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CTotal in parralel = C1 + C2 + C3 CTotal = 2 + 2 + 2 → 6 C Total in series = 1 / C\(_1\) + 1 / C\(_2\) 1 / C\(_T\) = 1/2 + 1/6 1 / C\(_T\) = \(\frac{4}{6}\) cross multiply C\(_T\) = \(\frac{3}{2}\) or 1.5mF

Resistance in series will be given as R = 10+10 → 20 Resistance in parallel will be given as \(\frac{1}{R}\) = \(\frac{1}{10}\) + \(\frac{1}{10}\) \(\frac{1}{R}\) = \(\frac{2}{10}\) → \(\frac{1}{R}\) = \(\frac{1}{5}\) Resistance in parallel = 5 Total resistance = 20 + 5 → 25ohms

-The radiography of cultural objects is the use of radiography to understand intrinsic details about objects. Most commonly this involves X-rays of paintings to reveal underdrawing, pentimenti alterations in the course of painting or by later restorers, and sometimes previous paintings on the support. -Hard X-rays are used for destroying cancerous cells -Soft X-rays are used in medicine to photograph human parts -. In the present work, the spatial elemental imaging capabilities of micro-X-ray fluorescence (MXRF) were used to visualize fingerprint patterns based on inorganic elements present in the prints. However; Cobalt-60 is widely used in industrial radiography for the rapid checking of faults in wields and castings.

All of the above (D)

2CO(g) + O2(g) ---> 2CO\(_2\)(g) 2vol 1vol 2vol Volume in cm\(^3\) Before sparking: 10 // 21 Volume in cm\(^3\) On sparking 10 // 5 // 10 Volume in cm\(^3\) After sparking: 0// 16 // 10 Total volume of resulting gas = Volume of [excess oxygen + CO\(_2\) produced + unreactive components] = 16 + 10 + 79 → 105cm\(^3\)

Alkali metals tend to lose one electron and form ions with a single positive charge. They form ionic compounds (salts) in reaction with the halogens (alkali halides)

Thus, with a dilute solution of sulphuric acid and platinum electrodes, hydrogen gas is evolved at the cathode

The gas liberated is H\(_2\)S which in the presence of lead acetate paper gives a black coloration

Sulfate ions in solution are detected using barium chloride solution. The test solution is acidified using a few drops of dilute hydrochloric acid, and then a few drops of barium chloride solution are added. A white precipitate of barium sulfate forms if sulfate ions are present.

\(CuSO_4+2NH_4OH→(NH_4)_2SO_4+Cu(OH)_2\)(pale−blue) The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper [II] sulphate, an azure blue(deep blue) soluble complex salt. \(Cu(OH)_2+(NH_4)_2SO_4+2NH_4OH→[Cu(NH_3)_4]SO_4+4H_2O\)

Various oxides of mercury: Mercury(I) oxide (mercurous oxide), Hg\(_2\)O. Mercury(II) oxide (mercuric oxide), HgO. No specific oxide of mercury is refered in the question.

- When sulfur (IV) oxide or sulfur dioxide is passed on with chlorine gas, mixture of acids are formed. - It leads to the formation of sulfuric acid and hydrochloric acid. - This is a redox reaction

\(\sqrt{41830} = 204.5238\) \(\approxeq 205\) (to three significant figures)

0.0052048 = 0.00520

\((2^{0} + 4^{-\frac{1}{2}})^{2}\) = \((1 + (\frac{1}{4})^{\frac{1}{2}})^{2}\) = \((1 + \frac{1}{2})^{2}\) = \((\frac{3}{2})^{2} = \frac{9}{4}\)