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Let n be the amount of 20% solution. Then: .08(12)+.2n=.16(12+n) .96+.2n=1.92+.16n .04n=0.96 n=24 liters of 20% solution in the new mix ☺☺☺☺

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Answer is: 3·10⁻⁴ mol of silver atoms. N(silver-Ag) = 1,8·10²⁰. n(Ag) = N÷Na. n(Ag) = 1,8·10²⁰ ÷ 6·10²³ 1/mol = 0,0003 mol = 3·10⁻⁴ mol. N - number of atoms of silver. Na - Avogadro number or Avogadro constant ( 6·10²³ mol⁻¹).¹ n - amount of substance.

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What are the countries?

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It is definitely A Explanation:

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Creo que la respuesta es ayuda

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An experiment was performed to compare the abrasive wear of two different laminated materials. twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. ten pieces of material 2 were similarly tested. in each case, the depth of wear was observed. the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4, while the samples of material 2 gave an average of 81 and a sample standard deviation of 5. assume the populations to be approximately normal with equal variances. a) can we conclude at the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units? b) repeat part (a) if n1=n2=60.

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