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).
at pluto -- 0.61m/s2 at earth -- 9.8m/s2 mean u can have much high compared to earth
Salts are the products of the reactions between acids and bases.
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.