To prepare N 80 thiosulfate, dissolve 80 grams of sodium thiosulfate pentahydrate (Na2S2O3·5H2O) in enough distilled water to make a final volume of 1 liter. Stir the solution until the thiosulfate is completely dissolved. Ensure that the solution is mixed well and store it in a properly labeled container. Always follow safety protocols when handling chemicals.
700 + 80 + n = 780 + n = 700 + 80 + n = 700 + 80 + n = 700 + 80 + n = 700 + 80 + n = 0 = 0
If: N/80 = 900 Then: N = 72,000
There are an infinite number of sets with mean 80. Here are some: {80, 80, 80}, {80, 80, 80, 80, 80, 80} {79, 80, 81}, {79, 79, 80, 81, 81}, {79, 79, 80, 82} (1, 80, 159}, {-40, 200} To produce a set of n numbers with mean 80, start with any set of n-1 numbers. Suppose their sum is S. Then add the number 80*n-S to the set. You will now have n numbers whose sum is S+80*n-S = 80*n So the mean of this set is 80.
58% : 766 = 80% : n 58%n/58% = (766 * 80%)/58% n = 1056.55
10n = 80.n = 8.
To prepare a 250 ml saturated solution of sodium thiosulphate, start by adding sodium thiosulphate (Na2S2O3·5H2O) to a clean container, gradually mixing it with distilled water until no more solid dissolves and the solution reaches saturation. Typically, this requires adding approximately 80 grams of sodium thiosulphate to the water. Stir the mixture thoroughly and allow it to settle for any undissolved solids to settle at the bottom. Finally, transfer the clear solution to a volumetric flask and make up the volume to 250 ml with distilled water.
Dissolve 2.48818g of Na2s2o3.5h2o in frishly boiled distilled water. and add 2ml of chloroform for stability, it gives 0.01N sodium thiosulphate sol.
Dissolve 31,62 g anhydrous thiosulfate in 1 L demineralized water at 20 oC.
700 + 80 + n = 780 + n = 700 + 80 + n = 700 + 80 + n = 700 + 80 + n = 700 + 80 + n = 0 = 0
If: N/80 = 900 Then: N = 72,000
100% is equal to 625000. 80% : 500000 = 100% : N 80%N/80%= (500000 * 100%)/80% N= 5000000/8 N= 625000
There are an infinite number of sets with mean 80. Here are some: {80, 80, 80}, {80, 80, 80, 80, 80, 80} {79, 80, 81}, {79, 79, 80, 81, 81}, {79, 79, 80, 82} (1, 80, 159}, {-40, 200} To produce a set of n numbers with mean 80, start with any set of n-1 numbers. Suppose their sum is S. Then add the number 80*n-S to the set. You will now have n numbers whose sum is S+80*n-S = 80*n So the mean of this set is 80.
To prepare 0.2M solution of anhydrous sodium thiosulfate (Na2S2O3), you dissolve 24.6g of anhydrous Na2S2O3 in distilled water and dilute it to 1 liter. This is the molar mass method, where molar mass of Na2S2O3 is 158.10 g/mol.
Sodium thiosulphate is typically white in color, appearing as a crystalline powder or solid.
Let the number be 'm' & 'n' Hence Product ; mn = 80 Sum ; m + n = 21 m = 21 - n Substitute (21 - n) n = 80 21n - n^2 = 80 n^2 - 21 n + 80 = 0 It is now in quadratic form to factor. ( n - 16)(n - 5) = 0 Hence the numbers are 5 & 16
162/2, 242/3, and, in general, (80*n + 2)/n for any integer n > 1.162/2, 242/3, and, in general, (80*n + 2)/n for any integer n > 1.162/2, 242/3, and, in general, (80*n + 2)/n for any integer n > 1.162/2, 242/3, and, in general, (80*n + 2)/n for any integer n > 1.
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