40°C = 40°C + 273 = 313K
Final temperature (T2) = 25°C = 25°C + 273 = 298K
Final pressure (P2) = 1 atm = 760torr
Final volume (V2) =?
Step 2:
Determination of the final volume i.e the volume of the gas outside Matt's body.
The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:
P1V1/T1 = P2V2/T2
780 x 400/313 = 760 x V2 /298
Cross multiply to express in linear form
313 x 760 x V2 = 780 x 400 x 298
Divide both side by 313 x 760
V2 = (780 x 400 x 298) /(313 x 760)
V2 = 390.85mL
Therefore, the volume of the gas outside Matt's body is 390.85mL
Here we have to calculate the amount of 
ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄ 
g of ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
Answer: C. variable depending on where its found
The chemical compositions of minerals is very necessary in order to define or describe a certain mineral. Some minerals can be written in simple chemical formula while others are also written in complicated chemical formulas.
