Q: How do you prepare N 80 thiosulphate?

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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.

Related questions

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.

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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58% : 766 = 80% : n 58%n/58% = (766 * 80%)/58% n = 1056.55

n * 4 = 80 Divide both sides by 4: n = 20