In chemistry, S2 typically refers to a sulfide ion, which is a negatively charged ion composed of two atoms of sulfur. Sulfide ions are common in compounds such as hydrogen sulfide (H2S) and metal sulfides (e.g. FeS, ZnS).
Ferrous is the traditional name of the iron (II) ion. (Fe 2+) The formula for ferrous sulfide is FeS since sulfur has a charge of -2.
The chemical formula for lithium is Li. The chemical formula for sulfide ion is S2-. Therefore, the chemical formula for lithium sulfide would be Li2S.
The symbol for sulfide, one of the many ions, is S2-.
A mole of sulfur (S) contains approximately 6.022 x 10^23 sulfur atoms. The molecular weight of sulfur (S) is about 32 g/mol. Therefore, a mole of sulfur (S2) would weigh 64 grams (32 g/mol x 2).
It is not possible to determine which one is larger in size without more context or information about what S and S2 refer to. Size comparison between two entities requires specific measurements or criteria to make a valid comparison.
What is a syllabus chemistery?" What is a syllabus chemistery?"
chemistery is the study of chemicals and how the react.
gllelo galle
chemist or a chemistery
nope
Yes, (provided you passed your BSc).
Salts are the products of the reactions between acids and bases.
at pluto -- 0.61m/s2 at earth -- 9.8m/s2 mean u can have much high compared to earth
Base is a solution that have ph over 7 and can accept hydrogen ions.
When mercury is found in an elemental form, it means that it exists in its pure, uncombined state as a single atom or a molecule of mercury. Elemental mercury is a liquid at room temperature and is known for its unique properties, such as being a good conductor of electricity.
s2= love
The sum of an arithmetic series of n terms is: S(n) = n(2a1 + (n - 1)d) / 2 where a1 is the first term and d is the difference between terms. The key to solving this is to realise that two sums are required, S1 and S2 such that: S1 = S(r) S2 = S(3r) - S(r). But that this second sum can also be stated as: S2 = sum of 2r terms starting at term ar+1 So that the problem then becomes: Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44. This gives: S1 = r (2a1 + (r - 1)d) / 2 = r(2(2x1 + 3) + (r - 1)2) / 2 = r(2r + 8) / 2 = r(r + 4) S2 = 2r(2ar+1 + (2r - 1)d) = 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2 = 2r(2(2r + 5) + (2r - 1)2) / 2 = 2r(2r + 5 +2r - 1) = 2r(4r + 4) = 8r(r + 1) As S1 : S2 = 7 : 44, 44 S1 = 7 S2 => 44 r(r + 4) = 7 x 8r(r + 1))0 => 11(r + 4) = 14(r + 1) => 11r + 44 = 14r + 14 => 3r = 30 => r = 10 As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed. To check: a1 = 2 x 1 + 3 = 5 S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020 S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140 Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880 Ratio of S1 : S2 = 140 : 880 = 14 : 88 = 7 : 44 Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25: S2 = 20(2 x 25 + 19 x 2)/2 = 20 x 44 = 880.