What molar quantity of the metal salt are present in a $0.667 \cdot g$ $0.667*g$ mass of sodium dichromate?

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What molar quantity of the metal salt are present in a $0.667 \cdot g$ $0.667*g$ mass of sodium dichromate?

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Answer 1 · 2016-05-11T04:26:28

$5.10 mmol$ $"5.10 mmol"$

Explanation:

$Moles of sodium dichromate = \frac{0.667 \cdot g}{261.97 \cdot g \cdot m o l} = 2.55 \times 10^{- 3} \cdot m o l$ $"Moles of sodium dichromate" = (0.667*g)/(261.97*g*mol )=2.55xx10^-3 *mol$ .

Given the composition of sodium dichromate, $N a_{2} C r_{2} O_{7}$ $Na_2Cr_2O_7$ , there are twice this molar quantity of sodium ions. How many oxygen atoms?

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What molar quantity of the metal salt are present in a $0.667 \cdot g$ $0.667*g$ mass of sodium dichromate?

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What molar quantity of the metal salt are present in a 0.667⋅g0.667*g mass of sodium dichromate?

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What molar quantity of the metal salt are present in a $0.667 \cdot g$ $0.667*g$ mass of sodium dichromate?